ROMANIAN MATHEMATICAL MAGAZINE

R M MROMANIAN MATHEMATICAL MAGAZINE

Founding EditorFounding EditorDANIEL SITARUDANIEL SITARU

Available onlineAvailable onlinewww.ssmrmh.rowww.ssmrmh.ro

ISSN-L 2501-0099ISSN-L 2501-0099

RMM - Calculus Marathon 1501 - RMM - Calculus Marathon 1501 - 16001600

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1 RMM-CALCULUS MARATHON 1501-1600

Proposed by

Daniel Sitaru – Romania,Vasile Mircea Popa-Romania

Asmat Qatea-Afghanistan,Kaushik Mahanta-Assam-India

Srinivasa Raghava-AIRMC-India,Sujeethan Balendran-SriLanka

Narendra Bhandari-Bajura-Nepal,Orxan Abasov-Azerbaijan

Abdul Mukhtar-Nigeria,Ty Halpen-Florida-SUA,Angad Singh-India,George

Apostolopoulos-Messolonghi-Greece,Amrit Awasthi-India,Surjeet Singhania-

India.Florică Anastase-Romania,Neculai Stanciu-Romania,Mohammad

Hamed Nasery-Afghanistan,Costel Florea-Romania,Mikael Bernardo-

Mozambique,Simon Peter-Madagascar,Durmuş Ogmen-Turkiye

Ajetunmobi Abdulqoyyum-Nigeria,Syed Shahabudeen-India

Probal Chakraborty-India,Tobi Joshua-Nigeria,Ose Favour-Nigeria

Onikoyi Adeboye-Nigeria,Marin Chirciu-Romania,Marian Ursărescu-Romania

Ruxandra Daniela Tonilă-Romania

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Solutions by

Daniel Sitaru – Romania,Rana Ranino-Setif-Algerie

Jose Ferreira Queiroz-Olinda-Brazil,Ty Halpen-Florida-USA,Amrit Awasthi-

Punjab-India,Syed Shahabudeen-Kerala-India,Akerele Olofin-Nigeria

Cornel Ioan Vălean-Romania,Ngulmun George Baite-India,Mohammad

Rostami-Afghanistan,Yen Tung Chung-Taichung-Taiwan,Orlando Irahola

Ortega-Bolivia,Ghuiam Naseri-Afghanistan,Marian Ursărescu-

Romania,Ajetunmobi Abdulquyyum-Nigeria,Muhammad Afzal-Pakistan

Adrian Popa-Romania,Angad Singh-India,Ruxandra Daniela Tonilă-Romania

Artan Ajredini-Presheva-Serbie,Naren Bhandari-Bajura-Nepal,Soumitra

Mandal-India,Ose Favour-Nigeria,Kartick Chandra Betal-India,Asmat Qatea-

Afghanistan,Ravi Prakash-New Delhi-India,Ahmed Yackoube Chach-

Mauritania,George Florin Șerban-Romania,Serlea Kabay-Liberia,Obaidullah

Jaihon-Afghanistan,Mikael Bernardo-Mozambique,Florică Anastase-Romania

Felix Marin-Romania,Kamel Gandouli Rezgui-Tunisia,Surjeet Singhania-India

Probal Chakraborty-Kolkata-India,Kaushik Mahanta-Assam-India

Hussain Reza Zadah-Afghanistan,Timson Azeez Folorunsho-Nigeria

Luca Paes Barreto-Pernambuco-Brazil,Abdul Mukhtar-Nigeria

Remus Florin Stanca-Romania,Mohammad Hamed Nasery-Afghanistan

Satyam Roy-India,Sujit Bhowmick-India,Sediqakbar Restheen-Afghanistan

Santiago Alvarez-Mexico,Almas Babirov-Azerbaijan,

Fayssal Abdelli-Bejaia-Algerie, Ajenikoko Gbolahan-Nigeria

Dawid Bialek-Poland

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1501. Prove that:

𝚿(𝟕

𝟖) − 𝚿(

𝟑

𝟖) = 𝝅√𝟐 − 𝟐√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)

where 𝚿(𝒙) is the digamma function.

Proposed by Vasile Mircea Popa-Romania

Solution 1 by Rana Ranino-Setif-Algerie

We know that:

∫𝒙𝒏

𝟏 + 𝒙𝒅𝒙

𝟏

𝟎

=𝟏

𝟐(𝝍(

𝒏

𝟐+ 𝟏) −𝝍 (

𝒏

𝟐+𝟏

𝟐))

𝛀 = 𝟐∫𝒙−𝟏𝟒

𝟏 + 𝒙𝒅𝒙

𝟏

𝟎

=𝒙=𝒙𝟒

𝟖∫𝒙𝟐

𝟏 + 𝒙𝟒𝒅𝒙

𝟏

𝟎

= 𝟒(∫𝒙𝟐 + 𝟏

𝟏 + 𝒙𝟒𝒅𝒙

𝟏

𝟎

+∫𝒙𝟐 − 𝟏

𝟏 + 𝒙𝟒𝒅𝒙

𝟏

𝟎

) =

= 𝟒(∫𝟏 +

𝟏𝒙𝟐

𝒙𝟐 +𝟏𝒙𝟐

𝒅𝒙𝟏

𝟎

+∫𝟏 −

𝟏𝒙𝟐

𝒙𝟐 +𝟏𝒙𝟐

𝒅𝒙𝟏

𝟎

)

𝒖 = 𝒙 −𝟏

𝒙; 𝒗 = 𝒙 +

𝟏

𝒙⇒ 𝛀 = 𝟒(∫

𝒅𝒖

𝒖𝟐 + 𝟐

𝟎

−∞

−∫𝒅𝒗

𝒗𝟐 − 𝟐

∞

𝟐

) =

= 𝟒(𝝅

𝟐√𝟐−𝟏

𝟐√𝟐𝐥𝐨𝐠 (

𝒗 − √𝟐

𝒗+ √𝟐)|𝟐

∞

= 𝝅√𝟐 − √𝟐 𝐥𝐨𝐠 (𝟐 + √𝟐

𝟐 − √𝟐) =

= 𝝅√𝟐 − √𝟐 𝐥𝐨𝐠 [(𝟐 + √𝟐

√𝟐)

𝟐

]

Therefore,

𝝍(𝟕

𝟖) −𝝍(

𝟑

𝟖) = 𝝅√𝟐 − 𝟐√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)

Solution 2 by Jose Ferreira Queiroz-Olinda-Brazil

𝝍(𝒓

𝒎) = −𝜸 − 𝐥𝐨𝐠𝟐𝒎 −

𝝅

𝟐⋅ 𝐜𝐨𝐭 (

𝝅𝒓

𝒎) + 𝟐 ∑ 𝐜𝐨𝐬 (

𝟐𝝅𝒏𝒗

𝒎)

[𝒎−𝟏𝟐]

𝒏=𝟏

𝐥𝐨𝐠 (𝐬𝐢𝐧 (𝝅𝒏

𝒎))

For 𝒓 = 𝟕,𝒎 = 𝟖, we have:

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𝝍(𝟕

𝟖) = −𝜸 − 𝐥𝐨𝐠 𝟏𝟔 −

𝝅

𝟐𝐜𝐨𝐭 (

𝟕𝝅

𝟖) + 𝟐∑𝐜𝐨𝐬 (

𝟒𝒏𝝅

𝟖) 𝐥𝐨𝐠 (𝐬𝐢𝐧 (

𝒏𝝅

𝟖))

𝟑

𝒏=𝟏

=

= −𝜸 − 𝟒 𝐥𝐨𝐠 𝟐 +𝝅

𝟐(√𝟐 + 𝟏) +

√𝟐

𝟐𝐥𝐨𝐠(𝟐 − √𝟐) −

√𝟐

𝟐𝐥𝐨𝐠(𝟐 + √𝟐)

For 𝒓 = 𝟑,𝒎 = 𝟖 we have:

𝝍(𝟑

𝟖) = −𝜸 − 𝐥𝐨𝐠 𝟏𝟔 −

𝝅

𝟐𝐜𝐨𝐭 (

𝟑𝝅

𝟖) + 𝟐∑𝐜𝐨𝐬 (

𝟔𝒏𝝅

𝟖) 𝐥𝐨𝐠 (𝐬𝐢𝐧 (

𝒏𝝅

𝟖))

𝟑

𝒏=𝟏

=

= −𝜸 − 𝟒 𝐥𝐨𝐠𝟐 −𝝅

𝟐(√𝟐 − 𝟏) −

√𝟐

𝟐𝐥𝐨𝐠(𝟐 − √𝟐) +

√𝟐

𝟐𝐥𝐨𝐠(𝟐 + √𝟐)

Now, 𝝍(𝟕

𝟖) − 𝝍(

𝟑

𝟖) = 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠(𝟐 − √𝟐) − √𝟐 𝐥𝐨𝐠(𝟐 + √𝟐) =

= 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠(𝟐 − √𝟐

𝟐 + √𝟐) = 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠(

𝟐

(𝟐 + √𝟐)𝟐) =

= 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠 (𝟏

𝟑 + 𝟐√𝟐) = 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠 (

𝟏

(𝟏 + √𝟐)𝟐)

Therefore,

𝝍(𝟕

𝟖) −𝝍(

𝟑

𝟖) = 𝝅√𝟐 − 𝟐√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)

1502. Prove that:

∏𝟐𝒏+ (−𝟏)

𝒏𝟐+𝒏𝟐

𝟐𝒏 + 𝐜𝐨𝐬 (𝒏𝝅𝟐)

∞

𝒏=𝟏

=√𝟒 − 𝟐√𝟐

𝟐

Proposed by Asmat Qatea-Afghanistan

Solution 1 by Ty Halpen-Florida-USA

∏𝟐𝒏+ (−𝟏)

𝒏𝟐+𝒏𝟐

𝟐𝒏 + 𝐜𝐨𝐬 (𝒏𝝅𝟐 )

∞

𝒏=𝟏

=

= (∏𝟐(𝟐𝒏) + (−𝟏)

(𝟐𝒏)𝟐+𝟐𝒏𝟐

𝟐(𝟐𝒏) + 𝐜𝐨𝐬(𝒏𝝅)

∞

𝒏=𝟏

)(∏𝟐(𝟐𝒏 − 𝟏) + (−𝟏)

(𝟐𝒏−𝟏)𝟐+𝟐𝒏−𝟏𝟐

𝟐(𝟐𝒏 − 𝟏) + 𝐬𝐢𝐧(𝒏𝝅)

∞

𝒏=𝟏

)

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= (∏𝟒𝒏 + (−𝟏)𝒏

𝟒𝒏 + (−𝟏)𝒏

∞

𝒏=𝟏

)(∏𝟒𝒏− 𝟐 + (−𝟏)𝒏

𝟒𝒏 − 𝟐

∞

𝒏=𝟏

) =∏(𝟏 +(−𝟏)𝒏

𝟒𝒏 − 𝟐)

∞

𝒏=𝟏

Now, use the result:

∏(𝟏+(−𝟏)𝒏𝒄

𝒂𝒏 + 𝒃)

∞

𝒏=𝟏

=𝟐−𝒃𝒂√𝝅𝚪(

𝒂 + 𝒃𝟐 )

𝚪 (𝒂 + 𝒃 − 𝒄𝟐𝒂 )𝚪 (

𝟐𝒂+ 𝒃 + 𝒄𝟐𝒂 )

Fro 𝒂 = 𝟒, 𝒃 = −𝟐, 𝒄 = 𝟏:

∏(𝟏+(−𝟏)𝒏

𝟒𝒏 − 𝟐)

∞

𝒏=𝟏

=√𝟐𝝅𝚪 (

𝟏𝟐)

𝚪(𝟏𝟖) 𝚪(

𝟕𝟖)= √𝟐𝐬𝐢𝐧 (

𝝅

𝟖) =

√𝟒 − 𝟐√𝟐

𝟐

Solution 2 by Amrit Awasthi-Punjab-India

Consider the following cases:

I) 𝒏 = 𝟒𝒌 + 𝟏: (−𝟏)𝒏𝟐+𝒏

𝟐 = −𝟏, 𝐜𝐨𝐬 (𝒏𝝅

𝟐) = 𝟎

II) 𝒏 = 𝟒𝒌 + 𝟐: (−𝟏)𝒏𝟐+𝒏

𝟐 = −𝟏, 𝐜𝐨𝐬 (𝒏𝝅

𝟐) = −𝟏

III) 𝒏 = 𝟒𝒌 + 𝟑: (−𝟏)𝒏𝟐+𝒏

𝟐 = 𝟏, 𝐜𝐨𝐬 (𝒏𝝅

𝟐) = 𝟎

IV) 𝒏 = 𝟒𝒌: (−𝟏)𝒏𝟐+𝒏

𝟐 = 𝟏, 𝐜𝐨𝐬 (𝒏𝝅

𝟐) = 𝟏

Hence, rewriting the product, we have:

𝛀 =∏𝟐(𝟒𝒌 + 𝟏) − 𝟏

𝟐(𝟒𝒌 + 𝟏) + 𝟎⋅𝟐(𝟒𝒌 + 𝟐) − 𝟏

𝟐(𝟒𝒌 + 𝟐) − 𝟏⋅𝟐(𝟒𝒌 + 𝟑) + 𝟏

𝟐(𝟒𝒌 + 𝟑) + 𝟎⋅𝟐(𝟒(𝒌 + 𝟏)) + 𝟏

𝟐(𝟒(𝒌 + 𝟏)) + 𝟏

∞

𝒌=𝟎

=

= 𝐥𝐢𝐦𝒏→∞

∏𝟖𝒌+ 𝟏

𝟖𝒌 + 𝟐⋅𝟖𝒌 + 𝟕

𝟖𝒌 + 𝟔

𝒏

𝒌=𝟎

=𝟏

𝟐⋅𝟕

𝟔𝐥𝐢𝐦𝒏→∞

∏𝒌+

𝟏𝟖

𝒌+𝟐𝟖

⋅𝒌 +

𝟕𝟖

𝒌 +𝟔𝟖

𝒏

𝒌=𝟏

=

=𝟏

𝟐⋅𝟕


𝚪(𝒏 +𝟏𝟖 + 𝟏)

𝚪(𝟏𝟖 + 𝟏)

𝚪(𝒏 +𝟐𝟖 + 𝟏)

𝚪(𝟐𝟖+ 𝟏)

⋅

𝚪 (𝒏 +𝟕𝟖 + 𝟏)

𝚪 (𝟕𝟖 + 𝟏)

𝚪 (𝒏 +𝟔𝟖 + 𝟏)

𝚪 (𝟔𝟖+ 𝟏)

=

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=𝟏

𝟐⋅𝟕


𝒏𝟏𝟖+𝟏−

𝟐𝟖+𝟕𝟖+𝟏−

𝟔𝟖−𝟏 ⋅

𝚪 (𝟐𝟖 + 𝟏)𝚪 (

𝟔𝟖 + 𝟏)

𝚪 (𝟕𝟖 + 𝟏)𝚪 (

𝟏𝟖 + 𝟏)

=

=𝟏

𝟐⋅𝟕

𝟔⋅

𝟑𝟏𝟔𝚪 (

𝟏𝟒)𝚪 (𝟏 −

𝟏𝟒)

𝟕𝟔𝟒𝚪 (

𝟏𝟖)𝚪 (𝟏 −

𝟏𝟖)=𝟏

𝟐⋅𝟕

𝟔⋅𝟑

𝟏𝟔⋅𝟔𝟒

𝟕⋅

𝝅

𝐬𝐢𝐧𝝅𝟒

𝝅

𝐬𝐢𝐧𝝅𝟖

=𝟏

𝟐⋅𝟕

𝟔⋅𝟏𝟐 𝐬𝐢𝐧

𝝅𝟖

𝟕 𝐬𝐢𝐧𝝅𝟒

= √𝟐 𝐬𝐢𝐧𝝅

𝟖

=√𝟒 − 𝟐√𝟐

𝟐

1503. Prove that:

∫ ∫ ∫ …∫𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏

√𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)(𝟏 + 𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏)

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

=

= 𝝅𝒏√𝝅𝒏+𝟐 𝑭𝒏𝒏+𝟏 (

𝟏

𝟐,𝟏

𝟐,… ,

𝟏

𝟐⏟ (𝒏+𝟏)−𝒕𝒊𝒎𝒆𝒔

; 𝟏, 𝟏, . . . , 𝟏⏞ 𝒏−𝒕𝒊𝒎𝒆𝒔

; −𝟏)

Proposed by Kaushik Mahanta-Assam-India

Solution 1 by Syed Shahabudeen-Kerala-India



𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

=

=∑(−𝟏)𝒌

𝟒𝒌(𝟐𝒌

𝒌)

∞

𝒌=𝟎

∫ ∫ ∫ …∫(𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏)

𝒌−𝟏𝟐 𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏

√(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

=

= ∑(−𝟏)𝒌

𝟒𝒌(𝟐𝒌

𝒌)𝚪𝒏 (𝒌 +

𝟏𝟐) 𝚪

𝒏 (𝟏𝟐)

𝚪𝒏(𝒌 + 𝟏)

∞

𝒌=𝟎

=

(𝟐𝒌)!

𝟒𝒌𝒌!=(𝟏𝟐)𝒌

=∑(−𝟏)𝒌 (𝟏

𝟐)𝒌

𝚪𝒏 (𝒌 +𝟏𝟐) 𝚪

𝒏 (𝟏𝟐)

𝚪𝒏(𝒌 + 𝟏)𝒌!

∞

𝒌=𝟎

=∑(−𝟏)𝒌

𝒌!(𝟏

𝟐)𝒌

𝚪𝒏 (𝒌 +𝟏𝟐) 𝚪

𝒏 (𝟏𝟐)

𝚪𝒏(𝒌 + 𝟏)

∞

𝒌=𝟎

=

= 𝚪𝟐𝒏 (𝟏

𝟐)∑(−𝟏)𝒌 (

𝟏

𝟐)𝒌

(𝟏𝟐)𝒌

𝒏

(𝟏)𝒌𝒏𝒌!

∞

𝒌=𝟎

= 𝝅𝒏∑(𝟏𝟐)𝒌

𝒏+𝟏

(−𝟏)𝒌

(𝟏)𝒌𝒏𝒌!

∞

𝒌=𝟎

=

= 𝝅𝒏√𝝅𝒏+𝟐 𝑭𝒏𝒏+𝟏 (𝟏

𝟐,𝟏

𝟐,… ,

𝟏


; 𝟏, 𝟏, . . . , 𝟏⏞ 𝒏−𝒕𝒊𝒎𝒆𝒔

; −𝟏)

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Solution 2 by Akerele Olofin-Nigeria



𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

=

= ∫ ∫ ∫ …∫𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏

√𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

𝟏

𝟏 + 𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏=

= ∫ ∫ ∫ …∫𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏

√𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

∑(

𝟏𝟐+ 𝒌 − 𝟏

𝒌) (−𝟏)𝒌 (∏𝒙𝒊

𝒏

𝒌=𝟏

)

𝒌∞

𝒌=𝟎

=∑(−𝟏)𝒌 (𝒌 −

𝟏𝟐

𝒌)

∞

𝒌=𝟎

∫ ∫ ∫ …∫(∏ 𝒙𝒊

𝒏𝒊=𝟏 )𝒌−

𝟏𝟐

√∏ (𝟏 − 𝒙𝒊)𝒏𝒊=𝟏

𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

𝟏

𝟎

⇒ 𝑱𝒏 =∑(−𝟏)𝒌(𝒌 −

𝟏𝟐

𝒌)∏∫ 𝒙

𝒊

𝒌−𝟏𝟐(𝟏 − 𝒙𝒊)

−𝟏𝟐𝒅𝒙𝒊

𝟏

𝟎

𝒏

𝒊=𝟏

∞

𝒌=𝟎

∵ ∫ 𝒙𝒊

𝒌−𝟏𝟐(𝟏 − 𝒙𝒊)

−𝟏𝟐

𝟏

𝟎

𝒅𝒙𝒊 = 𝜷(𝒌 +𝟏

𝟐,𝟏

𝟐) =

𝚪 (𝒌 +𝟏𝟐)𝚪 (

𝟏𝟐)

𝚪(𝒌 + 𝟏)

Therefore,

𝑱𝒏 =∑𝚪(𝒌 +

𝟏𝟐)

𝚪 (𝟏𝟐)𝚪

(𝒌 + 𝟏)(𝚪 (𝒌 +

𝟏𝟐)𝚪 (

𝟏𝟐)

𝚪(𝒌 + 𝟏))

𝒏∞

𝒌=𝟎

= 𝝅𝒏∑𝚪𝒏+𝟏 (𝒌 +

𝟏𝟐)

𝚪𝒏+𝟏 (𝟏𝟐)𝚪

𝒏(𝒌 + 𝟏)

(−𝟏)𝒌

𝚪(𝒌 + 𝟏)

∞

𝒌=𝟎

=

= 𝝅𝒏√𝝅𝒏+𝟐 𝑭𝒏𝒏+𝟏 (

𝟏

𝟐,𝟏

𝟐,… ,

𝟏


; 𝟏, 𝟏, . . . , 𝟏⏞ 𝒏−𝒕𝒊𝒎𝒆𝒔

; −𝟏)

1504. If we have:

∫ (𝐬𝐢𝐧𝝅𝒙

𝒙𝟑−𝐜𝐨𝐬𝝅𝒙

𝒙𝟐−𝐬𝐢𝐧𝐡 𝝅𝒙

𝒙𝟑+𝐜𝐨𝐬𝐡𝝅𝒙

𝒙𝟐)𝒆−𝝅𝒙

√𝒙

∞

𝟎

𝒅𝒙 = 𝝅𝟐𝜶 + 𝝅𝟑𝜷

then find the values of 𝜶 and 𝜷.

Proposed by Srinivasa Raghava-AIRMC-India

Solution by Syed Shahabudeen-Kerala-India

𝛀 = ∫ (𝐬𝐢𝐧 𝝅𝒙

𝒙𝟑−𝐜𝐨𝐬 𝝅𝒙

𝒙𝟐−𝐬𝐢𝐧𝐡𝝅𝒙

𝒙𝟑+𝐜𝐨𝐬𝐡𝝅𝒙

𝒙𝟐)𝒆−𝝅𝒙

√𝒙

∞

𝟎

𝒅𝒙 =

www.ssmrmh.ro


= ∫𝒆−𝝅𝒙 𝐬𝐢𝐧𝝅𝒙

𝒙𝟕𝟐

∞

𝟎

𝒅𝒙 +∫𝒆−𝝅𝒙 𝐜𝐨𝐬𝝅𝒙

𝒙𝟓𝟐

∞

𝟎

𝒅𝒙 −∫𝒆−𝝅𝒙 𝐬𝐢𝐧𝐡𝝅𝒙

𝒙𝟕𝟐

∞

𝟎

𝒅𝒙 +∫𝒆−𝝅𝒙 𝐜𝐨𝐬𝐡𝝅𝒙

𝒙𝟓𝟐

∞

𝟎

𝒅𝒙

= 𝑨 −𝑩− 𝑪+ 𝑫.

𝑨 = ∫𝒆−𝝅𝒙 𝐬𝐢𝐧𝝅𝒙

𝒙𝟕𝟐

∞

𝟎

𝒅𝒙 = 𝑰𝒎∫𝒆𝒊𝝅𝒙𝒆−𝒊𝝅𝒙

𝒙𝟕𝟐

∞

𝟎

𝒅𝒙 = 𝑰𝒎(𝓛 {𝒆𝒊𝝅𝒙𝒙−𝟕𝟐}) = 𝑰𝒎(

𝚪 (−𝟓𝟐)

(𝝅 − 𝒊𝝅)−𝟓𝟐

)

= −𝟖𝝅𝟑

𝟏𝟓𝑰𝒎(

𝟏

(𝟏 − 𝒊)−𝟓𝟐

) = −𝟖𝝅𝟑

𝟏𝟓𝑰𝒎(

(𝟏 + 𝒊)−𝟓𝟐

𝟐−𝟓𝟐

) = −𝟖𝝅𝟑

𝟏𝟓𝑰𝒎

(

(√𝟐𝒆

𝒊𝝅𝟒 )−𝟓𝟐

𝟐−𝟓𝟐

)

=

=𝟖𝝅𝟑𝟐

𝟓𝟒

𝟏𝟓𝐬𝐢𝐧𝟓𝝅

𝟖

𝑩 = ∫𝒆−𝝅𝒙 𝐜𝐨𝐬 𝝅𝒙

𝒙𝟓𝟐

∞

𝟎

𝒅𝒙 = 𝑹𝒆∫𝒆𝒊𝝅𝒙𝒆−𝒊𝝅𝒙

𝒙𝟓𝟐

∞

𝟎

𝒅𝒙 = 𝑹𝒆 (𝓛 {𝒆𝒊𝝅𝒙𝒙−𝟓𝟐}) = 𝑹𝒆(

𝚪 (−𝟑𝟐)

(𝝅 − 𝒊𝝅)−𝟑𝟐

)

=𝟒𝝅𝟐

𝟑𝑹𝒆

(

(√𝟐𝒆

𝒊𝝅𝟒 )

−𝟑𝟐

𝟐−𝟑𝟐

)

=𝟒𝝅𝟐𝟐

𝟑𝟒

𝟑𝐜𝐨𝐬

𝟑𝝅

𝟖

𝑪 = ∫𝒆−𝝅𝒙 𝐬𝐢𝐧𝐡𝝅𝒙

𝒙𝟕𝟐

∞

𝟎

𝒅𝒙 =𝟏

𝟐(𝓛 {𝒆𝝅𝒙𝒙−

𝟕𝟐} − 𝓛 {𝒆−𝝅𝒙𝒙−

𝟕𝟐}) =

=𝟏

𝟐𝐥𝐢𝐦𝒔→𝝅

(𝚪 (−

𝟓𝟐)

(𝒔 − 𝝅)−𝟓𝟐

−𝚪(−

𝟓𝟐)

(𝒔 + 𝝅)−𝟓𝟐

) =−𝚪(−

𝟓𝟐)

𝟐(𝟐𝝅)−𝟓𝟐

=𝟏𝟔𝝅𝟑√𝟐

𝟏𝟓

𝑫 = ∫𝒆−𝝅𝒙 𝐜𝐨𝐬𝐡𝝅𝒙

𝒙𝟓𝟐

∞

𝟎

𝒅𝒙 =𝟏

𝟐(𝓛{𝒆𝝅𝒙𝒙−

𝟓𝟐} + 𝓛 {𝒆−𝝅𝒙𝒙−

𝟓𝟐}) =

=𝟏

𝟐𝐥𝐢𝐦𝒔→𝝅

(𝚪 (−

𝟑𝟐)

(𝒔 − 𝝅)−𝟑𝟐

+𝚪 (−

𝟑𝟐)

(𝒔 + 𝝅)−𝟑𝟐

) =𝚪(−

𝟑𝟐)

𝟐(𝟐𝝅)−𝟑𝟐

=𝟒𝝅𝟐√𝟐

𝟑

𝛀 =𝟖𝝅𝟑𝟐

𝟓𝟒


𝟖−𝟒𝝅𝟐𝟐

𝟑𝟒

𝟑𝐜𝐨𝐬

𝟑𝝅

𝟖−𝟏𝟔𝝅𝟑√𝟐

𝟏𝟓+𝟒𝝅𝟐√𝟐

𝟑=

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= 𝝅𝟐 (𝟒√𝟐

𝟑−𝟐𝟏𝟏𝟒

𝟑𝐜𝐨𝐬

𝟑𝝅

𝟖) + 𝝅𝟑 (

𝟐𝟏𝟕𝟒


𝟖−𝟏𝟔√𝟐

𝟏𝟓)

Therefore, 𝜶 =𝟒√𝟐

𝟑−𝟐𝟏𝟏𝟒

𝟑𝐜𝐨𝐬

𝟑𝝅

𝟖, 𝜷 =

𝟐𝟏𝟕𝟒

𝟏𝟓𝐬𝐢𝐧

𝟓𝝅

𝟖−𝟏𝟔√𝟐

𝟏𝟓

1505. Evaluate the integral in a closed-form

𝛀 = ∫ (𝐭𝐚𝐧𝟐 𝒙

𝐜𝐨𝐬𝟑𝒙𝟐

+𝟗√𝟐 𝐜𝐨𝐬

𝟑𝒙𝟒

𝟐 𝐬𝐢𝐧𝟑𝒙𝟒+ 𝟏

+𝟏𝟔 𝐬𝐢𝐧 𝒙

𝟒 𝐜𝐨𝐬 𝒙 + 𝟏)

𝝅𝟑

𝟎

𝒅𝒙


Solution by Rana Ranino-Setif-Algerie

𝛀 = ∫ (𝐭𝐚𝐧𝟐 𝒙


+𝟗√𝟐𝐜𝐨𝐬

𝟑𝒙𝟒

𝟐 𝐬𝐢𝐧𝟑𝒙𝟒 + 𝟏

+𝟏𝟔𝐬𝐢𝐧𝒙

𝟒 𝐜𝐨𝐬 𝒙 + 𝟏)

𝝅𝟑

𝟎

𝒅𝒙 =

= ∫𝐭𝐚𝐧𝟐 𝒙


𝝅𝟑

𝟎

𝒅𝒙 + ∫𝟗√𝟐𝐜𝐨𝐬

𝟑𝒙𝟒

𝟐 𝐬𝐢𝐧𝟑𝒙𝟒 + 𝟏

𝝅𝟑

𝟎

𝒅𝒙 +∫𝟏𝟔𝐬𝐢𝐧 𝒙

𝟒 𝐜𝐨𝐬 𝒙 + 𝟏

𝝅𝟑

𝟎

𝒅𝒙 = 𝑨 +𝑩 + 𝑪

𝑨 = ∫𝐭𝐚𝐧𝟐 𝒙


𝝅𝟑

𝟎

𝒅𝒙 = 𝟐∫𝐭𝐚𝐧𝟐 𝟐𝒙

𝐜𝐨𝐬𝟑 𝒙

𝝅𝟔

𝟎

𝒅𝒙 = 𝟖∫𝐭𝐚𝐧𝟐 𝒂𝐬𝐞𝐜𝟑 𝒙

(𝟏 − 𝐭𝐚𝐧𝟐 𝒙)𝟐𝐜𝐨𝐭𝟒 𝒙

𝐜𝐨𝐭𝟒 𝒙

𝝅𝟔

𝟎

𝒅𝒙 =

= 𝟖∫𝐜𝐬𝐜𝟐 𝒙 𝐬𝐞𝐜𝒙

(𝐜𝐨𝐭𝟐 𝒙 − 𝟏)𝟐

𝝅𝟔

𝟎

𝒅𝒙 =𝒕=𝐜𝐬𝐜 𝒙

𝟖∫𝒕𝟐

(𝒕𝟐 − 𝟐)(𝒕𝟐 − 𝟏)

∞

𝟎

𝒅𝒕 =

= ∫ (𝟑√𝟐

𝒕 + √𝟐−𝟑√𝟐

𝒕 − √𝟐+

𝟒

𝒕 − 𝟏−

𝟒

𝒕 + 𝟏+

𝟐

(𝒕 + √𝟐)𝟐 +

𝟐

(𝒕 − √𝟐)𝟐)

∞

𝟐

𝒅𝒕

𝑨 = [𝟑√𝟐 𝐥𝐨𝐠 (𝒕 + √𝟐

𝒕 − √𝟐) + 𝟒 𝐥𝐨𝐠 (

𝒕 − 𝟏

𝒕 + 𝟏) −

𝟒𝒕

𝒕𝟐 − 𝟐]𝟐

∞

=

= 𝟒 + 𝟒 𝐥𝐨𝐠𝟑 − 𝟔√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)

𝑩 = ∫𝟗√𝟐𝐜𝐨𝐬

𝟑𝒙𝟒

𝟐 𝐬𝐢𝐧𝟑𝒙𝟒 + 𝟏

𝝅𝟑

𝟎

𝒅𝒙 =𝒕=𝟐 𝐬𝐢𝐧

𝟑𝒙𝟒𝟔√𝟐∫

𝒅𝒕

𝒕

𝟏+√𝟐

𝟏

= 𝟔√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)

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𝑪 = ∫𝟏𝟔𝐬𝐢𝐧𝒙

𝟒 𝐜𝐨𝐬 𝒙 + 𝟏

𝝅𝟑

𝟎

𝒅𝒙 =𝒕=𝟒 𝐜𝐨𝐬 𝒙+𝟏

𝟒∫𝒅𝒕

𝒕

𝟓

𝟑

= 𝟒 𝐥𝐨𝐠 𝟓 − 𝟒 𝐥𝐨𝐠𝟑

Therefore,

𝛀 = ∫ (𝐭𝐚𝐧𝟐 𝒙


+𝟗√𝟐𝐜𝐨𝐬

𝟑𝒙𝟒

𝟐 𝐬𝐢𝐧𝟑𝒙𝟒 + 𝟏

+𝟏𝟔𝐬𝐢𝐧 𝒙

𝟒 𝐜𝐨𝐬 𝒙 + 𝟏)

𝝅𝟑

𝟎

𝒅𝒙 = 𝟒(𝟏 + 𝐥𝐨𝐠 𝟓)

1506. Find:

𝛀 = ∫𝐥𝐨𝐠𝟐 𝒙 𝐥𝐨𝐠 (

(𝟏 + 𝒙)𝟐

𝟒𝒙)

𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 − 𝟐 𝐥𝐨𝐠(𝟐)∫𝐥𝐨𝐠𝟐(𝒙)

𝟏 + 𝒙

𝟏

𝟎

𝒅𝒙 =𝝅𝟒

𝟕𝟐

without using harmonic series, beta function.

Proposed by Sujeethan Balendran-SriLanka

Solution by Cornel Ioan Vălean-Romania

It is known that:

(𝟏): (−𝟏)𝒎

(𝒎 − 𝟏)!∫𝐥𝐨𝐠𝒎−𝟏(𝒙) 𝐥𝐨𝐠 (

𝟏 + 𝒙𝟐 )

𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 =

=𝟏

𝟐(𝒎𝜻(𝒎+ 𝟏) − 𝟐 𝐥𝐨𝐠(𝟐) (𝟏 − 𝟐𝟏−𝒎)𝜻(𝒎)− ∑(𝟏− 𝟐−𝒌)(𝟏− 𝟐𝟏+𝒌−𝒎)𝜻(𝒌 + 𝟏)𝜻(𝒎− 𝒌))

𝒎−𝟐

𝒌=𝟏

Based on (𝟏), we get that:

∫𝐥𝐨𝐠𝟐 𝒙 𝐥𝐨𝐠 (

𝟏 + 𝒙𝟐 )

𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 =𝟑

𝟐𝐥𝐨𝐠(𝟐) 𝜻(𝟑) −

𝟏𝟗

𝟕𝟐𝟎𝝅𝟒,

which we need in our calculations below.

We also need that:

∫𝐥𝐨𝐠𝟐 𝒙

𝟏 + 𝒙

𝟏

𝟎

𝒅𝒙 = ∑(−𝟏)𝒏−𝟏∫ 𝒙𝒏−𝟏𝟏

𝟎

𝐥𝐨𝐠𝟐(𝒙)

∞

𝒏=𝟏

𝒅𝒙 = 𝟐∑(−𝟏)𝒏−𝟏𝟏

𝒏𝟑

∞

𝒏=𝟏

=𝟑

𝟐𝜻(𝟑) 𝐚𝐧𝐝

∫𝐥𝐨𝐠𝟑(𝒙)

𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 = ∑∫ 𝒙𝒏−𝟏 𝐥𝐨𝐠𝟑(𝒙)𝒅𝒙𝟏

𝟎

∞

𝒏=𝟏

= −𝟔∑𝟏

𝒏𝟒

∞

𝒏=𝟏

= −𝟔𝜻(𝟒).

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Returning to the main result, cleverly splitting and using the auxiliary results above, we

get:

𝛀 = 𝟐∫𝐥𝐨𝐠𝟐(𝒙) 𝐥𝐨𝐠 (

𝟏 + 𝒙𝟐 )

𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 −∫𝐥𝐨𝐠𝟑(𝒙)

𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 − 𝟐 𝐥𝐨𝐠(𝟐)∫𝐥𝐨𝐠𝟐(𝒙)

𝟏 + 𝒙

𝟏

𝟎

𝒅𝒙 =𝝅𝟒

𝟕𝟐

1507. Prove that:

∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐

𝟏

𝟎

(−𝒙 +𝒙𝟐

𝟐𝟐−𝒙𝟑

𝟑𝟐+⋯)𝒅𝒙 =

𝑮𝜻(𝟐)

𝟖

where 𝑮 − is Catalan’s constant.

Proposed by Narendra Bhandari-Bajura-Nepal

Solution 1 by Ngulmun George Baite-India

𝑰 = ∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐

𝟏

𝟎

(−𝒙 +𝒙𝟐

𝟐𝟐−𝒙𝟑

𝟑𝟐+⋯)𝒅𝒙

𝐋𝐞𝐭 𝛇 = −𝒙 +𝒙𝟐

𝟐𝟐−𝒙𝟑

𝟑𝟐+⋯ =∑(−𝟏)𝒏

𝒙𝒏

𝒏𝟐

∞

𝒏=𝟏

= ∑(−𝒙)𝒏

𝒏𝟐

∞

𝒏=𝟏

⇒ 𝜻 = 𝑳𝒊𝟐(−𝒙)

𝑰 = ∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐𝑳𝒊𝟐(−𝒙)

𝟏

𝟎

𝒅𝒙

∵ 𝑳𝒊𝟐(−𝟐) = ∫𝟐 𝐥𝐨𝐠 𝒕

𝟏 + 𝒕𝟐

𝟏

𝟎

𝒅𝒕

𝑰 = ∫ ∫𝒙 𝐥𝐨𝐠𝒙 𝐥𝐨𝐠 𝒚

(𝟏 + 𝒙𝟐)(𝟏 + 𝒙𝒚)𝒅𝒚

𝟏

𝟎

𝒅𝒙𝟏

𝟎

=

=𝟏

𝟐[∫ (∫

𝒙 𝐥𝐨𝐠𝒙 𝐥𝐨𝐠𝒚

(𝟏 + 𝒙𝟐)(𝟏 + 𝒙𝒚)𝒅𝒚

𝟏

𝟎

)𝒅𝒙𝟏

𝟎

+∫ (∫𝒚 𝐥𝐨𝐠 𝒙 𝐥𝐨𝐠𝒚

(𝟏 + 𝒚𝟐)(𝟏 + 𝒙𝒚)𝒅𝒚

𝟏

𝟎

)𝒅𝒙𝟏

𝟎

] =

=𝟏

𝟐∫ ∫ (

𝒙

(𝟏 + 𝒙𝟐)(𝟏 + 𝒙𝒚)+

𝒚

(𝟏 + 𝒚𝟐)(𝟏 + 𝒙𝒚)) 𝐥𝐨𝐠 𝒙 𝐥𝐨𝐠 𝒚𝒅𝒚

𝟏

𝟎

𝒅𝒙𝟏

𝟎

=

=𝟏

𝟐∫ ∫

𝒙 + 𝒚 + 𝒙𝟐𝒚 + 𝒙𝒚𝟐

(𝟏 + 𝒙𝟐)(𝟏 + 𝒚𝟐)(𝟏 + 𝒙𝒚)𝐥𝐨𝐠𝒙 𝐥𝐨𝐠 𝒚𝒅𝒚

𝟏

𝟎

𝟏

𝟎

𝒅𝒙

=𝟏

𝟐∫ ∫

(𝒙 + 𝒚)(𝟏 + 𝒙𝒚)

(𝟏 + 𝒙𝟐)(𝟏 + 𝒚𝟐)(𝟏 + 𝒙𝒚)

𝟏

𝟎

𝐥𝐨𝐠 𝒙 𝐥𝐨𝐠𝒚 𝒅𝒚𝟏

𝟎

𝒅𝒙 =

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=𝟏

𝟐∫ ∫

(𝒙 + 𝒚)

(𝟏 + 𝒙𝟐)(𝟏 + 𝒚𝟐)𝐥𝐨𝐠 𝒙 𝐥𝐨𝐠𝒚

𝟏

𝟎

𝒅𝒚𝟏

𝟎

𝒅𝒙 = ∫𝒙 𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

∫𝐥𝐨𝐠 𝒚

𝟏 + 𝒚𝟐𝒅𝒚

𝟏

𝟎

=

= (∫ ∑(−𝟏)𝒌−𝟏𝒙𝟐𝒌−𝟏 𝐥𝐨𝐠 𝒙𝒅𝒙

∞

𝒌=𝟏

𝟏

𝟎

)(∫ ∑(−𝟏)𝒌−𝟏𝒚𝟐𝒌−𝟐 𝐥𝐨𝐠 𝒚𝒅𝒚

∞

𝒌=𝟏

=𝟏

𝟎

)

= (∑(−𝟏)𝒌−𝟏∫ 𝒙𝟐𝒌−𝟏 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

∞

𝒌=𝟏

)(∑(−𝟏)𝒌−𝟏∫ 𝒚𝟐𝒌−𝟐 𝐥𝐨𝐠 𝒚𝒅𝒚𝟏

𝟎

∞

𝒌=𝟏

) =

= (∑(−𝟏)𝒌−𝟏 [𝒙𝟐𝒌

𝟐𝒌]𝟎

𝟏

− ∫𝒙𝟐𝒌

𝟐𝒌

𝟏

𝒙𝒅𝒙

𝟏

𝟎

∞

𝒌=𝟏

)(∑(−𝟏)𝒌−𝟏 [𝒚𝟐𝒌−𝟏

𝟐𝒌 − 𝟏]𝟎

𝟏

−∫𝒚𝟐𝒌−𝟏

𝟐𝒌 − 𝟏

𝟏

𝒚𝒅𝒚

𝟏

𝟎

∞

𝒌=𝟏

) =

= (∑(−𝟏)𝒌−𝟏 ⋅−𝟏

(𝟐𝒌)𝟐

∞

𝒌=𝟏

)(∑(−𝟏)𝒌−𝟏 ⋅−𝟏

(𝟐𝒌 − 𝟏)𝟐

∞

𝒌=𝟏

) =

= (𝟏

𝟒𝜼(𝟐))𝑮 =

𝟏

𝟒𝜼(𝟐)𝑮 =

𝟏

𝟒⋅𝜻(𝟐)

𝟐⋅ 𝑮 =

𝟏

𝟖𝜻(𝟐)𝑮

Therefore,

∫𝐥𝐨𝐠𝒙

𝟏 + 𝒙𝟐

𝟏

𝟎

(−𝒙 +𝒙𝟐

𝟐𝟐−𝒙𝟑

𝟑𝟐+⋯)𝒅𝒙 =

𝑮𝜻(𝟐)

𝟖

Solution 2 by Mohammad Rostami-Afghanistan

𝛀 = ∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐∑(−𝒙)𝒏

𝒏𝟐

∞

𝒏=𝟏

𝒅𝒙𝟏

𝟎

= ∫(−𝟏)𝒏𝒙𝒏

𝒏𝟐∑(−𝒙𝟐)𝒌

𝝏

𝝏𝒂|𝒂=𝟎

∞

𝒌=𝟎

𝟏

𝟎

𝒙𝒂𝒅𝒙 =

= ∑(−𝟏)𝒏

𝒏𝟐

∞

𝒏=𝟏

∑(−𝟏)𝒌 𝝏

𝝏𝒂|𝒂=𝟎

∞

𝒌=𝟎

∫ 𝒙𝒏+𝟐𝒌+𝒂𝟏

𝟎

𝒅𝒙 =

= ∑(−𝟏)𝒏

𝒏𝟐∑(−𝟏)𝒌 [

𝟏

𝒏 + 𝟐𝒌 + 𝒂 + 𝒂]𝒂=𝟎

′∞

𝒌=𝟎

=

∞

𝒏=𝟏

= ∑(−𝟏)𝒏

𝒏𝟐∑(−𝟏)𝒌 ⋅

−𝟏

(𝟐𝒌 + 𝒏 + 𝟏)𝟐

∞

𝒌=𝟎

∞

𝒏=𝟏

; {𝒌 → 𝒌 −𝒏

𝟐}

𝛀 = ∑(−𝟏)𝒏

𝒏𝟐∑

(−𝟏)𝒌−𝒏𝟐

(−𝟏) [𝟐 (𝒌 −𝒏𝟐) + 𝒏 + 𝟏

]𝟐

∞+𝒏𝟐

𝒌=𝒏𝟐

∞

𝒏=𝟏

=

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= ∑(−𝟏)

𝒏𝟐−𝟏

𝒏𝟐

∞

𝒏=𝟏

∑(−𝟏)𝒌

(𝟐𝒌 + 𝟏)𝟐

∞

𝒌=𝒏𝟐

{𝒌 =𝒏

𝟐= 𝒕 ∈ 𝑷 = {𝟎, 𝟏, 𝟐, 𝟑, 𝟒… } ⇒ 𝒏 = 𝟐𝒕}

𝛀 =∑(−𝟏)𝒕−𝟏

𝟒𝒕𝟐

∞𝟐

𝒕=𝟏𝟐

∑(−𝟏)𝒕

(𝟐𝒕 + 𝟏)𝟐

∞

𝒕=𝟎

𝒕∈𝑷,𝒕≥𝟏⇒ 𝛀 =

𝟏

𝟒∑(−𝟏)𝒕−𝟏

𝒕𝟐⋅ 𝑮

∞

𝒕=𝟏

⇒

𝛀 =𝟏

𝟒⋅ 𝜼(𝟐) ⋅ 𝑮 =

𝟏

𝟒(𝟏 − 𝟐𝟏−𝟐)𝜻(𝟐)𝑮

Therefore,

∫𝐥𝐨𝐠𝒙

𝟏 + 𝒙𝟐

𝟏

𝟎

(−𝒙 +𝒙𝟐

𝟐𝟐−𝒙𝟑

𝟑𝟐+⋯)𝒅𝒙 =

𝑮𝜻(𝟐)

𝟖

1508.

𝑰𝒏 = ∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠(𝚪(𝒙))𝟏

𝟎

𝒅𝒙, 𝑷𝒏 = ∫𝒙𝒏

𝟏 + 𝒙𝒅𝒙

𝟏

𝟎

Prove:

𝟐√𝝅 ⋅ 𝑰𝒏 ⋅ 𝚪 (𝒏 + 𝟐

𝟐) = 𝚪 (

𝒏 + 𝟏

𝟐) (𝐥𝐨𝐠𝝅 + 𝑷𝒏)



𝑰𝒏 = ∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠(𝚪(𝒙))𝟏

𝟎

𝒅𝒙 =𝒙→𝟏−𝒙

∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠 𝚪(𝟏 − 𝒙)𝟏

𝟎

𝒅𝒙 =

=𝟏

𝟐∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠(𝚪(𝒙)𝚪(𝟏 − 𝒙))𝟏

𝟎

𝒅𝒙 =

=𝟏

𝟐𝐥𝐨𝐠𝝅∫ 𝐬𝐢𝐧𝒏(𝝅𝒙)𝒅𝒙

𝟏

𝟎

−𝟏

𝟐∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠(𝐬𝐢𝐧(𝝅𝒙))𝟏

𝟎

𝒅𝒙 =𝒕=𝝅𝒙 𝐥𝐨𝐠 𝝅

𝟐𝝅∫ 𝐬𝐢𝐧𝒏 𝒕 𝒅𝒕𝝅

𝟎

−

−𝟏

𝟐𝝅∫ 𝐬𝐢𝐧𝒏 𝒕 𝐥𝐨𝐠(𝐬𝐢𝐧 𝒕)𝝅

𝟎

=𝐥𝐨𝐠𝝅

𝝅∫ 𝐬𝐢𝐧𝒏 𝒕 𝒅𝒕

𝝅𝟐

𝟎

−𝟏

𝝅∫ 𝐬𝐢𝐧𝒏 𝒕

𝝅𝟐

𝟎

𝐥𝐨𝐠(𝐬𝐢𝐧 𝒕) 𝒅𝒕 =

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=𝐥𝐨𝐠𝝅

𝝅∫ 𝐬𝐢𝐧𝒏 𝒕 𝒅𝒕

𝝅𝟐

𝟎

−𝟏

𝝅

𝝏

𝝏𝒔|𝒔=𝟎

∫ 𝐬𝐢𝐧𝒏+𝒔 𝒕 𝒅𝒕

𝝅𝟐

𝟎

∫ 𝐬𝐢𝐧𝒏 𝒕 𝒅𝒕

𝝅𝟐

𝟎

=𝟏

𝟐𝜷 (𝒏 + 𝟏

𝟐,𝟏

𝟐) =

𝚪(𝒏 + 𝟏𝟐 ) 𝚪(

𝟏𝟐)

𝚪(𝒏 + 𝟐𝟐 )

= √𝝅𝚪(𝒏 + 𝟏𝟐 )

𝚪(𝒏 + 𝟐𝟐 )

𝑰𝒏 =𝐥𝐨𝐠𝝅

𝟐√𝝅

𝚪(𝒏 + 𝟏𝟐 )

𝚪(𝒏 + 𝟐𝟐)−

𝟏

𝟐√𝝅

𝝏

𝝏𝒔|𝒔=𝟎

𝚪(𝒏 + 𝒔 + 𝟏

𝟐 )

𝚪(𝒏 + 𝒔 + 𝟐

𝟐)=

=𝐥𝐨𝐠𝝅

𝟐√𝝅

𝚪 (𝒏 + 𝟏𝟐 )

𝚪 (𝒏 + 𝟐𝟐 )

−𝟏

𝟒√𝝅

𝐥𝐨𝐠𝝅

𝟐√𝝅

𝚪 (𝒏 + 𝟏𝟐 )

𝚪 (𝒏 + 𝟐𝟐 )

{𝝍(𝒏 + 𝟏

𝟐) −𝝍(

𝒏 + 𝟐

𝟐)}

𝟐√𝝅𝚪(𝒏 + 𝟐

𝟐) 𝑰𝒏 = 𝚪(

𝒏 + 𝟏

𝟐) {𝐥𝐨𝐠𝝅 +

𝟏

𝟐(𝝍(

𝒏 + 𝟐

𝟐) − 𝝍(

𝒏 + 𝟏

𝟐))

Therefore,

𝟐√𝝅 ⋅ 𝑰𝒏 ⋅ 𝚪 (𝒏 + 𝟐

𝟐) = 𝚪 (

𝒏 + 𝟏

𝟐) (𝐥𝐨𝐠𝝅 + 𝑷𝒏)

1509. Find:

𝛀 = ∫ 𝒙 𝐜𝐨𝐭 𝒙 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)

𝝅𝟐

𝟎

𝒅𝒙


Solution by Cornel Ioan Vălean-Romania

∵ ∑(∫ 𝒕𝟐𝒏−𝟏𝟏 − 𝒕

𝟏 + 𝒕𝒅𝒕

𝟏

𝟎

)𝐬𝐢𝐧𝟐(𝟐𝒏𝒙)

𝒏

∞

𝒏=𝟏

= 𝐥𝐨𝐠(𝐬𝐢𝐧𝒙) 𝐥𝐨𝐠(𝐜𝐨𝐬 𝒙) , 𝟎 < 𝑥 <𝝅

𝟐

We also have the trivial results,

𝒂𝒏 = ∫ 𝒙 𝐭𝐚𝐧𝒙 𝐬𝐢𝐧𝟐(𝟐𝒏𝒙)𝒅𝒙

𝝅𝟐

𝟎

=𝝅

𝟒𝑯𝟐𝒏 −

𝝅

𝟏𝟔⋅𝟏

𝒏 𝐚𝐧𝐝

𝒃𝒏 = ∫ 𝐥𝐨𝐠(𝐜𝐨𝐬 𝒙) 𝐬𝐢𝐧𝟐(𝟐𝒏𝒙)𝒅𝒙

𝝅𝟐

𝟎

=𝝅

𝟏𝟔⋅𝟏

𝒏−𝝅

𝟒𝐥𝐨𝐠 𝟐

where both results are easily derived by exploiting the differences 𝒂𝒏+𝟏 − 𝒂𝒏 and

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𝒃𝒏+𝟏 − 𝒃𝒏 or using Fourrier series.

Returning to the main integrals where we use integration by parts and then exploit the

auxiliary results above, we have:

𝑰 = 𝟐∫ 𝒙 𝐥𝐨𝐠(𝐬𝐢𝐧 𝒙) 𝐥𝐨𝐠(𝐜𝐨𝐬 𝒙) 𝐭𝐚𝐧 𝒙𝒅𝒙

𝝅𝟐

𝟎

−∫ 𝐥𝐨𝐠(𝐬𝐢𝐧 𝒙) 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)

𝝅𝟐

𝟎

𝒅𝒙 =

= 𝟐∑𝟏

𝒏(∫ 𝒕𝟐𝒏−𝟏

𝟏 − 𝒕

𝟏 + 𝒕𝒅𝒕

𝟏

𝟎

)𝒂𝒏

∞

𝒏=𝟏

−∑𝟏

𝒏(∫ 𝒕𝟐𝒏−𝟏

𝟏 − 𝒕

𝟏 + 𝒕𝒅𝒕

𝟏

𝟎

)𝒃𝒏

∞

𝒏=𝟏

=

= −𝝅𝟑

𝟒𝟖𝐥𝐨𝐠 𝟐 −

𝝅

𝟒𝐥𝐨𝐠 𝟐∫

𝐥𝐨𝐠(𝟏 − 𝒕𝟐)

𝒕𝒅𝒕

𝟏

𝟎⏟ −𝝅𝟐/𝟏𝟐

+𝝅

𝟒∫𝐥𝐨𝐠𝟐(𝟏 − 𝒕)

𝒕𝒅𝒕

𝟏

𝟎⏟ 𝟐𝜻(𝟑)

+𝝅

𝟏𝟔∫𝑳𝒊𝟐(𝒕

𝟐)

𝒕𝒅𝒕

𝟏

𝟎⏟ 𝜻(𝟑)/𝟐

+

+𝝅

𝟐𝐥𝐨𝐠 𝟐∫

𝐥𝐨𝐠(𝟏 − 𝒕𝟐)

𝟏 + 𝒕𝒅𝒕

𝟏

𝟎⏟

𝐥𝐨𝐠𝟐 𝟐−𝝅𝟐

𝟏𝟐

−𝝅

𝟐∫𝐥𝐨𝐠𝟐(𝟏 + 𝒕)

𝟏 + 𝒕𝒅𝒕

𝟏

𝟎⏟ 𝟏

𝟑 𝐥𝐨𝐠𝟑 𝟐

−𝝅

𝟐∫𝐥𝐨𝐠𝟐(𝟏 − 𝒕)

𝟏 + 𝒕𝒅𝒕

𝟏

𝟎⏟

𝟐𝑳𝒊𝟑(𝟏𝟐)

−

−𝝅

𝟒∫𝐥𝐨𝐠(𝟏 − 𝒕) 𝐥𝐨𝐠(𝟏 + 𝒕)

𝒕𝒅𝒕

𝟏

𝟎⏟ −𝟓/𝟖𝜻(𝟑)

=𝝅𝟑

𝟐𝟒𝐥𝐨𝐠 𝟐 +

𝝅

𝟔𝐥𝐨𝐠𝟑 𝟐 −

𝟑

𝟏𝟔𝝅𝜻(𝟑)

1510. For 𝒏 > 1, we have:

∫ ∫𝒙 𝐬𝐢𝐧 𝒕𝒙

𝐜𝐨𝐬𝐡𝝅𝒙𝒏

∞

𝟎

𝒅𝒕∞

𝟎

𝒅𝒙 =𝒏

𝟐


Solution by Ngulmun George Baite-India

𝑰 = ∫ ∫𝒙𝐬𝐢𝐧 𝒕𝒙


∞

𝟎

𝒅𝒕∞

𝟎

𝒅𝒙 =𝐜𝐨𝐬𝐡 𝒙=

𝟏𝟐(𝒆𝒙+𝒆−𝒙)


𝟏𝟐 (𝒆

𝝅𝒙𝒏 + 𝒆−

𝝅𝒙𝒏 )

∞

𝟎

𝒅𝒕∞

𝟎

𝒅𝒙 =

= 𝟐∫ ∫𝒙𝐬𝐢𝐧 𝒕𝒙

𝒆𝝅𝒙𝒏 + 𝒆−

𝝅𝒙𝒏

∞

𝟎

∞

𝟎

𝒅𝒕𝒅𝒙 = 𝟐∫ ∫𝒙𝐬𝐢𝐧 𝒕𝒙

𝒆𝝅𝒙𝒏 (𝟏 + 𝒆−

𝟐𝝅𝒙𝒏 )

∞

𝟎

𝒅𝒕∞

𝟎

𝒅𝒙 =

= 𝟐∫ ∫𝒆−𝝅𝒙𝒏 𝒙 𝐬𝐢𝐧 𝒕𝒙

𝟏 + 𝒆−𝟐𝝅𝒙𝒏

∞

𝟎

𝒅𝒕∞

𝟎

𝒅𝒙 = 𝟐∑(−𝟏)𝒌∫ [∫ 𝒙𝒆−𝝅𝒏(𝟐𝒌+𝟏)𝒙 𝐬𝐢𝐧 𝒕𝒙𝒅𝒙

∞

𝟎

] 𝒅𝒕∞

𝟎

∞

𝒌=𝟎

=

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𝐍𝐨𝐰, 𝓛{𝐬𝐢𝐧𝝎𝒙} = ∫ 𝒆−𝒔𝒙 𝐬𝐢𝐧𝝎𝒙∞

𝟎

𝒅𝒙 =𝝎

𝝎𝟐 + 𝒔𝟐

Differentiating respect to 𝒔, we have:

−∫ 𝒙𝒆−𝒔𝒙 𝐬𝐢𝐧𝝎𝒙∞

𝟎

𝒅𝒙 =−𝟐𝒔𝝎

(𝒔𝟐 + 𝝎𝟐)𝟐⇒ ∫ 𝒙𝒆−𝒔𝒙 𝐬𝐢𝐧𝝎𝒙

∞

𝟎

𝒅𝒙 =𝟐𝒔𝝎

(𝒔𝟐 +𝝎𝟐)𝟐

Put 𝒔 =𝝅

𝒏(𝟐𝒌 + 𝟏) and 𝝎 = 𝒕 ⇒

∫ 𝒙𝒆−𝝅𝒏(𝟐𝒌+𝟏)𝒙

∞

𝟎

𝐬𝐢𝐧 𝒕𝒙𝒅𝒙 =

𝟐𝝅𝒏(𝟐𝒌 + 𝟏)𝒕

(𝝅𝟐

𝒏𝟐(𝟐𝒌 + 𝟏)𝟐 + 𝒕𝟐)

𝟐

𝑰 = 𝟐∑(−𝟏)𝒌∫

𝟐𝝅𝒏(𝟐𝒌 + 𝟏)𝒕

(𝝅𝟐

𝒏𝟐(𝟐𝒌 + 𝟏)𝟐 + 𝒕𝟐)

𝟐

∞

𝟎

𝒅𝒕

∞

𝒌=𝟎

=

=𝟒𝝅

𝒏∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)

∞

𝒌=𝟎

∫𝒕

(𝝅𝟐

𝒏𝟐(𝟐𝒌 + 𝟏)𝟐+ 𝒕𝟐)

𝟐

∞

𝟎

𝒅𝒕 =𝒕=𝝅𝒙𝒏

=𝟒𝝅

𝒏∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)

∞

𝒌=𝟎

∫

𝝅𝒙𝒏

𝝅𝟒

𝒏𝟒((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐

∞

𝟎

⋅𝝅

𝒏𝒅𝒙 =

=𝟒𝝅

𝒏⋅𝒏𝟐

𝝅𝟐∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)∫

𝒙

((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐

∞

𝟎

𝒅𝒙

∞

𝒌=𝟎

=

=𝟒𝒏

𝝅∫ ∑(−𝟏)𝒌 ⋅

𝒙(𝟐𝒌 + 𝟏)

((𝟐𝒌 + 𝟏) + 𝒙𝟐)𝟐 𝒅𝒙

∞

𝒌=𝟎

∞

𝟎

; (𝟏)

∵ 𝐬𝐞𝐜𝐡𝒙 = 𝝅∑(−𝟏)𝒌 ⋅𝟐𝒌 + 𝟏

(𝟏𝟐 + 𝒌)

𝟐

𝝅𝟐 + 𝒙𝟐

∞

𝒌=𝟎

𝐩𝐮𝐭 (𝒙 →𝝅𝒙

𝟐) ⇒

𝐬𝐞𝐜𝐡 (𝝅𝒙

𝟐) = 𝝅∑(−𝟏)𝒌 ⋅

𝟐𝒌 + 𝟏

(𝟐𝒌 + 𝟏)𝟐𝝅𝟐

𝟒 + (𝝅𝒙𝟒 )

𝟐

∞

𝒌=𝟎

=𝟒

𝝅∑(−𝟏)𝒌

𝟐𝒌 + 𝟏

(𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐

∞

𝒌=𝟎

Differentiating respect to 𝒙, we have:

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𝒅

𝒅𝒙𝐬𝐞𝐜𝐡 (

𝝅𝒙

𝟐) =

𝟒

𝝅∑(−𝟏)𝒌

−𝟐𝒙(𝟐𝒌 + 𝟏)

((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐

∞

𝒌=𝟎

= −𝟖

𝝅∑(−𝟏)𝒌

𝒙(𝟐𝒌 + 𝟏)

((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐

∞

𝒌=𝟎

∑(−𝟏)𝒌𝒙(𝟐𝒌 + 𝟏)

((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐

∞

𝒌=𝟎

= −𝝅

𝟖

𝒅


𝝅𝒙

𝟐)

From (𝟏) we have:

𝑰 =𝟒𝒏

𝝅∫ −

𝝅

𝟖

𝒅


𝝅𝒙

𝟐)

∞

𝟎

𝒅𝒙 = −𝒏

𝟐∫

𝒅

𝒅𝒙𝐬𝐞𝐜𝐡

𝝅𝒙

𝟐

∞

𝟎

𝒅𝒙 =

= −𝒏

𝟐|𝐬𝐞𝐜𝐡 (

𝝅𝒙

𝟐)|𝟎

∞

= −𝒏

𝟐[𝐥𝐢𝐦𝒙→∞

𝐬𝐞𝐜𝐡 (𝝅𝒙

𝟐) − 𝐥𝐢𝐦

𝒙→𝟎𝐬𝐞𝐜𝐡 (

𝝅𝒙

𝟐)] =

𝒏

𝟐

Therefore,



∞

𝟎

𝒅𝒕∞

𝟎

𝒅𝒙 =𝒏

𝟐

1511. Find:

𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛


Solution 1 by Yen Tung Chung-Taichung-Taiwan

𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛

𝟑𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛 +

+∫ ∫ ∫𝒚𝟐 − 𝒛𝒙

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛 +

+∫ ∫ ∫𝒛𝟐 − 𝒙𝒚

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛 =

= ∫ ∫ ∫𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛 = ∫ ∫ ∫𝟏

𝒙 + 𝒚 + 𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛 =

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= ∫ ∫ 𝐥𝐨𝐠(𝒙 + 𝒚 + 𝒛)|𝟏𝟐

𝟐

𝟏

𝒅𝒚𝒅𝒙𝟐

𝟏

=

= ∫ ∫ (𝐥𝐨𝐠(𝟐 + 𝒚 + 𝒛)𝟐

𝟏

− 𝐥𝐨𝐠(𝟏 + 𝒚 + 𝒛)) 𝒅𝒚𝟐

𝟏

𝒅𝒛 =

= ∫ [(𝟐 + 𝒚 + 𝒛)(𝐥𝐨𝐠(𝟐 + 𝒚 + 𝒛) − 𝟏) − (𝟏 + 𝒚 + 𝒛)(𝐥𝐨𝐠(𝟏 + 𝒚 + 𝒛) − 𝟏)]𝟏𝟐𝒅𝒛

𝟐

𝟏

=

= ∫ ((𝟒 + 𝒛)(𝐥𝐨𝐠(𝟒 + 𝒛) − 𝟏) − (𝟑 + 𝒛)(𝐥𝐨𝐠(𝟑 + 𝒛) − 𝟏) − (𝟑 + 𝒛)(𝐥𝐨𝐠(𝟑 + 𝒛) − 𝟏)𝟐

𝟏

+ (𝟐 + 𝒛)(𝐥𝐨𝐠(𝟐 + 𝒛) − 𝟏))𝒅𝒛 =

= ∫ (𝟒 + 𝒛) 𝐥𝐨𝐠(𝟒 + 𝒛)𝒅𝒙 −𝟐

𝟏

𝟐∫ (𝟑 + 𝒛) 𝐥𝐨𝐠(𝟑 + 𝒛)𝒅𝒛𝟐

𝟏

+∫ (𝟐 + 𝒛) 𝐥𝐨𝐠(𝟐 + 𝒛)𝒅𝒛𝟐

𝟏

=

= ∫ 𝒖 𝐥𝐨𝐠 𝒖𝒅𝒖𝟔

𝟓

− 𝟐∫ 𝒖 𝐥𝐨𝐠𝒖𝒅𝒖𝟓

𝟒

+∫ 𝒖𝐥𝐨𝐠 𝒖𝒅𝒖𝟒

𝟑

=

= (𝟏

𝟐𝒖𝟐 𝐥𝐨𝐠𝒖 −

𝟏

𝟒𝒖𝟐)|

𝟓

𝟔

− 𝟐(𝟏

𝟐𝒖𝟐 𝐥𝐨𝐠𝒖 −

𝟏

𝟒𝒖𝟐)|

𝟒

𝟓

+ (𝟏

𝟐𝒖𝟐 𝐥𝐨𝐠 𝒖 −

𝟏

𝟒𝒖𝟐)|

𝟑

𝟒

=

= (𝟏𝟖 𝐥𝐨𝐠 𝟔 −𝟐𝟓

𝟐𝐥𝐨𝐠𝟓 −

𝟏𝟏

𝟒) − 𝟐(

𝟐𝟓

𝟐𝐥𝐨𝐠 𝟓 − 𝟏𝟔 𝐥𝐨𝐠 𝟐 −

𝟗

𝟒) + (𝟏𝟔 𝐥𝐨𝐠 𝟐 −

𝟗

𝟐𝐥𝐨𝐠𝟑 −

𝟗

𝟒)

= 𝟔𝟔 𝐥𝐨𝐠𝟐 +𝟐𝟕

𝟐𝐥𝐨𝐠𝟑 −

𝟕𝟓

𝟐𝐥𝐨𝐠 𝟓

Therefore,

𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛 = 𝟐𝟐 𝐥𝐨𝐠 𝟐 +𝟗

𝟐𝐥𝐨𝐠 𝟑 −

𝟐𝟓

𝟐𝐥𝐨𝐠𝟓

Solution 2 by Syed Shahabudeen-Kerala-india

𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛

𝟑𝛀 = ∫ ∫ ∫ ∑𝒙𝟐 − 𝒚𝒛

𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛𝒄𝒚𝒄

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛 = ∫ ∫ ∫𝟏

𝒙 + 𝒚 + 𝒛

𝟐

𝟏

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛 =

= ∫ ∫ ∫ ∫ 𝒕𝒙+𝒚+𝒛−𝟏𝟏

𝟎

𝒅𝒙𝟐

𝟏

𝒅𝒚𝟐

𝟏

𝒅𝒛𝟐

𝟏

𝒅𝒕 = ∫ 𝒕−𝟏𝟏

𝟎

∫ 𝒕𝒙𝟐

𝟏

𝒅𝒙∫ 𝒕𝒚𝒅𝒚𝟐

𝟏

∫ 𝒕𝒛𝟐

𝟏

𝒅𝒛𝒅𝒕 =

= ∫ 𝒕−𝟏 (𝒕𝟑 − 𝒕

𝐥𝐨𝐠𝒚)

𝟑𝟏

𝟎

𝒅𝒕 = ∫𝒕𝟐(𝒕 − 𝟏)𝟑

𝐥𝐨𝐠𝟑 𝒕

𝟏

𝟎

𝒅𝒕

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𝛀 = −𝟏

𝟑∫𝒕𝟐(𝟏 − 𝒕)𝟑


𝟏

𝟎

𝒅𝒕

𝐋𝐞𝐭: 𝛀(𝒂) = −𝟏

𝟑∫𝒕𝒂(𝟏 − 𝒕)𝟑


𝟏

𝟎

𝒅𝒕 ⇒𝝏𝟑𝛀

𝝏𝒂𝟑= −

𝟏

𝟑∫ 𝒕𝒂(𝟏 − 𝒕)𝟑𝟏

𝟎

𝒅𝒕 =

= −𝟏

𝟑(𝚪(𝒂 + 𝟏)𝚪(𝟒)

𝚪(𝒂 + 𝟓)) = −𝟐 ⋅

𝟏

(𝒂 + 𝟏)(𝒂 + 𝟐)(𝒂 + 𝟑)(𝒂 + 𝟒)

𝛀(𝒂) = −𝟐∫𝟏

(𝒂 + 𝟏)(𝒂 + 𝟐)(𝒂 + 𝟑)(𝒂 + 𝟒)𝒅𝟑𝒂 =

= −𝟐∫(𝟏

𝟔(𝒂 + 𝟏)−

𝟏

𝟐(𝒂 + 𝟐)+

𝟏

𝟐(𝒂 + 𝟑)−

𝟏

𝟔(𝒂 + 𝟒))𝒅𝟑𝒂

∫𝟏

𝒂 + 𝒏𝒅𝟑𝒂 =

𝟏

𝟐(𝒂 + 𝒏)𝟐 𝐥𝐨𝐠(𝒂 + 𝒏) −

𝟑𝒂

𝟒(𝒂 + 𝟐)

𝑨(𝒂) =𝟏

𝟔∫(

𝟏

𝒂 + 𝟏−

𝟏

𝒂 + 𝟒)𝒅𝟑𝒂 =

𝟏

𝟏𝟐((𝒂 + 𝟏)𝟐 𝐥𝐨𝐠(𝒂 + 𝟏) − (𝒂 + 𝟒)𝟐 𝐥𝐨𝐠(𝒂 + 𝟒)

𝑩(𝒂) =𝟏

𝟐∫(

𝟏

𝒂 + 𝟑−

𝟏

𝒂 + 𝟐)𝒅𝟑𝒂 =

𝟏

𝟒((𝒂 + 𝟑)𝟐 𝐥𝐨𝐠(𝒂 + 𝟑) − (𝒂 + 𝟐)𝟐 𝐥𝐨𝐠(𝒂 + 𝟐))

𝑨(𝟐) =𝟏

𝟏𝟐(𝟗 𝐥𝐨𝐠 𝟑 − 𝟑𝟔 𝐥𝐨𝐠 𝟔),𝑩(𝟐) =

𝟏

𝟒(𝟐𝟓 𝐥𝐨𝐠 𝟓 − 𝟏𝟔 𝐥𝐨𝐠 𝟒)

𝛀(𝟐) = −𝟐(𝑨(𝟐) + 𝑩(𝟐)) =

= −𝟐(𝟏

𝟏𝟐(𝟗 𝐥𝐨𝐠𝟑 − 𝟑𝟔 𝐥𝐨𝐠 𝟔) +

𝟏

𝟒(𝟐𝟓 𝐥𝐨𝐠 𝟓 − 𝟏𝟔 𝐥𝐨𝐠 𝟒)

Therefore,

𝛀(𝟐) = 𝟐𝟐 𝐥𝐨𝐠𝟐 +𝟗

𝟐𝐥𝐨𝐠 𝟑 +

𝟐𝟓

𝟐𝐥𝐨𝐠 𝟓

1512. Find:

𝛀 = ∫𝟏 − 𝐬𝐢𝐧𝟒 𝒙

(𝟏 + 𝐬𝐢𝐧𝟒 𝒙)√𝟏 + 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙

𝝅𝟔

𝟎


Solution 1 by Cornel Ioan Vălean-Romania

All we need is a clever variable change and a well-known established integral result,

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∫√𝐭𝐚𝐧𝒙𝒅𝒙 =𝟏

√𝟐(𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧 𝒙 − 𝐜𝐨𝐬 𝒙) − 𝐜𝐨𝐬𝐡−𝟏(𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙) + 𝑪

-one way to evaluate it is by calculating ∫(√𝐭𝐚𝐧 𝒙 + √𝐜𝐨𝐭 𝒙)𝒅𝒙 and ∫(√𝐭𝐚𝐧 𝒙 −

√𝐜𝐨𝐭 𝒙)𝒅𝒙

By letting the variable change 𝟏−𝐬𝐢𝐧𝟒 𝒙

𝟏+𝐬𝐢𝐧𝟒 𝒙→ 𝐬𝐢𝐧 𝒙, we get

𝛀 =𝟏

𝟐√𝟐∫ √𝐭𝐚𝐧𝒙

𝝅𝟐

𝐬𝐢𝐧−𝟏(𝟏𝟓𝟏𝟕)

𝒅𝒙 =𝝅

𝟖−𝟏

𝟒𝐬𝐢𝐧−𝟏 (

𝟕

𝟏𝟕) +

𝟏

𝟒𝐜𝐨𝐬𝐡−𝟏 (

𝟐𝟑

𝟏𝟕) =

=𝟏

𝟒𝐥𝐨𝐠 (

𝟐𝟑 + 𝟒√𝟏𝟓

𝟏𝟕) +

𝟏

𝟐𝐜𝐨𝐭−𝟏 (𝟐√

𝟑

𝟓)

Solution 2 by Cornel Ioan Vălean-Romania

𝛀 = ∫𝟏 − 𝐬𝐢𝐧𝟒 𝒙

(𝟏 + 𝐬𝐢𝐧𝟒 𝒙)√𝟏 + 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙

𝝅𝟔

𝟎

= ∫

𝟏𝒕𝟑− 𝒕

(𝒕𝟐 +𝟏𝒕𝟐)√𝟏𝒕𝟐− 𝒕𝟐

𝟏𝟐

𝟎

𝒅𝒕 =√𝟏

𝒕𝟐−𝒕𝟐→𝒕

= ∫𝒕𝟐

𝟒 + 𝒕𝟒

∞

√𝟏𝟓𝟐

𝒅𝒕 =𝟏

𝟒𝐥𝐨𝐠 (

𝟐𝟑 + 𝟒√𝟏𝟓

𝟏𝟕) +

𝟏

𝟐𝐜𝐨𝐭−𝟏 (𝟐√

𝟑

𝟓)

1513. Find without any software:

𝛀 = ∫√𝐬𝐢𝐧 𝒙 ⋅ 𝐭𝐚𝐧 𝒙 𝒅𝒙

Proposed by Orxan Abasov-Azerbaijan


𝛀 = ∫√𝐬𝐢𝐧𝒙 ⋅ 𝐭𝐚𝐧 𝒙𝒅𝒙 = ∫(√𝐬𝐢𝐧𝒙)

𝟑

𝟏 − 𝐬𝐢𝐧𝟐 𝒙⋅ 𝐜𝐨𝐬 𝒙 𝒅𝒙 =

𝒚=√𝐬𝐢𝐧𝒙,𝟐𝒚𝒅𝒚=𝐜𝐨𝐬 𝒙𝒅𝒙

= ∫𝒚𝟑

𝟏 − 𝒚𝟒⋅ 𝟐𝒚𝒅𝒚 = 𝟐∫

𝒚𝟒

𝟏 − 𝒚𝟒𝒅𝒚 = 𝟐∫(−𝟏 +

𝟏

𝟏 − 𝒚𝟒)𝒅𝒚 =

= −𝟐𝒚 +∫(𝟏

𝟏− 𝒚𝟐+

𝟏

𝟏 + 𝒚𝟐)𝒅𝒚 = −𝟐𝒚 + 𝐭𝐚𝐧𝐡−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝑪 =

= −𝟐√𝐬𝐢𝐧𝒙 + 𝐭𝐚𝐧𝐡−𝟏(√𝐬𝐢𝐧𝒙) + 𝐭𝐚𝐧−𝟏(√𝐬𝐢𝐧 𝒙) + 𝑪 =

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= −𝟐√𝐬𝐢𝐧𝒙 +𝟏

𝟐𝐥𝐨𝐠 (

𝟏 + √𝐬𝐢𝐧𝒙

𝟏 − √𝐬𝐢𝐧𝒙) + 𝐭𝐚𝐧−𝟏(√𝐬𝐢𝐧 𝒙) + 𝑪

Solution 2 by Orlando Irahola Ortega-Bolivia

𝛀 = ∫√𝐬𝐢𝐧 𝒙 ⋅ 𝐭𝐚𝐧𝒙 𝒅𝒙 = ∫(√𝐬𝐢𝐧𝒙)

𝟑

𝟏 − 𝐬𝐢𝐧𝟐 𝒙⋅ 𝐜𝐨𝐬 𝒙𝒅𝒙 =

𝒕𝟐=𝐬𝐢𝐧𝒙 ,𝟐𝒕𝒅𝒕=𝐜𝐨𝐬 𝒙𝒅𝒙

= ∫𝟐𝒕𝟒

𝟏 − 𝒕𝟒𝒅𝒕 = −𝟐∫

(𝒕𝟒 − 𝟏) + 𝟏

𝒕𝟒 − 𝟏𝒅𝒕 ⇒

−𝟏

𝟐𝛀 = 𝒕 −∫

𝒅𝒕

𝒕𝟒 − 𝟏= 𝒕 − 𝑰,

𝑰 = ∫𝒅𝒕

𝒕𝟒 − 𝟏=𝒕=𝟏𝒚− ∫

𝒚𝟐

𝟏 − 𝒚𝟒𝒅𝒚 = −

𝟏

𝟐∫𝟐𝒚𝟐

𝟏 − 𝒚𝟒𝒅𝒚

=𝟏

𝟐∫(𝒚𝟐 − 𝟏) + (𝒚𝟐 + 𝟏)

(𝒚𝟐 − 𝟏)(𝒚𝟐 + 𝟏)𝒅𝒚 =

𝟏

𝟐∫

𝒅𝒚

𝒚𝟐 + 𝟏+𝟏

𝟐∫

𝒅𝒚

𝒚𝟐 − 𝟏=

=𝟏

𝟐𝐭𝐚𝐧−𝟏 𝒚 +

𝟏

𝟒𝐥𝐨𝐠 (

𝒚 − 𝟏

𝒚+ 𝟏) =

𝟏

𝟐𝐭𝐚𝐧−𝟏 (

𝟏

𝒕) +

𝟏

𝟒𝐥𝐨𝐠 (

𝟏 − 𝒕

𝟏 + 𝒕) =

=𝟏

𝟐𝐭𝐚𝐧−𝟏 𝒕 +

𝟏

𝟒𝐥𝐨𝐠 (

𝟏 − 𝒕

𝟏 + 𝒕) ⇒

−𝟏

𝟐𝛀 = 𝒕 −∫

𝒅𝒕

𝒕𝟒 − 𝟏= 𝒕 − 𝑰 = 𝒕 −

𝟏

𝟐𝐭𝐚𝐧−𝟏 𝒕 −

𝟏

𝟒𝐥𝐨𝐠 (

𝟏 − 𝒕

𝟏 + 𝒕)

Therefore,

𝛀 = −𝟐√𝐬𝐢𝐧𝒙 +𝟏

𝟐𝐥𝐨𝐠(

𝟏 + √𝐬𝐢𝐧 𝒙

𝟏 − √𝐬𝐢𝐧 𝒙) + 𝐭𝐚𝐧−𝟏(√𝐬𝐢𝐧𝒙) + 𝑪

Solution 3 by Ghuiam Naseri-Afghanistan

𝛀 = ∫√𝐬𝐢𝐧𝒙 ⋅ 𝐭𝐚𝐧 𝒙𝒅𝒙 = ∫𝐬𝐢𝐧𝒙

𝐜𝐨𝐬 𝒙√𝐬𝐢𝐧𝒙𝒅𝒙 = ∫

(𝐬𝐢𝐧𝒙)𝟑𝟐

𝐜𝐨𝐬 𝒙𝒅𝒙 =

= ∫𝐜𝐨𝐬𝒙 ⋅ (𝐬𝐢𝐧 𝒙)

𝟑𝟐

𝐜𝐨𝐬𝟐 𝒙𝒅𝒙 = ∫

𝐜𝐨𝐬𝒙 ⋅ (𝐬𝐢𝐧 𝒙)𝟑𝟐

𝟏 − 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙 =

𝒖=√𝐬𝐢𝐧𝒙∫

𝟐𝒖𝟒

−𝒖𝟒 + 𝟏𝒅𝒖 = 𝟐∫

𝒖𝟒

−𝒖𝟒 + 𝟏𝒅𝒖

= 𝟐∫(−𝟏

𝟐⋅

𝟏

𝒖𝟐 + 𝟏−𝟏

𝟒⋅𝟏

𝒖 + 𝟏+𝟏

𝟒⋅𝟏

𝒖 − 𝟏+ 𝟏)𝒅𝒖 =

= −∫𝟏

𝒖𝟐 + 𝟏𝒅𝒖 −

𝟏

𝟐∫

𝟏

𝒖 + 𝟏𝒅𝒖 +

𝟏

𝟐∫

𝟏

𝒖 − 𝟏𝒅𝒖 + 𝟐𝒖 =

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= − 𝐭𝐚𝐧−𝟏 𝒖 −𝟏

𝟐𝐥𝐨𝐠(𝒖 + 𝟏) +

𝟏

𝟐𝐥𝐨𝐠(𝒖 − 𝟏) + 𝟐𝒖 + 𝑪

Therefore,

𝛀 = −𝟐√𝐬𝐢𝐧𝒙 +𝟏

𝟐𝐥𝐨𝐠(

𝟏 + √𝐬𝐢𝐧 𝒙

𝟏 − √𝐬𝐢𝐧 𝒙) + 𝐭𝐚𝐧−𝟏(√𝐬𝐢𝐧𝒙) + 𝑪


𝛀 = ∫𝐥𝐨𝐠(𝟗𝒙 − 𝟒)

𝟑𝒙𝟐 + 𝟐

𝟐

𝟏

𝒅𝒙

Proposed by Daniel Sitaru-Romania

Solution by Marian Ursărescu-Romania

Put 𝟗𝒙 − 𝟒 = 𝒕 ⇒ 𝒙 =𝒕+𝟒

𝟗, 𝒅𝒙 =

𝟏

𝟗𝒅𝒕

𝛀 = ∫𝐥𝐨𝐠(𝟗𝒙 − 𝟒)

𝟑𝒙𝟐 + 𝟐

𝟐

𝟏

𝒅𝒙 =𝟏

𝟗∫

𝐥𝐨𝐠 𝒕

𝟑 (𝒕 + 𝟒𝟗 )

𝟐

+ 𝟐

𝟏𝟒

𝟓

𝒅𝒕 =𝟏

𝟗∫

𝐥𝐨𝐠 𝒕

𝟑(𝒕𝟐 + 𝟖𝒕 + 𝟏𝟔)𝟖𝟏 + 𝟐

𝟏𝟒

𝟓

𝒅𝒕 =

= 𝟑∫𝐥𝐨𝐠 𝒕

𝒕𝟐 + 𝟖𝒕 + 𝟕𝟎

𝟏𝟒

𝟓

𝒅𝒕; (𝟏)

𝐋𝐞𝐭: 𝑰 = ∫𝐥𝐨𝐠 𝒕

𝒕𝟐 + 𝟖𝒕 + 𝟕𝟎

𝟏𝟒

𝟓

𝒅𝒕 =𝒕=𝟕𝟎𝒚− 𝟕𝟎∫

𝐥𝐨𝐠 (𝟕𝟎𝒚 )

𝟕𝟎𝟐

𝒚𝟐+ 𝟖 ⋅

𝟕𝟎𝒚 + 𝟕𝟎

𝟓

𝟏𝟒

𝒅𝒚 =

= ∫𝐥𝐨𝐠𝟕𝟎 − 𝐥𝐨𝐠𝒚

𝒚𝟐 + 𝟖𝒚 + 𝟕𝟎

𝟒

𝟓

𝒅𝒚 = 𝐥𝐨𝐠𝟕𝟎∫𝟏

𝒚𝟐 + 𝟖𝒚 + 𝟕𝟎𝒅𝒚

𝟏𝟒

𝟓

− ∫𝐥𝐨𝐠 𝒚

𝒚𝟐 + 𝟖𝒚 + 𝟕𝟎

𝟏𝟒

𝟓

𝒅𝒚 =

=𝟏

𝟐𝐥𝐨𝐠𝟕𝟎∫

𝟏

(𝒚 + 𝒚)𝟒 + (√𝟓𝟒)𝟐 𝒅𝒚

𝟏𝟒

𝟓

=𝟏

𝟐√𝟓𝟒𝐥𝐨𝐠 𝟕𝟎 𝐭𝐚𝐧−𝟏 (

𝒚 + 𝟒

√𝟓𝟒)|𝟓

𝟏𝟒

=

=𝟏

𝟐√𝟓𝟒𝐥𝐨𝐠 𝟕𝟎 (𝐭𝐚𝐧−𝟏 (

𝟏𝟖

√𝟓𝟒) − 𝐭𝐚𝐧−𝟏 (

𝟗

√𝟓𝟒)) =

=𝟏

𝟐√𝟓𝟒𝐥𝐨𝐠𝟕𝟎 (𝐭𝐚𝐧−𝟏 √𝟔 − 𝐭𝐚𝐧−𝟏 (

𝟑

√𝟔)) ; (𝟐)

From (1),(2) it follows that:

𝛀 = ∫𝐥𝐨𝐠(𝟗𝒙 − 𝟒)

𝟑𝒙𝟐 + 𝟐

𝟐

𝟏

𝒅𝒙 =𝟏

𝟐√𝟔𝐥𝐨𝐠 𝟕𝟎 (𝐭𝐚𝐧−𝟏 √𝟔 − 𝐭𝐚𝐧−𝟏 (

𝟑

√𝟔))

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1515. Find a closed form:

𝛀 = ∫𝒙 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑

𝟏

𝟎

𝒅𝒙

Proposed by Abdul Mukhtar-Nigeria


𝛀 = ∫𝒙 𝐥𝐨𝐠𝒙

𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑

𝟏

𝟎

𝒅𝒙 = ∫𝒙 𝐥𝐨𝐠 𝒙

(𝟏 − 𝒙)(𝟏 + 𝒙𝟐)

𝟏

𝟎

𝒅𝒙 = −∫(𝟏 − 𝒙 − 𝟏) 𝐥𝐨𝐠 𝒙

(𝟏 − 𝒙)(𝟏 + 𝒙𝟐)𝒅𝒙

𝟏

𝟎

=

= −∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐

𝟏

𝟎

𝒅𝒙 +∫𝐥𝐨𝐠 𝒙

(𝟏 − 𝒙)(𝟏 + 𝒙𝟐)𝒅𝒙

𝟏

𝟎

=

= −∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

+𝟏

𝟐∫𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝒅𝒙

𝟏

𝟎

+𝟏

𝟐∫

𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

+𝟏

𝟐∫𝒙 𝐥𝐨𝐠𝒙

𝟏 + 𝒙𝟐

𝟏

𝟎

𝒅𝒙 =

= −𝟏

𝟐∫

𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐

𝟏

𝟎

𝒅𝒙 +𝟏

𝟐∫𝐥𝐨𝐠 𝒙


𝟏

𝟎

+𝟏

𝟐∫𝒙 𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

=

= −𝟏

𝟐∫ ∑(−𝒙𝟐)

𝝏

𝝏𝒂|𝒂=𝟎𝒙𝒂𝒅𝒙

∞

𝒌=𝟎

𝟏

𝟎

+𝟏

𝟐∫ ∑𝒙𝒏

𝝏

𝝏𝒃|𝒃=𝟎𝒙𝒃𝒅𝒙

∞

𝒏=𝟎

𝟏

𝟎

+𝟏

𝟐∫ 𝒙 ∑(−𝒙𝟐)𝒎

𝝏

𝝏𝒄|𝒄=𝟎𝒙𝒄

∞

𝒎=𝟎

𝒅𝒙𝟏

𝟎

=

= −𝟏

𝟐∑(−𝟏)𝒌

𝝏

𝝏𝒂|𝒂=𝟎

∞

𝒌=𝟎

∫ 𝒙𝟐𝒌+𝒂𝒅𝒙𝟏

𝟎

+𝟏

𝟐∑

𝝏

𝝏𝒃|𝒃=𝟎

∫ 𝒙𝒏+𝒃𝒅𝒙𝟏

𝟎

∞

𝒏=𝟎

+

+𝟏

𝟐∑(−𝟏)𝒎

𝝏

𝝏𝒄|𝒄=𝟎

∫ 𝒙𝟐𝒎+𝒄+𝟏𝟏

𝟎

𝒅𝒙

∞

𝒎=𝟎

=

= −𝟏

𝟐∑(−𝟏)𝒌 [

𝟏

𝟐𝒌 + 𝒂 + 𝟏]𝒂=𝟎

′∞

𝒌=𝟎

+𝟏

𝟐∑[

𝟏

𝒏 + 𝒃 + 𝟏]𝒃=𝟎

′∞

𝒏=𝟎

+𝟏

𝟐∑(−𝟏)𝒎 [

𝟏

𝟐𝒎+ 𝒄 + 𝟐]𝒄=𝟎

′∞

𝒎=𝟎

=𝟏

𝟐∑

(−𝟏)𝒌

(𝟐𝒌 + 𝟏)𝟐

∞

𝒏=𝟎

−𝟏

𝟐∑

𝟏

(𝒏 + 𝟏)𝟐

∞

𝒏=𝟎

−𝟏

𝟖∑

(−𝟏)𝒎

(𝒎 + 𝟏)𝟐

∞

𝒎=𝟎

=

=𝟏

𝟐𝑮 −

𝟏

𝟐∑

𝟏

𝒏𝟐

∞

𝒏=𝟏

−𝟏

𝟖∑

(−𝟏)𝒎−𝟏

𝒎𝟐

∞

𝒎=𝟏

=𝟏

𝟐𝑮 −

𝟏

𝟐𝜻(𝟐) −

𝟏

𝟖𝜼(𝟐) =

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=𝟏

𝟐𝑮 −

𝟏

𝟐𝜻(𝟐) −

𝟏

𝟏𝟔𝜻(𝟐) =

𝟏

𝟐𝑮 −

𝟗

𝟏𝟔⋅𝝅𝟐

𝟔=𝟏

𝟐𝑮 −

𝟑𝝅𝟐

𝟑𝟐

Therefore,

𝛀 =𝟏

𝟐𝑮 −

𝟑𝝅𝟐

𝟑𝟐



𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑

𝟏

𝟎

𝒅𝒙 = ∫𝒙(𝟏 + 𝒙) 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟒

𝟏

𝟎

𝒅𝒙 = ∑∫ (𝒙𝟒𝒏+𝟏 + 𝒙𝟒𝒏+𝟐) 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

∞

𝒏=𝟎

=

= −𝟏

𝟏𝟔∑[

𝟏

(𝒏 +𝟏𝟐)𝟐 +

𝟏

(𝒏 +𝟑𝟒)𝟐]

∞

𝒏=𝟎

= −𝟏

𝟏𝟔{𝝍(𝟏) (

𝟏

𝟐) + 𝝍(𝟏) (

𝟑

𝟒)} = −

𝟏

𝟏𝟔(𝟑𝝅𝟐

𝟐− 𝟖𝑮)

Therefore,

𝛀 =𝟏

𝟐𝑮 −

𝟑𝝅𝟐

𝟑𝟐

Solution 3 by Ajetunmobi Abdulquyyum-Nigeria


𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑

𝟏

𝟎

𝒅𝒙 = ∫𝒙 𝐥𝐨𝐠𝒙

(𝟏 + 𝒙𝟐)(𝟏 − 𝒙)𝒅𝒙

𝟏

𝟎

=

= ∫ (𝒙 − 𝟏

𝟐(𝒙𝟐 + 𝟏)−

𝟏

𝟐(𝒙 − 𝟏)) 𝐥𝐨𝐠 𝒙𝒅𝒙

𝟏

𝟎

=𝟏

𝟐{∫

𝒙 𝐥𝐨𝐠 𝒙

𝒙𝟐 + 𝟏𝒅𝒙

𝟏

𝟎

−∫𝐥𝐨𝐠𝒙

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

+∫𝐥𝐨𝐠𝒙

𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙}

=𝟏

𝟐{𝑨 − 𝑩 + 𝑪}

𝑨 = ∫𝒙 𝐥𝐨𝐠 𝒙

𝒙𝟐 + 𝟏𝒅𝒙

𝟏

𝟎

=; |𝐥𝐨𝐠 𝒙 = −𝒕 ⇒ 𝒅𝒙 = 𝒆−𝒕(−𝒅𝒕)

𝒙𝟐 = 𝒆−𝟐𝒕| ; = −∫

𝒆−𝟐𝒕

𝟏—𝒆−𝟐𝒕

∞

𝟎

𝒅𝒕 =

= −∑(−𝟏)𝒏∫ 𝒕𝒆−(𝟐𝒏+𝟐)𝒕𝒅𝒕∞

𝟎𝒏≥𝟎

= −∑(−𝟏)𝒏 ⋅𝟏

𝟒(𝒏 + 𝟏)𝟐𝒏≥𝟎

= −𝟏

𝟒∑

(−𝟏)𝒏

(𝒏 + 𝟏)𝟐𝒏≥𝟎

=

= −𝟏

𝟒∑(−𝟏)𝒏−𝟏

𝒏𝟐𝒏≥𝟏

= −𝟏

𝟒𝜼(𝟐) = −

𝝅𝟐

𝟒𝟖

Also,

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𝑩 = ∫𝐥𝐨𝐠𝒙

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

=𝒙=𝐭𝐚𝐧 𝒙

∫ 𝐥𝐨𝐠(𝐭𝐚𝐧𝒙)

𝝅𝟒

𝟎

𝒅𝒙 = −𝑮

𝑪 = ∫𝐥𝐨𝐠𝒙

𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 =∑∫ 𝒙𝒏 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎𝒏≥𝟎

=∑𝟏

(𝒏 + 𝟏)𝟐𝒏≥𝟎

= −𝝅𝟐

𝟔

Hence,

𝛀 =𝟏

𝟐(𝑨 − 𝑩 + 𝑪) =

𝟏

𝟐(−𝝅𝟐

𝟒𝟖+ 𝑮 −

𝝅𝟐

𝟔)

Therefore,

𝛀 =𝟏

𝟐𝑮 −

𝟑𝝅𝟐

𝟑𝟐


𝛀(𝒏) = ∫𝒙𝒏

√𝟏 − 𝐥𝐨𝐠 𝒙

𝒆

𝟎

𝒅𝒙,𝒏 > 𝟎


Solution 1 by Ajentunmobi Abdulqoyuum-Nigeria


√𝟏 − 𝐥𝐨𝐠𝒙

𝒆

𝟎

𝒅𝒙 =; |𝒕 = √𝟏 − 𝐥𝐨𝐠 𝒙 ;𝒅𝒕 =

𝟏

𝟐𝒙√𝟏 − 𝐥𝐨𝐠 𝒙𝒅𝒙

𝒅𝒙 = −𝟐𝒕𝒆𝟏−𝒕𝟐, 𝒙 = 𝒆𝟏−𝒕

𝟐| ;

= ∫𝟐𝒕𝒆𝟏−𝒕

𝟐𝒆𝟏−𝒕

𝟐

𝒕𝒅𝒕

∞

𝟎

= 𝟐∫ 𝒆(𝒏+𝟏)−(𝒏+𝟏)𝒕𝟐𝒅𝒕

∞

𝟎

=

= 𝟐𝒆𝒏+𝟏∫ 𝒆−(𝒏+𝟏)𝒕𝟐

∞

𝟎

𝒅𝒕;|

|

(𝒏 + 𝟏)𝒕𝟐 = 𝒛; 𝒅𝒕 =𝒅𝒛

𝟐(𝒏 + 𝟏)√𝒛

𝒏 + 𝟏

𝒕 = √𝒛

𝒏 + 𝟏

|

|

𝛀 = 𝟐𝒆𝒏+𝟏∫ 𝒆−𝒛 ⋅𝒅𝒛

𝟐(𝒏 + 𝟏)√𝒛

𝒏+ 𝟏

∞

𝟎

=𝒆𝒏+𝟏

√𝒏 + 𝟏∫ 𝒛

𝟏𝟐−𝟏𝒆−𝒛

∞

𝟎

𝒅𝒛 =√𝝅𝒆𝒏+𝟏

√𝒏 + 𝟏

𝑵𝒐𝒕𝒆: ∫ 𝒛𝟏𝟐−𝟏𝒆−𝒛𝒅𝒛

∞

𝟎

= 𝚪 (𝟏

𝟐) = √𝝅

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𝒆

𝟎

𝒅𝒙 =

𝒕=𝟏−𝐥𝐨𝐠 𝒙;

𝒅𝒕=−𝒆𝟏−𝒕𝒅𝒕𝒆𝒏+𝟏∫

𝒆−𝒕(𝒏+𝟏)

√𝒕𝒅𝒕

∞

𝟎

𝓛{𝒕𝒏} = ∫ 𝒆−𝒔𝒕𝒕𝒏𝒅𝒕∞

𝟎

=𝚪(𝒏 + 𝟏)

𝒔𝒏+𝟏, 𝐰𝐡𝐞𝐧 𝒏 = −

𝟏

𝟐; 𝒔 = 𝒏 + 𝟏 ⇒

∫𝒆−𝒕(𝒏+𝟏)

√𝒕

∞

𝟎

𝒅𝒕 =𝚪(𝟏𝟐)

𝒔𝟏𝟐

=√𝝅

√𝒏 + 𝟏⇒ 𝛀(𝒏) =

√𝝅𝒆𝒏+𝟏

√𝒏 + 𝟏

Therefore,



𝒆

𝟎

𝒅𝒙 =√𝝅𝒆𝒏+𝟏

√𝒏 + 𝟏

Solution 3 by Muhammad Afzal-Pakistan



𝒆

𝟎

𝒅𝒙 =𝒖=𝟏−𝐥𝐨𝐠 𝒙

∫𝒆(𝟏−𝒖)(𝒏+𝟏)

√𝒖

∞

𝟎

𝒅𝒖 = 𝒆𝒏+𝟏∫𝒆−𝒖(𝒏+𝟏)

√𝒖𝒅𝒖

∞

𝟎

=

=𝒆𝒏+𝟏

𝒏 + 𝟏∫ 𝒖−

𝟏𝟐𝒆−𝒖(𝒏+𝟏)𝒅𝒖

∞

𝟎

=𝒚=𝒖(𝒏+𝟏) 𝒆𝒏+𝟏

𝒏 + 𝟏∫

𝒚−𝟏𝟐

(𝒏 + 𝟏)−𝟏𝟐

⋅ 𝒆−𝒚∞

𝟎

𝒅𝒚 =

=𝒆𝒏+𝟏

√𝒏 + 𝟏∫ 𝒚

𝟏𝟐−𝟏𝒆−𝒚𝒅𝒚

∞

𝟎

=𝒆𝒏+𝟏

√𝒏 + 𝟏𝚪(𝟏

𝟐)

Therefore,



𝒆

𝟎


√𝒏 + 𝟏

Solution 4 by Adrian Popa-Romania



𝒆

𝟎

𝒅𝒙 =√𝟏−𝐥𝐨𝐠 𝒙=𝒕

𝟐∫ 𝒆𝒏+𝟏 ⋅ 𝒆−𝒕𝟐(𝒏+𝟏)𝒅𝒕

∞

𝟎

= 𝟐𝒆𝒏+𝟏∫ 𝒆−𝒕𝟐(𝒏+𝟏)𝒅𝒕

∞

𝟎

=

=(𝒏+𝟏)𝒕𝟐=𝒖

𝟐𝒆𝒏+𝟏∫ 𝒆−𝒖 ⋅𝟏

𝟐√𝒏 + 𝟏𝒖−

𝟏𝟐

∞

𝟎

𝒅𝒖 =𝒆𝒏+𝟏

√𝒏 + 𝟏∫ 𝒖−

𝟏𝟐𝒆−𝒖

∞

𝟎

𝒅𝒖 =

=𝒆𝒏+𝟏

√𝒏 + 𝟏𝚪(𝟏

𝟐) =

𝒆𝒏+𝟏√𝝅

√𝒏 + 𝟏

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𝒆

𝟎

𝒅𝒙,𝒏 > 𝟎


Solution 1 by Ajentunmobi Abdulqoyuum-Nigeria



𝒆

𝟎

𝒅𝒙 =; |𝒕 = √𝟏 − 𝐥𝐨𝐠 𝒙 ;𝒅𝒕 =

𝟏

𝟐𝒙√𝟏 − 𝐥𝐨𝐠 𝒙𝒅𝒙

𝒅𝒙 = −𝟐𝒕𝒆𝟏−𝒕𝟐, 𝒙 = 𝒆𝟏−𝒕

𝟐| ;

= ∫𝟐𝒕𝒆𝟏−𝒕

𝟐𝒆𝟏−𝒕

𝟐

𝒕𝒅𝒕

∞

𝟎

= 𝟐∫ 𝒆(𝒏+𝟏)−(𝒏+𝟏)𝒕𝟐𝒅𝒕

∞

𝟎

=

= 𝟐𝒆𝒏+𝟏∫ 𝒆−(𝒏+𝟏)𝒕𝟐

∞

𝟎

𝒅𝒕;|

|

(𝒏 + 𝟏)𝒕𝟐 = 𝒛; 𝒅𝒕 =𝒅𝒛

𝟐(𝒏 + 𝟏)√𝒛

𝒏 + 𝟏

𝒕 = √𝒛

𝒏 + 𝟏

|

|

𝛀 = 𝟐𝒆𝒏+𝟏∫ 𝒆−𝒛 ⋅𝒅𝒛

𝟐(𝒏 + 𝟏)√𝒛

𝒏+ 𝟏

∞

𝟎

=𝒆𝒏+𝟏

√𝒏 + 𝟏∫ 𝒛

𝟏𝟐−𝟏𝒆−𝒛

∞

𝟎

𝒅𝒛 =√𝝅𝒆𝒏+𝟏

√𝒏 + 𝟏

𝑵𝒐𝒕𝒆: ∫ 𝒛𝟏𝟐−𝟏𝒆−𝒛𝒅𝒛

∞

𝟎

= 𝚪 (𝟏

𝟐) = √𝝅




𝒆

𝟎

𝒅𝒙 =

𝒕=𝟏−𝐥𝐨𝐠 𝒙;

𝒅𝒕=−𝒆𝟏−𝒕𝒅𝒕𝒆𝒏+𝟏∫

𝒆−𝒕(𝒏+𝟏)

√𝒕𝒅𝒕

∞

𝟎

𝓛{𝒕𝒏} = ∫ 𝒆−𝒔𝒕𝒕𝒏𝒅𝒕∞

𝟎

=𝚪(𝒏 + 𝟏)

𝒔𝒏+𝟏, 𝐰𝐡𝐞𝐧 𝒏 = −

𝟏

𝟐; 𝒔 = 𝒏 + 𝟏 ⇒

∫𝒆−𝒕(𝒏+𝟏)

√𝒕

∞

𝟎

𝒅𝒕 =𝚪(𝟏𝟐)

𝒔𝟏𝟐

=√𝝅

√𝒏 + 𝟏⇒ 𝛀(𝒏) =

√𝝅𝒆𝒏+𝟏

√𝒏 + 𝟏

Therefore,

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𝒆

𝟎


√𝒏 + 𝟏




𝒆

𝟎

𝒅𝒙 =𝒖=𝟏−𝐥𝐨𝐠 𝒙

∫𝒆(𝟏−𝒖)(𝒏+𝟏)

√𝒖

∞

𝟎

𝒅𝒖 = 𝒆𝒏+𝟏∫𝒆−𝒖(𝒏+𝟏)

√𝒖𝒅𝒖

∞

𝟎

=

=𝒆𝒏+𝟏

𝒏 + 𝟏∫ 𝒖−

𝟏𝟐𝒆−𝒖(𝒏+𝟏)𝒅𝒖

∞

𝟎

=𝒚=𝒖(𝒏+𝟏) 𝒆𝒏+𝟏

𝒏 + 𝟏∫

𝒚−𝟏𝟐

(𝒏 + 𝟏)−𝟏𝟐

⋅ 𝒆−𝒚∞

𝟎

𝒅𝒚 =

=𝒆𝒏+𝟏

√𝒏 + 𝟏∫ 𝒚

𝟏𝟐−𝟏𝒆−𝒚𝒅𝒚

∞

𝟎

=𝒆𝒏+𝟏

√𝒏 + 𝟏𝚪(𝟏

𝟐)

Therefore,



𝒆

𝟎


√𝒏 + 𝟏




𝒆

𝟎

𝒅𝒙 =√𝟏−𝐥𝐨𝐠 𝒙=𝒕

𝟐∫ 𝒆𝒏+𝟏 ⋅ 𝒆−𝒕𝟐(𝒏+𝟏)𝒅𝒕

∞

𝟎

= 𝟐𝒆𝒏+𝟏∫ 𝒆−𝒕𝟐(𝒏+𝟏)𝒅𝒕

∞

𝟎

=

=(𝒏+𝟏)𝒕𝟐=𝒖

𝟐𝒆𝒏+𝟏∫ 𝒆−𝒖 ⋅𝟏

𝟐√𝒏 + 𝟏𝒖−

𝟏𝟐

∞

𝟎

𝒅𝒖 =𝒆𝒏+𝟏

√𝒏 + 𝟏∫ 𝒖−

𝟏𝟐𝒆−𝒖

∞

𝟎

𝒅𝒖 =

=𝒆𝒏+𝟏

√𝒏 + 𝟏𝚪(𝟏

𝟐) =

𝒆𝒏+𝟏√𝝅

√𝒏 + 𝟏

1518. Prove that:

∫𝒅𝒙

(𝟒 𝐥𝐨𝐠𝟐 𝒙 + 𝝅𝟐)𝟐(𝒙𝟐 + 𝟏)

∞

𝟎

=𝐥𝐨𝐠𝟐

𝟒𝝅𝟑+

𝟏

𝟗𝟔𝝅

Proposed by Ty Halpen-Florida-SUA


𝛀 = ∫𝒅𝒙

(𝟒 𝐥𝐨𝐠𝟐 𝒙 + 𝝅𝟐)𝟐(𝒙𝟐 + 𝟏)

∞

𝟎

=𝒙=𝒆

−(𝝅𝒕𝟐) 𝟏

𝟐𝝅𝟑∫

𝒆−𝝅𝒕𝟐

(𝒕𝟐 + 𝟐)𝟐(𝟏 + 𝒆−𝝅𝒕)𝒅𝒕

∞

∞

=

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=𝟏

𝟒𝝅𝟑∫

𝐬𝐞𝐜𝐡 (𝝅𝒕𝟐 )

(𝒕𝟐 + 𝟏)𝟐𝒅𝒕

∞

∞

=𝟏

𝟐𝝅𝟑∫

𝐬𝐞𝐜𝐡 (𝝅𝒕𝟐 )

(𝒕𝟐 + 𝟏)𝟐

∞

𝟎

𝒅𝒕

𝐔𝐬𝐢𝐧𝐠: 𝐬𝐞𝐜𝐡 (𝝅𝒕

𝟐) =

𝟒

𝝅∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)

𝒕𝟐 + (𝟐𝒌 + 𝟏)𝟐

∞

𝒌=𝟎

𝛀 =𝟐

𝝅𝟒∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)

∞

𝒌=𝟎

∫𝒅𝒕

(𝒕𝟐 + 𝟏)𝟐(𝒕𝟐 + (𝟐𝒌 + 𝟏)𝟐)

∞

𝟎

𝑰(𝒂) = ∫𝒅𝒕

(𝒕𝟐 + 𝟏)𝟐(𝒕𝟐 + 𝒂𝟐)

∞

𝟎

=

=𝟏

𝒂𝟐 − 𝟏∫

𝒅𝒕

(𝟏 + 𝒕𝟐)𝟐

∞

𝟎

−𝟏

(𝒂𝟐 − 𝟏)𝟐∫ (

𝟏

𝟏 + 𝒕𝟐−

𝟏

𝒕𝟐 + 𝒂𝟐)

∞

𝟎

𝒅𝒕 =

=𝝅

𝟒(𝒂𝟐 − 𝟏)−

𝝅(𝒂− 𝟏)

𝟐𝒂(𝒂𝟐 − 𝟏)𝟐

𝑰(𝒂) =𝝅(𝒂𝟑 − 𝟑𝒂 + 𝟐)

𝟒𝒂(𝒂𝟐 − 𝟏)𝟐=𝝅(𝒂− 𝟏)𝟐(𝒂 + 𝟐)

𝟒𝒂(𝒂 − 𝟏)𝟐(𝒂 + 𝟏)𝟐=𝝅(𝒂 + 𝟐)

𝟒𝒂(𝒂 + 𝟏)𝟐

𝛀 =𝟏

𝟐𝝅𝟑∑(−𝟏)𝒌

𝟐𝒌 + 𝟑

(𝟐𝒌 + 𝟐)𝟐

∞

𝒌=𝟎

=𝟏

𝟐𝝅𝟑∑[

(−𝟏)𝒌

𝟐𝒌 + 𝟐+

(−𝟏)𝒌

(𝟐𝒌 + 𝟐)𝟐]

∞

𝒌=𝟎

=

=𝟏

𝟒𝝅𝟑∑(−𝟏)𝒌−𝟏

𝒌

∞

𝒌=𝟏

+𝟏

𝟖𝝅𝟑∑(−𝟏)𝒌−𝟏

𝒌𝟐

∞

𝒌=𝟏

=𝐥𝐨𝐠 𝟐

𝟒𝝅𝟑+

𝟏

𝟗𝟔𝝅

Therefore,

∫𝒅𝒙

(𝟒 𝐥𝐨𝐠𝟐 𝒙 + 𝝅𝟐)𝟐(𝒙𝟐 + 𝟏)

∞

𝟎

=𝐥𝐨𝐠 𝟐

𝟒𝝅𝟑+

𝟏

𝟗𝟔𝝅

1519. Prove that:

𝟏

𝒆< |∫ 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙) 𝑱𝟎(𝟐√𝒙)

∞

𝟎

𝒅𝒙| <𝝅𝟐

𝟔

where 𝑱𝒏(𝒙) is the Bessel function of order 𝒏.

Proposed by Angad Singh-India

Solution 1 by proposer

We know from the definition of Bessel function, that

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𝑱𝟎(𝒙) = ∑(−𝟏)𝒌

𝒌!𝟐

∞

𝒌=𝟎

(𝒙

𝟐)𝟐𝒌 𝒏>𝟎⇒ ∫ 𝒆−𝒏𝒙𝑱𝟎(𝟐√𝒙)

∞

𝟎

𝒅𝒙 =𝒆−𝟏𝒏

𝒏⇒

∫ ∑𝒆−𝒏𝒙

𝒏𝑱𝟎(𝟐√𝒙)

∞

𝒏=𝟎

𝒅𝒙∞

𝟎

= ∑𝒆−𝟏𝒏

𝒏𝟐

∞

𝒏=𝟏

Observe that:

∑𝒆−𝟏𝒏

𝒏𝟐

∞

𝒏=𝟏

>𝟏

𝒆; ∑

𝒆−𝟏𝒏

𝒏𝟐

∞

𝒏=𝟏

<∑𝟏

𝒏𝟐

∞

𝒏=𝟏

=𝝅𝟐

𝟔 𝐚𝐧𝐝 ∑

𝒆−𝒙

𝒏

∞

𝒏=𝟏

= − 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙)


𝑱𝒏(𝒙) = (𝒙

𝟐)𝒏

∑(−𝟏)𝒌𝒙𝟐𝒌

𝟐𝒌𝒌! 𝚪(𝒏 + 𝒌 + 𝟏)

∞

𝒌=𝟎

⇒ 𝑱𝟎(𝟐√𝒙) =∑(−𝟏)𝒌(𝟐√𝒙)

𝟐𝒌

𝟐𝒌𝒌! 𝚪(𝒌 + 𝟏)

∞

𝒌=𝟎

=∑(−𝟏)𝒌

𝒌!𝟐𝒙𝒌

∞

𝒌=𝟎

⇒ |∫ 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙) 𝑱𝟎(𝟐√𝒙)∞

𝟎

𝒅𝒙| = |∑(−𝟏)𝒌

𝒌!𝟐

∞

𝒌=𝟎

∫ 𝒙𝒌 ⋅ 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙)∞

𝟎

𝒅𝒙| =

= |∑(−𝟏)𝒌+𝟏

𝒌!𝟐

∞

𝒌=𝟎

𝑳𝒊𝒌+𝟐(𝟏)(𝒌!)| = |∑(−𝟏)𝒌𝑳𝒊𝒌+𝟐(𝟏)

𝒌!

∞

𝒌=𝟎

| = |∑(−𝟏)𝒌+𝟏

𝒌!𝜻(𝒌 + 𝟐)

∞

𝒌=𝟎

| =

= |∑𝟏

𝒏𝟐

∞

𝒌=𝟏

∑(−𝟏)𝒌+𝟏

𝒏𝒌(𝒌!)

∞

𝒏=𝟎

| = |−∑𝟏

𝒏𝟐𝒆−𝟏𝒏

∞

𝒌=𝟎

| = ∑ |−𝒆−

𝟏𝒏

𝒏𝟐|

𝒏

𝒌=𝟏

= ∑𝒆−𝟏𝒏

𝒏𝟐

∞

𝒏=𝟏

≅ 𝟎.𝟖𝟒𝟔𝟒𝟐

𝟏

𝒆≅ 𝟎. 𝟑𝟔𝟕𝟖𝟕;

𝝅𝟐

𝟔≅ 𝟏. 𝟔𝟒𝟒𝟗 ⇒ ∑

𝒆−𝟏𝒏

𝒏𝟐

∞

𝒏=𝟏

>𝟏

𝒆; ∑

𝒆−𝟏𝒏

𝒏𝟐

∞

𝒏=𝟏

<𝝅𝟐

𝟔

Therefore,

𝟏

𝒆< |∫ 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙) 𝑱𝟎(𝟐√𝒙)

∞

𝟎

𝒅𝒙| <𝝅𝟐

𝟔

1520. Let 𝒃 > 𝒂 > 𝟏 and 𝒏 be a positive integer. Prove that:

∫√𝒆𝒏𝒙

𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏

𝐥𝐨𝐠𝒃

𝐥𝐨𝐠𝒂

𝒅𝒙 ≤ 𝐥𝐨𝐠( √𝒃

𝒂

𝒏+𝟏

) , 𝒙 ∈ ℝ

Proposed by George Apostolopoulos-Messolonghi-Greece

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𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏 ≥ (𝒏 + 𝟏) √𝒆𝒏𝒙+(𝒏−𝟏)𝒙+⋯+𝒙+𝟎𝒏+𝟏

=

= (𝒏 + 𝟏) √𝒆𝒏𝒙(𝒏+𝟏)

𝟐

𝒏+𝟏

= (𝒏 + 𝟏)𝒆𝒏𝒙𝟐 = (𝒏 + 𝟏)√𝒆𝒏𝒙 ⇒

∫√𝒆𝒏𝒙

𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 ≤𝟏

𝒏 + 𝟏∫

√𝒆𝒏𝒙

√𝒆𝒏𝒙

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 =

=𝟏

𝒏 + 𝟏𝒙|𝐥𝐨𝐠 𝒂

𝐥𝐨𝐠 𝒃

=𝟏

𝒏 + 𝟏(𝐥𝐨𝐠 𝒃 − 𝐥𝐨𝐠𝒂) =

𝟏

𝒏 + 𝟏𝐥𝐨𝐠 (

𝒃

𝒂) = 𝐥𝐨𝐠( √

𝒃

𝒂

𝒏+𝟏

)

Therefore,

∫√𝒆𝒏𝒙

𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂


𝒂

𝒏+𝟏

) , 𝒙 ∈ ℝ

Solution 2 by Ruxandra Daniela Tonilă-Romania

∫√𝒆𝒏𝒙

𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 = ∫√𝒆𝒏𝒙

𝒆(𝒏+𝟏)𝒙 − 𝟏𝒆𝒙 − 𝟏

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 =

= ∫𝒆𝒏𝒙(𝒆𝒙 − 𝟏)

𝒆(𝒏+𝟏)𝒙 − 𝟏⋅𝟏

√𝒆𝒏𝒙

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 = ∫𝒆(𝒏+𝟏)𝒙 − 𝒆𝒏𝒙

𝒆(𝒏+𝟏)𝒙 − 𝟏⋅𝟏

√𝒆𝒏𝒙

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 =

= ∫𝒆(𝒏+𝟏)𝒙 (𝟏 − (

𝟏𝒆)𝒙

)

𝒆(𝒏+𝟏)𝒙(𝟏 − [(𝟏𝒆)𝒙

]𝒏+𝟏 ⋅

𝟏

√𝒆𝒏𝒙

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 =

= ∫𝟏 − (

𝟏𝒆)𝒙

(𝟏 − (𝟏𝒆)𝒙

) (𝟏 + (𝟏𝒆)𝒙

+ (𝟏𝒆)𝟐𝒙

+⋯+ (𝟏𝒆)𝒏𝒙

)

⋅𝟏

√𝒆𝒏𝒙

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 =

=𝟏

𝒏 + 𝟏∫

𝒏 + 𝟏

𝟏 + (𝟏𝒆)𝒙

+ (𝟏𝒆)𝟐𝒙

+⋯+ (𝟏𝒆)𝒏𝒙 ⋅

𝟏

√𝒆𝒏𝒙𝒅𝒙

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

≤𝑯𝑮𝑴

≤𝟏

𝒏 + 𝟏∫

𝟏

√𝒆𝒏𝒙√𝟏 ⋅ 𝒆𝒙 ⋅ 𝒆𝟐𝒙 ⋅ … ⋅ 𝒆𝒏𝒙

𝒏+𝟏𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

𝒅𝒙 =𝟏

𝒏 + 𝟏∫

𝟏

√𝒆𝒏𝒙⋅ √𝒆

𝒏(𝒏+𝟏)𝒙𝟐

𝒏+𝟏

𝒅𝒙𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂

=

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=𝟏

𝒏 + 𝟏(𝐥𝐨𝐠𝒃 − 𝐥𝐨𝐠 𝒂) = 𝐥𝐨𝐠( √

𝒃

𝒂

𝒏+𝟏

)

Therefore,

∫√𝒆𝒏𝒙

𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏

𝐥𝐨𝐠 𝒃

𝐥𝐨𝐠 𝒂


𝒂

𝒏+𝟏

) , 𝒙 ∈ ℝ

1521. If for 𝒏 ≥ 𝒌 and 𝒏, 𝒌, 𝒂 > 𝟎; 𝒏, 𝒌 ∈ ℕ;

𝝃𝒂(𝒌, 𝒏) =∑√𝒂𝒓

𝒏

𝒓=𝒌

= √𝒂𝒌 + √𝒂

𝒌+𝟏+⋯+ √𝒂

𝒏 𝐚𝐧𝐝 𝜻𝒏(𝒔) = ∑𝟏

𝒌𝒔

𝒏

𝒌=𝟏

=𝟏

𝟏𝒔+𝟏

𝟐𝒔+⋯+

𝟏

𝒏𝒔

then prove:

𝟐(√𝒏 + 𝟏 − 𝟏) + 𝝃𝒆(𝟐, 𝒏 + 𝟏) < 𝜻𝒏(𝟎) + 𝜻𝒏 (𝟏

𝟐) + 𝜻𝒏(𝟏) < 𝝃𝒆(𝟏, 𝒏) + 𝟐√𝒏

𝐰𝐡𝐞𝐫𝐞, 𝒆 = ∑𝟏

𝒏!

∞

𝒏=𝟎

Proposed by Amrit Awasthi-India

Solution by proposer

We can write 𝐥𝐨𝐠 𝒙 as 𝐥𝐨𝐠 𝒙 = ∫𝟏

𝒕𝒅𝒕

𝒙

𝟏 and 𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌) = ∫

𝟏

𝒕𝒅𝒕

𝟏+𝟏

𝒌𝟏

.

Now, as 𝒕 ∈ [𝟏, 𝟏 +𝟏

𝒌] : 𝟏 ≤ 𝒕 ≤ 𝟏 +

𝟏

𝒌⇒

𝟏

𝟏+𝟏

𝒌

≤𝟏

𝒕≤ 𝟏

Integrating under the same interval we have:

∫𝒌

𝒌 + 𝟏

𝟏+𝟏𝒌

𝟏

𝒅𝒕 ≤ ∫𝟏

𝒕

𝟏+𝟏𝒌

𝟏

𝒅𝒕 ≤ ∫ 𝒅𝒕𝟏+𝟏𝒌

𝟏

⇔

𝒌

𝒌 + 𝟏(𝟏 +

𝟏

𝒌− 𝟏) ≤ 𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌) ≤ (𝟏 +

𝟏

𝒌− 𝟏) ⇔

𝟏

𝒌 + 𝟏≤ 𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌) ≤

𝟏

𝒌

Now, summing up from 𝒌 = 𝟏 to 𝒌 = 𝒏 we have:

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∑𝒆𝟏𝒌+𝟏

𝒏

𝒌=𝟏

≤∑(𝟏 +𝟏

𝒌)

𝒏

𝒌=𝟏

≤∑𝒆𝟏𝒌

𝒏

𝒌=𝟏

⇔∑ √𝒆𝒌

𝒏+𝟏

𝒌=𝟐

≤∑𝟏

𝒌𝟎

𝒏

𝒌=𝟏

+∑𝟏

𝒌

𝒏

𝒌=𝟏

≤∑ √𝒆𝒌

𝒏

𝒌=𝟏

𝝃𝒆(𝟐, 𝒏 + 𝟏) ≤ 𝜻𝒏(𝟎) + 𝜻𝒏(𝟏) + 𝝃𝒆(𝟏, 𝒏); (∗)

Now, consider the following

𝟐(√𝒎+ 𝟏 − √𝒎) = 𝟐(√𝒎 + 𝟏 − √𝒎)(√𝒎+ 𝟏 + √𝒎)

√𝒎+ 𝟏 + √𝒎=

𝟐

√𝒎+ 𝟏 + √𝒎<

<𝟐

√𝒎+ √𝒎=𝟏

√𝒎

Thus, 𝟐(√𝒎+ 𝟏 − √𝒎) <𝟏

√𝒎

Proceeding in a similar manner we can prove that:

𝟐(√𝒎 − √𝒎− 𝟏) >𝟏

√𝒎

Combining both results, we have

𝟐(√𝒎+ 𝟏 − √𝒎) <𝟏

√𝒎< 𝟐(√𝒎− √𝒎 − 𝟏)

Summing from 𝒎 = 𝟏 to 𝒎 = 𝒏 we have

∑ 𝟐(√𝒎+ 𝟏 − √𝒎)

𝒏

𝒎=𝟏

< ∑𝟏

√𝒎

𝒏

𝒎=𝟏

< ∑ 𝟐(√𝒎− √𝒎− 𝟏)

𝒏

𝒎=𝟏

Now, both the upper and lower bounds are telescopic sums, hence after canceling the terms we are left with

𝟐(√𝒏 + 𝟏 − 𝟏) < 𝜻𝒏 (𝟏

𝟐) < 𝟐√𝒏; (∗∗)

Adding (∗), (∗∗) the final inequality becomes strict as the second inequality is strict. Hence,

𝟐(√𝒏 + 𝟏 − 𝟏) + 𝝃𝒆(𝟐, 𝒏 + 𝟏) < 𝜻𝒏(𝟎) + 𝜻𝒏 (𝟏

𝟐) + 𝜻𝒏(𝟏) < 𝝃𝒆(𝟏, 𝒏) + 𝟐√𝒏

1522. For 𝒏 ≥ 𝟎 prove or disprove:

𝐥𝐢𝐦𝒌→∞

𝐥𝐢𝐦𝒏→∞

∏(𝟏+∫ (𝟏 − 𝐭𝐚𝐧 𝒙

𝟏 + 𝐭𝐚𝐧 𝒙)𝒏𝝅

𝟒

𝟎

𝒅𝒙)

𝒏𝒎𝒌

𝒎=𝟏

𝟏

√𝒌= 𝒆

𝜸𝟐

where 𝜸 is Euler-Mascheroni constant.

Proposed by Naren Bhandari-Bajura-Nepal

www.ssmrmh.ro


Solution by Artan Ajredini-Presheva-Serbie

Let 𝒇𝒏(𝒙) = (𝟏−𝐭𝐚𝐧 𝒙

𝟏+𝐭𝐚𝐧 𝒙)𝒏

. For 𝒙 ∈ [𝟎,𝝅

𝟒] we have |𝒇𝒏(𝒙)| ≤ 𝟏 = 𝒈(𝒙), and

∫ 𝒈(𝒙)

𝝅𝟒

𝟎

𝒅𝒙 =𝝅

𝟒. 𝐍𝐨𝐰,


𝒇𝒏(𝒙) = {𝟎, 𝒙 ∈ (𝟎,

𝝅

𝟒]

𝟏, 𝒙 = 𝟎

By Lebesgue Dominated Convergence Theorem, we have that:


∫ 𝒇𝒏(𝒙)

𝝅𝟒

𝟎

𝒅𝒙 = 𝟎 ⋅ 𝝌(𝟎,𝝅𝟒]+ 𝟏 ⋅ 𝝌{𝟏} = 𝟎; (𝟏)

On the other hand, substituting 𝒖 =𝟏−𝐭𝐚𝐧 𝒙

𝟏+𝐭𝐚𝐧 𝒙 in the integral 𝑰 = ∫ 𝒏 (

𝟏−𝐭𝐚𝐧 𝒙

𝟏+𝐭𝐚𝐧 𝒙)𝒏𝝅

𝟒𝟎

𝒅𝒙 and

integrating by parts, we get

𝑰 =𝟏

𝟐−∫

𝒖𝒏(𝟏 − 𝒖𝟐)

𝟏 + 𝒖𝟐

𝟏

𝟎

𝒅𝒖

Again, let 𝒈𝒏(𝒖) =𝒖𝒏(𝟏−𝒖𝟐)

𝟏+𝒖𝟐. For 𝒖 ∈ [𝟎, 𝟏] we have that 𝒈𝒏(𝒖) ≤

𝟏−𝒖𝟐

𝟏+𝒖𝟐= 𝒉(𝒖), and

∫ 𝒉(𝒖)𝟏

𝟎

𝒅𝒖 =𝝅 − 𝟐

𝟐.𝐍𝐨𝐰, 𝐥𝐢𝐦

𝒏→∞𝒈𝒏(𝒖) = 𝟎, ∀𝒖 ∈ [𝟎, 𝟏], 𝐚𝐧𝐝

using Lebesgue Dominated Convergence Theorem, we obtain


∫ 𝒈𝒏(𝒖)𝟏

𝟎

𝒅𝒖 = 𝟎. 𝐓𝐡𝐞𝐫𝐞𝐟𝐨𝐫𝐞,


∫ 𝒏(𝟏 − 𝐭𝐚𝐧 𝒙

𝟏 + 𝐭𝐚𝐧 𝒙)𝒏

𝝅𝟒

𝟎

𝒅𝒙 =𝟏

𝟐; (𝟐)

By using (𝟏), (𝟐), we get

𝐥𝐢𝐦𝒌→∞


∏(𝟏+∫ (𝟏 − 𝐭𝐚𝐧 𝒙


𝝅𝟒

𝟎

𝒅𝒙)

𝒏𝒎𝒌

𝒎=𝟏

𝟏

√𝒌=

= 𝐥𝐢𝐦𝒌→∞

∏(𝐥𝐢𝐦𝒏→∞

(𝟏 + ∫ (𝟏− 𝐭𝐚𝐧 𝒙


𝒅𝒙

𝝅𝟒

𝟎

)

𝒏𝒎𝒌

𝒎=𝟏

𝟏

√𝒌=

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∏(𝐞𝐱𝐩 {𝐥𝐢𝐦𝒏→∞

𝟏

𝒎∫ 𝒏(

𝟏− 𝐭𝐚𝐧 𝒙


𝒅𝒙

𝝅𝟒

𝟎

})

𝒌

𝒎=𝟏

𝟏

√𝒌= 𝐥𝐢𝐦𝒌→∞

∏𝟏

√𝒌𝒆𝟏𝟐𝒎

𝒌

𝒎=𝟏

=


∏𝒆𝟏𝟐(𝟏𝒎−𝐥𝐨𝐠 𝒌)

𝒌

𝒎=𝟏

= 𝒆𝜸𝟐

1523.

𝝓𝒏 =𝟒

𝝅∫

𝐜𝐨𝐭𝐡(𝒏𝒙−𝟏) − 𝒙𝒏−𝟏

𝒏(𝟏 + 𝒙𝟐)𝟐𝒅𝒙

∞

𝟎

, 𝚽𝒏 =𝐜𝐨𝐬(𝒏𝝅)

𝒏; ∀𝒏 ∈ ℕ

Prove that:

𝐥𝐢𝐦𝒎→∞


∏∏(𝟏 +𝝓𝒏)𝒏𝒌−𝒓𝚽𝒓

𝒎

𝒓=𝟐

∞

𝒌=𝟏

= 𝒆𝜸

where 𝜸 is Euler-Mascheroni constant.

Proposed by Surjeet Singhania-India

Solution by Naren Bhandari-Bajura-Nepal

Since we have two convergent integrals so we have validity for the linearity of integrals for

𝝓𝒏 that is

𝝅

𝟒𝝓𝒏 = ∫

𝐜𝐨𝐭𝐡 (𝒏𝒙)

𝒏(𝟏 + 𝒙𝟐)𝟐

∞

𝟎

𝒅𝒙⏟

𝑰𝟏

− ∫𝒙𝒅𝒙

𝒏𝟐(𝟏 + 𝒙𝟐)𝟐

∞

𝟎⏟ 𝑰𝟐

Now, let 𝝀𝒏(𝒙) =𝐜𝐨𝐭𝐡(

𝒏

𝒙)

𝒏(𝟏+𝒙𝟐)𝟐 ≤ |

𝟏

𝒏(𝟏+𝒙𝟐)𝟐| ≤ |

𝟏

𝟐𝒏(𝟏+𝒙𝟐)| = 𝑽𝒏(𝒙) and hence

∫ 𝑽𝒏(𝒙)∞

𝟎

𝒅𝒙 =𝝅

𝟒𝒏 𝐚𝐧𝐝 𝐥𝐢𝐦

𝒏→∞𝑽𝒏(𝒙) = 𝟎.

By Lebesgue Dominating convergence theorem, we have:


∫ 𝝀𝒏(𝒙)∞

𝟎

𝒅𝒙 = 𝐥𝐢𝐦𝒏→∞

∫ 𝑽𝒏(𝒙)∞

𝟎

𝒅𝒙 = 𝟎; (𝟏)

The integral 𝑰𝟐 has primitive which is

∫𝒙𝒅𝒙

𝒏𝟐(𝟏 + 𝒙𝟐)

∞

𝟎

= ∫𝟐𝒙𝒅𝒙

𝟐𝒏𝟐(𝟏 + 𝒙𝟐)𝟐

∞

𝟎

= −𝟏

𝟐𝒏𝟐(𝟏 + 𝒙𝟐)𝟐|𝟎

∞

=𝟏

𝟐𝒏𝟐

www.ssmrmh.ro


Giving us 𝐥𝐢𝐦𝒏→∞

𝑰𝟐 = 𝟎 (which we can even perform to show that

𝑰𝟐 = 𝟎 as 𝒏 → ∞ even by LTCD) ;(2)

So, from (1),(2)we have:


(𝟏 + 𝝓𝒏)𝒏 ≤ 𝐥𝐢𝐦

𝒏→∞(𝟏 + 𝑰𝟏 − 𝑰𝟐)

𝒏 = 𝐥𝐢𝐦𝒏→∞

(𝟏 +𝟏

𝒏−𝟐

𝒏𝝅𝟐)𝒏

= 𝒆

Now, we have just to evaluate

𝐞𝐱𝐩 ( 𝐥𝐢𝐦𝒎→∞

∑∑𝐜𝐨𝐬(𝝅𝒓)

𝒌𝒓𝒓

𝒎

𝒓=𝟐

∞

𝒌=𝟏

) = 𝐞𝐱𝐩 (𝟏 −∑𝜻(𝒌) − 𝟏

𝒌

∞

𝒌=𝟐

)

= 𝐞𝐱𝐩 (𝟏 −∑𝒙𝒌

𝒌!⋅

𝒅𝒙

𝒙𝒆𝒙(𝒆𝒙 − 𝟏)

∞

𝒌=𝟐

)

Interchanging sum and integral signs we have

∫𝒆𝒙 − 𝒙 − 𝟏

𝒙𝒆𝒙(𝒆𝒙 − 𝟏)

∞

𝟎

𝒅𝒙 = ∫ 𝒆−𝒙𝒅𝒙∞

𝟎

−∫ (𝟏

𝒆𝒙 − 𝟏−𝟏

𝒙𝒆𝒙)

∞

𝟎

𝒅𝒙 = 𝟏 − 𝜸

Since it is well known that ∫ (𝟏

𝒆𝒙−𝟏−

𝟏

𝒙𝒆𝒙)

∞

𝟎𝒅𝒙 = 𝜸 and hence we have desired result 𝒆𝜸.

In other way we calculate

∑𝜻(𝒌) − 𝟏

𝒌

∞

𝒌=𝟐

= ∑∑𝟏

𝒌𝒎𝒌

𝒌≥𝟐𝒎≥𝟐

= 𝐥𝐢𝐦𝑵→∞

∑ (𝟏

𝒎− 𝐥𝐨𝐠𝒎 + 𝐥𝐨𝐠(𝒎+ 𝟏))

𝑵

𝒎=𝟐

=

= − 𝐥𝐢𝐦𝑵→∞

(𝑯𝑵 − 𝟏 − 𝐥𝐨𝐠𝑵) =𝟏 − 𝜸.

1524. Find:

𝛀 = 𝐥𝐢𝐦𝛆→𝟎𝛆>0

∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝜺

𝒅𝒙


Solution 1 by Soumitra Mandal-India

𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝟎

𝒅𝒙 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

𝟏

𝟎

𝒅𝒙 +∫𝒙𝟐 𝐥𝐨𝐠𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝟏

𝒅𝒙 =𝒙=𝟏𝒚

= ∫𝒙𝟐(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔

𝟏

𝟎

𝒅𝒙 −∫𝐥𝐨𝐠𝒚

𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚

𝟏

𝟎

=

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= ∫𝒙𝟐(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔

𝟏

𝟎

𝒅𝒙 − ∫(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔

𝟏

𝟎

=

= 𝟐∫𝒙𝟐 𝐥𝐨𝐠𝒙

𝟏 − 𝒙𝟔

𝟏

𝟎

𝒅𝒙 −∫𝒙𝟒 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔

𝟏

𝟎

𝒅𝒙 − ∫𝐥𝐨𝐠𝒙

𝟏 − 𝒙𝟔

𝟏

𝟎

𝒅𝒙 =

𝟐∑∫ 𝒙𝟔𝒏+𝟐 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

∞

𝒏=𝟎

−∑∫ 𝒙𝟔𝒏+𝟒 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

∞

𝒏=𝟎

−∑∫ 𝒙𝟔𝒏 𝐥𝐨𝐠 𝒙𝟏

𝟎

𝒅𝒙

∞

𝒏=𝟎

=

= −𝟐∑𝟏

(𝟔𝒏 + 𝟑)𝟐

∞

𝒏=𝟎

+∑𝟏

(𝟔𝒏 + 𝟓)𝟐

∞

𝒏=𝟎

+∑𝟏

(𝟔𝒏 + 𝟏)𝟐

∞

𝒏=𝟎

=

= −𝝍(𝟏) (

𝟏𝟐)

𝟏𝟖+𝝍(𝟏) (

𝟓𝟔)

𝟑𝟔+𝝍(𝟏) (

𝟏𝟔)

𝟑𝟔= −

𝝅𝟐

𝟑𝟔+𝝍(𝟏) (𝟏 −

𝟏𝟔) + 𝝍

(𝟏) (𝟏𝟔)

𝟑𝟔=𝝅𝟐

𝟗−𝝅𝟐

𝟑𝟔=𝝅𝟐

𝟏𝟐

Solution 2 by Ajetunmobi Abdulquyyum-Nigeria

𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝟎


𝒙𝟒 + 𝒙𝟐 + 𝟏

𝟏

𝟎

𝒅𝒙 + ∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝟏

𝒅𝒙⏟

𝑨

𝑨 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝟏

𝒅𝒙 =𝒙=𝟏𝒕−∫

𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐 + 𝒙𝟒

𝟏

𝟎

𝒅𝒙 ⇒


𝒙𝟒 + 𝒙𝟐 + 𝟏

𝟏

𝟎

𝒅𝒙 − ∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐 + 𝒙𝟒

𝟏

𝟎

𝒅𝒙 =

= ∫𝒙𝟐(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

−∫(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐 + 𝒙𝟒𝒅𝒙

𝟏

𝟎

=

= ∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔

𝟏

𝟎

𝒅𝒙 − ∫𝒙𝟒 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

−∫𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

+ ∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

=

= 𝟐∫𝒙𝟐 𝐥𝐨𝐠𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

−∫𝒙𝟒 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

− ∫𝐥𝐨𝐠𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

= 𝑩− 𝑪 −𝑫

𝑩 = 𝟐∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

=𝒙𝟔=𝒛 𝟏

𝟏𝟖∫𝒛𝟏𝟐−𝟏 𝐥𝐨𝐠 𝒛

𝟏 − 𝒛

𝟏

𝟎

𝒅𝒛 = −𝟏

𝟏𝟖𝝍(𝟏) (

𝟏

𝟐)

𝑪 = ∫𝒙𝟒 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

= ∫𝒛𝟐𝟑 𝐥𝐨𝐠 (𝒛

𝟏𝟔)

𝟏 − 𝒛⋅𝟏

𝟔𝒛−𝟓𝟔

𝟏

𝟎

𝒅𝒛 =𝟏

𝟑𝟔∫𝒛𝟓𝟔−𝟏 𝐥𝐨𝐠 𝒛

𝟏 − 𝒛𝒅𝒛

𝟏

𝟎

= −𝟏

𝟑𝟔𝝍(𝟏) (

𝟓

𝟔)

𝑫 = ∫𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

=𝟏

𝟑𝟔∫𝒛−𝟓𝟔 𝐥𝐨𝐠 𝒛

𝟏 − 𝒛𝒅𝒛

𝟏

𝟎

=𝟏

𝟑𝟔∫𝒛𝟏𝟔−𝟏 𝐥𝐨𝐠 𝒛

𝟏 − 𝒛

𝟏

𝟎

𝒅𝒛 = −𝟏

𝟑𝟔𝝍(𝟏) (

𝟏

𝟔)

Thus,

𝛀 = 𝑩 − 𝑪 −𝑫 = −𝟐

𝟑𝟔𝝍(𝟏) +

𝟏

𝟑𝟔𝝍(𝟏) (

𝟓

𝟔) +

𝟏

𝟑𝟔𝝍(𝟏) (

𝟏

𝟔) =

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𝝅𝟐

𝟗−𝝅𝟐

𝟑𝟔=𝝅𝟐

𝟏𝟐

Therefore,


∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝜺

𝒅𝒙 =𝝅𝟐

𝟏𝟐

Solution 3 by Ose Favour-Nigeria


∫𝒙𝟐 𝐥𝐨𝐠𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝜺


𝒙𝟒 + 𝒙𝟐 + 𝟏

𝟏

𝟎

𝒅𝒙 +∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝟏

𝒅𝒙 = 𝚽 +𝚿

𝚿 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏𝒅𝒙

∞

𝟏

=𝒖=𝟏𝒙− ∫

𝐥𝐨𝐠 𝒖

𝒖𝟒 + 𝒖𝟐 + 𝟏𝒅𝒖

𝟏

𝟎

𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙 − 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏𝒅𝒙

𝟏

𝟎

= ∫(𝟏 − 𝒙𝟐)(𝒙𝟐 − 𝟏) 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

= −∫(𝒙𝟐 − 𝟏)𝟐 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

=𝒕=𝒙𝟔

−𝟏

𝟑𝟔∫(𝒕𝟒𝟔−𝟓𝟔 − 𝟐𝒕

𝟐𝟔−𝟓𝟔 + 𝒕−

𝟓𝟔) 𝐥𝐨𝐠 𝒕

𝟏 − 𝒕𝒅𝒕

𝟏

𝟎

𝐋𝐞𝐭: 𝛀(𝒏) = ∫𝒕−𝟏𝟔+𝒏 − 𝟐𝒕−

𝟏𝟐+𝒏 + 𝒕−

𝟓𝟔+𝒏


𝟏

𝟎

𝛀 = −𝟏

𝟑𝟔𝛀′(𝟎)

𝛀(𝒏) = ∫𝒕−𝟏𝟔+𝒏 − 𝟐𝒕−

𝟏𝟐+𝒏 + 𝒕−

𝟓𝟔+𝒏


𝟏

𝟎

= −𝝍(𝟎) (𝒏 +𝟏

𝟔) −𝝍(𝟎) (𝒏 +

𝟓

𝟔) + 𝟐𝝍(𝟎) (𝒏 +

𝟏

𝟐)

𝛀′(𝒏) = −𝝍(𝟎)′(𝒏 +

𝟏

𝟔) −𝝍(𝟎)

′(𝒏 +

𝟓

𝟔) + 𝟐𝝍(𝟎)

′(𝒏 +

𝟏

𝟐)

𝛀′(𝟎) = −𝝍(𝟎)′(𝟏

𝟔) − 𝝍(𝟎)

′(𝟓

𝟔) + 𝟐𝝍(𝟎)

′(𝟏

𝟐) = −(𝝅𝟐 𝐜𝐬𝐜𝟐 (

𝝅

𝟔)) + 𝟐

𝝅𝟐

𝟐= −𝟑𝝅𝟐

𝛀 = −𝟏

𝟑𝟔𝛀′(𝟎) =

𝟏

𝟑𝟔⋅ 𝟑𝝅𝟐 =

𝝅𝟐

𝟏𝟐



𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝟎

𝒅𝒙 =𝒙=𝟏𝒚∫

−𝐥𝐨𝐠 𝒚

𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚

∞

𝟎

= ∫− 𝐥𝐨𝐠 𝒚

𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚

𝟏

𝟎

+∫− 𝐥𝐨𝐠 𝒚

𝒚𝟒 + 𝒚𝟐 + 𝟏

∞

𝟏

= 𝑰𝟏 + 𝑰𝟐

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𝑰𝟏 = ∫− 𝐥𝐨𝐠𝒚

𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚

𝟏

𝟎

= ∫−(𝟏 − 𝒚𝟐) 𝐥𝐨𝐠𝒚

𝟏 − 𝒚𝟔

𝟏

𝟎

𝒅𝒚 = −∫𝐥𝐨𝐠 𝒚

𝟏 − 𝒚𝟔𝒅𝒚

𝟏

𝟎

+∫𝒚𝟐 𝐥𝐨𝐠 𝒚

𝟏 − 𝒚𝟔𝒅𝒚

𝟏

𝟎

=

= −∫ ∑𝒚𝟔𝒏 𝝏

𝝏𝒂|𝒂=𝟎𝒚𝒂𝒅𝒚

∞

𝒏=𝟎

𝟏

𝟎

+∫ 𝒚𝟐∑𝒚𝟔𝒌𝝏

𝝏𝒃|𝒃=𝟎𝒚𝒃

∞

𝒌=𝟎

𝒅𝒚𝟏

𝟎

= −∑𝝏

𝝏𝒂|𝒂=𝟎

∞

𝒏=𝟎

∫ 𝒚𝟔𝒏+𝒂𝒅𝒚𝟏

𝟎

+

+∑𝝏

𝝏𝒃|𝒃=𝟎

∞

𝒌=𝟎

∫ 𝒚𝟔𝒌+𝒃+𝟐𝒅𝒚𝟏

𝟎

= −∑ [𝟏

𝟔𝒏 + 𝒂 + 𝟏]𝒂=𝟎

′∞

𝒏=𝟎

+∑[𝟏

𝟔𝒌 + 𝒃 + 𝟑]𝒃=𝟎

′∞

𝒌=𝟎

=

= ∑𝟏

(𝟔𝒏 + 𝟏)𝟐

∞

𝒏=𝟎

−∑𝟏

(𝟔𝒌 + 𝟑)𝟐

∞

𝒌=𝟎

=𝟏

𝟑𝟔[∑

𝟏

(𝒏 +𝟏𝟔)𝟐

∞

𝒏=𝟎

−∑𝟏

(𝒌 +𝟏𝟐)𝟐

∞

𝒌=𝟎

] =

=𝟏

𝟑𝟔[𝝍(𝟏) (

𝟏

𝟔) − 𝝍(𝟏) (

𝟏

𝟐)]

𝑰𝟐 = −∫𝐥𝐨𝐠 𝒚

𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚

∞

𝟏

=𝒚=𝟏𝒙∫

𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟒 + 𝒙𝟐 + 𝟏𝒅𝒙

𝟏

𝟎

= ∫𝒙𝟐(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

=

= ∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

−∫𝒙𝟒 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟔𝒅𝒙

𝟏

𝟎

=

= ∫ 𝒙𝟐∑𝒙𝟔𝒏𝝏

𝝏𝒂|𝒂=𝟎𝒙𝒂𝒅𝒙

∞

𝒏=𝟎

𝟏

𝟎

−∫ 𝒙𝟒∑𝒙𝟔𝒌𝝏

𝝏𝒃|𝒃=𝟎𝒙𝒃

∞

𝒌=𝟎

𝒅𝒙𝟏

𝟎

=

= ∑𝝏

𝝏𝒂|𝒂=𝟎

∫ 𝒙𝟔𝒏+𝒂+𝟐𝒅𝒙𝟏

𝟎

∞

𝒏=𝟎

−∑𝝏

𝝏𝒃|𝒃=𝟎

∞

𝒌=𝟎

∫ 𝒙𝟔𝒌+𝒃+𝟒𝟏

𝟎

𝒅𝒙 =

= ∑[𝟏

𝟔𝒏 + 𝒂 + 𝟑]𝒂=𝟎

′∞

𝒏=𝟎

−∑[𝟏

𝟔𝒌 + 𝒃 + 𝟓]𝒃=𝟎

′∞

𝒌=𝟎

= −∑𝟏

(𝟔𝒏 + 𝟑)𝟐

∞

𝒏=𝟎

+∑𝟏

(𝟔𝒏 + 𝟓)𝟐

∞

𝒌=𝟎

=

=𝟏

𝟑𝟔[∑

𝟏

(𝒌 +𝟓𝟔)𝟐

∞

𝒌=𝟎

−∑𝟏

(𝒏 +𝟏𝟐)𝟐

∞

𝒏=𝟎

] =𝟏

𝟑𝟔[𝝍(𝟏) (

𝟓

𝟔) − 𝝍(𝟏) (

𝟏

𝟐)]

Therefore,

𝛀 = 𝑰𝟏 + 𝑰𝟐 =𝟏

𝟑𝟔[𝝍(𝟏) (

𝟓

𝟔) ∓ 𝝍(𝟏) (

𝟓

𝟔) − 𝟐𝝍(𝟏) (

𝟏

𝟐)]

Solution 5 by Kartick Chandra Betal-India


𝒙𝟒 + 𝒙𝟐 + 𝟏

∞

𝟎

𝒅𝒙 = −∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐 + 𝒙𝟒𝒅𝒙

∞

𝟎

𝟐𝛀 = 𝟐∫(𝒙𝟐 − 𝟏) 𝐥𝐨𝐠 𝒙

𝟏 + 𝒙𝟐 + 𝒙𝟒𝒅𝒙

𝟏

𝟎

= 𝟐∫(𝟏 −

𝟏𝒙𝟐) 𝐥𝐨𝐠 𝒙

(𝒙 +𝟏𝒙)𝟏

− 𝟏

𝟏

𝟎

𝒅𝒙

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𝛀 =𝟏

𝟐𝐥𝐨𝐠(

𝒙𝟐 − 𝒙 + 𝟏

𝒙𝟐 + 𝒙 + 𝟏) 𝐥𝐨𝐠 𝒙|

𝟎

𝟏

−𝟏

𝟐∫ 𝐥𝐨𝐠(

𝒙𝟐 − 𝒙 + 𝟏

𝒙𝟐 + 𝒙 + 𝟏)𝒅𝒙

𝒙

𝟏

𝟎

=

= −𝟏

𝟐∫ {𝐥𝐨𝐠(

𝟏 + 𝒙𝟑

𝟏 − 𝒙𝟑) + 𝐥𝐨𝐠 (

𝟏 − 𝒙

𝟏+ 𝒙)}𝒅𝒙

𝟏

𝟎

=

=𝟏

𝟔∫ 𝐥𝐨𝐠 (

𝟏 − 𝒙

𝟏+ 𝒙)𝒅𝒙

𝒙

𝟏

𝟎

−𝟏

𝟐∫𝐥𝐨𝐠 (

𝟏 − 𝒙𝟏 + 𝒙)

𝒙𝒅𝒙

𝟏

𝟎

=

=𝟏

𝟑∫𝐥𝐨𝐠(𝟏 + 𝒙)

𝒙𝒅𝒙

𝟏

𝟎

−𝟏

𝟑∫𝐥𝐨𝐠(𝟏 − 𝒙)

𝒙𝒅𝒙

𝟏

𝟎

=

=𝟏

𝟑𝜼(𝟐) +

𝟏

𝟑𝜻(𝟐) =

𝟏

𝟐𝜻(𝟐) =

𝝅𝟐

𝟏𝟐


𝛀 = ∑ 𝐭𝐚𝐧−𝟏 ((𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟓) ⋅ 𝒏!

𝟏 + (𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ (𝒏!)𝟐)

∞

𝒏=𝟎


Solution 1 by Asmat Qatea-Afghanistan

∵ 𝐭𝐚𝐧−𝟏 𝒙 − 𝐭𝐚𝐧−𝟏 𝒚 = 𝐭𝐚𝐧−𝟏 (𝒙 − 𝒚

𝟏+ 𝒙𝒚),

∵ 𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔 = (𝒏 + 𝟏)(𝒏 + 𝟐)(𝒏 + 𝟑)

𝛀 = ∑𝐭𝐚𝐧−𝟏 ((𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟓) ⋅ 𝒏!

𝟏 + (𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ (𝒏!)𝟐)

∞

𝒏=𝟎

=

= ∑𝐭𝐚𝐧−𝟏 ((𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ 𝒏! − 𝒏!

𝟏 + (𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ (𝒏!)𝟐)

∞

𝒏=𝟎

=

= ∑𝐭𝐚𝐧−𝟏 ((𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ 𝒏!)

∞

𝒏=𝟎

−∑𝐭𝐚𝐧−𝟏(𝒏!)

∞

𝒏=𝟎

=

= ∑𝐭𝐚𝐧−𝟏((𝒏 + 𝟑)!) −

∞

𝒏=𝟎

∑𝐭𝐚𝐧−𝟏(𝒏!)

∞

𝒏=𝟎

=

= ∑ 𝐭𝐚𝐧−𝟏((𝒏 + 𝟑)!)

∞

𝒏=𝟎

− 𝟐 𝐭𝐚𝐧−𝟏(𝟏) − 𝐭𝐚𝐧−𝟏(𝟐) −∑𝐭𝐚𝐧−𝟏(𝒏!)

∞

𝒏=𝟑

= 𝝅− 𝐭𝐚𝐧−𝟏(𝟐)

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Solution 2 by Naren Bhandari-Bajura-Nepal

Note the 𝑮(𝒏) = 𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔 = (𝒏 + 𝟏)(𝒏+ 𝟐)(𝒏 + 𝟑) and it is easy to see that

∑𝒕𝒂𝒏−𝟏 ((𝑮(𝒏) − 𝟏)𝒏!

𝟏 + 𝑮(𝒏)(𝒏!)𝟐)

∞

𝒏=𝟎

= ∑(𝒕𝒂𝒏−𝟏((𝒏 + 𝟑)!) − 𝒕𝒂𝒏−𝟏(𝒏!))

∞

𝒏=𝟎

We have telescoping series giving us

−𝒕𝒂𝒏−𝟏(𝟎!) − 𝒕𝒂𝒏−𝟏(𝟏!) − 𝒕𝒂𝒏−𝟏(𝟐!) + 𝐥𝐢𝐦𝒏→∞

(𝒕𝒂𝒏−𝟏((𝒏 + 𝟑)!) + 𝒕𝒂𝒏−𝟏((𝒏 + 𝟐)!) +

𝒕𝒂𝒏−𝟏((𝒏 + 𝟏)!)) .For all 𝒏 > 𝟐 and on the further solving we have:

−𝝅

𝟐− 𝒕𝒂𝒏−𝟏(𝟐) +

𝟑𝝅

𝟐= 𝝅− 𝒕𝒂𝒏−𝟏(𝟐)

1526. 𝒙𝟎 = 𝟏, 𝒙𝟏 = 𝟎, 𝒙𝒏 = (𝒏 − 𝟏)(𝒙𝒏−𝟏 + 𝒙𝒏−𝟐), 𝒏 ≥ 𝟐, 𝒏 ∈ ℕ. Find:

𝛀 = 𝐥𝐢𝐦𝒏→∞

𝒙𝒏𝒏!


Solution 1 by Ravi Prakash-New Delhi-India

𝒙𝟎 = 𝟏, 𝒙𝟏 = 𝟎, 𝒙𝒏 = (𝒏 − 𝟏)(𝒙𝒏−𝟏 + 𝒙𝒏−𝟐), 𝒏 ≥ 𝟐,𝒏 ∈ ℕ; (𝟏) ⇒

𝒙𝒏 − 𝒏𝒙𝒏−𝟏 = −[𝒙𝒏−𝟏 − (𝒏 −)𝒙𝒏−𝟐]. Put: 𝒂𝒏 = 𝒙𝒏 − 𝒏𝒙𝒏−𝟏, ∀𝒏 ≥ 𝟏, 𝒂𝟏 = −𝟏

Also, (1) gives 𝒂𝒏 = −𝒂𝒏−𝟏, ∀𝒏 ≥ 𝟐 ⇒ (𝒂𝒏)𝒏≥𝟐 −geometric progression with ratio 𝒒 = −𝟏.

Thus, 𝒂𝒏 = (−𝟏)𝒏−𝟏𝒂𝟏, ∀𝒏 ≥ 𝟏 ⇒ 𝒙𝒏 − 𝒏𝒙𝒏−𝟏 = (−𝟏)

𝒏, ∀𝒏 ≥ 𝟏

⇒𝒙𝒏𝒏!−

𝒙𝒏−𝟏(𝒏 − 𝟏)!

=(−𝟏)𝒏

𝒏!⇒∑(

𝒙𝒓𝒓!−

𝒙𝒓−𝟏(𝒓 − 𝟏)!

)

𝒏

𝒓=𝟏

=∑(−𝟏)𝒓

𝒓!

𝒏

𝒓=𝟏

, ∀𝒏 ≥ 𝟏

⇒𝒙𝒏𝒏!=∑

(−𝟏)𝒓

𝒓!

𝒏

𝒓=𝟏

+𝒙𝟎𝟎!


𝒙𝒏𝒏!= 𝐥𝐢𝐦𝒏→∞

∑(−𝟏)𝒓

𝒓!

𝒏

𝒓=𝟎

=𝟏

𝒆

Solution 2 by Naren Bhandari-Bajura-Nepal

Since 𝒙𝟎 = 𝟏 and 𝒙𝟏 = 𝟎 and given recurrence relation 𝒙𝒏 = (𝒏 − 𝟏)(𝒙𝒏−𝟐 + 𝒙𝒏−𝟏) and

on expanding the recurrence relation we can observe that

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𝑷 = {𝒙𝒏|𝒏 ∈ ℤ≥𝟎} = {𝟏, 𝟎, 𝟏, 𝟐, 𝟗, 𝟒𝟒, 𝟐𝟔𝟓, 𝟏𝟖𝟓𝟒,… }

Since the sequence we have well know from Derangement so in other word we can

rewrite the recurrence in terms subfactorial, that is

! 𝒏 = (𝒏 − 𝟏)(! (𝒏 − 𝟐) + ! (𝒏 − 𝟏)), ∀𝒏 ≥ 𝟐 and hence required limit to be evaluated is


𝒙𝒏𝒏!= 𝐥𝐢𝐦𝐧→∞

! 𝒏

𝒏!= 𝐥𝐢𝐦𝒏→∞

∑(−𝟏)𝒌

𝒌!

𝒏

𝒌=𝟎

=𝟏

𝒆

𝐚𝐬 ! 𝒏 = 𝒏! ∑(−𝟏)𝒌

𝒌!

𝒏

𝒌=𝟎

1527. 𝑮(𝒏) −Barnes 𝑮-function, 𝑲(𝒏) − 𝑲 function. Find:

𝛀 =∑ √𝒏!

𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐)

𝒏∞

𝒏=𝟐



𝑮(𝒏) =(𝚪(𝒏))

𝒏−𝟏

𝑲(𝒏); 𝑮(𝒏 + 𝟐) =

((𝒏 + 𝟏)!)𝒏+𝟏

𝑲(𝒏 + 𝟐)

𝑲(𝒏 + 𝟏) = 𝟏𝟏 ⋅ 𝟐𝟐 ⋅ 𝟑𝟑 ⋅ … ⋅ 𝒏𝒏 ⇒𝑲(𝒏 + 𝟏)

𝑲(𝒏 + 𝟐)=

𝟏

(𝒏 + 𝟏)𝒏+𝟏

√𝒏!

𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐)

𝒏

=√

𝒏!

𝑲(𝒏 + 𝟏) ⋅((𝒏 + 𝟏)!)

𝒏+𝟏

𝑲(𝒏 + 𝟐)

𝒏 =√

𝒏!

((𝒏 + 𝟏)!)𝒏+𝟏

(𝒏 + 𝟏)𝒏+𝟏

𝒏 =

=√

𝒏!

(𝒏 + 𝟏)𝒏+𝟏 ⋅ (𝒏!)𝒏+𝟏

(𝒏 + 𝟏)𝒏+𝟏

𝒏 =𝟏

𝒏!

Therefore,

𝛀 = ∑ √𝒏!

𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐)

𝒏∞

𝒏=𝟐

= ∑𝟏

𝒏!

∞

𝒏=𝟐

= 𝒆 − 𝟐

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Solution 2 by Amrit Awasthi-India

We know: 𝑲(𝒏 + 𝟏) = 𝟏𝟏 ⋅ 𝟐𝟐 ⋅ 𝟑𝟑 ⋅ … ⋅ 𝒏𝒏 and 𝑮(𝒏+ 𝟐) = 𝟏! ⋅ 𝟐! ⋅ 𝟑! ⋅ … ⋅ 𝒏!

⇒ 𝑮(𝒏 + 𝟐) = 𝟏𝒏 ⋅ 𝟐𝒏−𝟏 ⋅ 𝟑𝒏−𝟐 ⋅ … ⋅ (𝒏 − 𝟏)𝟐 ⋅ 𝒏𝟏

𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐) = (𝟏𝟏 ⋅ 𝟐𝟐 ⋅ 𝟑𝟑 ⋅ … ⋅ 𝒏𝒏) ⋅ (𝟏𝒏 ⋅ 𝟐𝒏−𝟏 ⋅ 𝟑𝒏−𝟐 ⋅ … ⋅ (𝒏 − 𝟏)𝟐 ⋅ 𝒏𝟏) =

= 𝟏𝒏+𝟏 ⋅ 𝟐𝒏+𝟏 ⋅ … ⋅ (𝒏 − 𝟏)𝒏+𝟏 ⋅ 𝒏𝒏+𝟏 = (𝒏!)𝒏+𝟏

Therefore,

𝛀 = ∑ √𝒏!

𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐)

𝒏∞

𝒏=𝟐

= ∑ √𝒏!

(𝒏!)𝒏+𝟏𝒏

∞

𝒏=𝟐

= ∑ √𝟏

(𝒏!)𝒏𝒏

∞

𝒏=𝟐

=

= ∑𝟏

𝒏!

∞

𝒏=𝟐

= 𝒆 − 𝟐

1528. Find:


√(𝟐𝒏)! ⋅ (𝟐∑𝟏

(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!

𝒏

𝒌=𝟎

−𝟒𝒏

(𝟐𝒏)!)

𝒏



(𝟐𝒏

𝒏 − 𝒌) =

(𝟐𝒏)!

(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!⇒

𝟏

(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!=( 𝟐𝒏𝒏−𝒌)

(𝟐𝒏)!

∑(𝟐𝒏

𝒏− 𝒌)

𝒏

𝒌=𝟎

= (𝟐𝒏

𝒏) + (

𝟐𝒏

𝒏 − 𝟏) + (

𝟐𝒏

𝒏 − 𝟐) +⋯+ (

𝟐𝒏

𝟎)

= (𝟐𝒏

𝒏 + 𝟏) + (

𝟐𝒏

𝒏 + 𝟐) +⋯+ (

𝟐𝒏

𝟐𝒏) = 𝑺

∵ (𝟐𝒏

𝟎) + (

𝟐𝒏

𝟏) + (

𝟐𝒏

𝟐) +⋯+ (

𝟐𝒏

𝟐𝒏) = 𝟐𝟐𝒏 ⇒ 𝟐𝑺 + (

𝟐𝒏

𝒏) = 𝟐𝟐𝒏 ⇒ 𝟐𝑺 = 𝟒𝒏 − (

𝟐𝒏

𝒏)

⇒ 𝑺 =𝟒𝒏 − (𝟐𝒏

𝒏)

𝟐⇒∑(

𝟐𝒏

𝒏 − 𝒌)

𝒏

𝒌=𝟎

= 𝑺 + (𝟐𝒏

𝒏) =

(𝟒𝒏 + (𝟐𝒏𝒏))

𝟐

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√(𝟐𝒏)! ⋅ (𝟐∑𝟏

(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!

𝒏

𝒌=𝟎

−𝟒𝒏

(𝟐𝒏)!)

𝒏

=


√(𝟐𝒏)! ⋅ (𝟒𝒏 + (𝟐𝒏

𝒏)

𝟐(𝟐𝒏)!⋅ 𝟐 −

𝟒𝒏

(𝟐𝒏)!)

𝒏


√(𝟐𝒏

𝒏)

𝒏


√(𝟐𝒏)!

(𝒏!)𝟐𝒏

=𝑪−𝑫


(𝟐𝒏 + 𝟐)!

((𝒏 + 𝟏)!)𝟐 ⋅(𝒏!)𝟐

(𝟐𝒏)!= 𝐥𝐢𝐦𝒏→∞

(𝟐𝒏)! (𝟐𝒏 + 𝟏)(𝟐𝒏+ 𝟐)(𝒏!)𝟐

(𝒏!)𝟐(𝒏 + 𝟏)𝟐(𝟐𝒏)!= 𝟒

Solution 2 by Ahmed Yackoube Chach-Mauritania

𝑰 = ∑𝟏

(𝒏 − 𝒌)! (𝒏 + 𝒌)!

𝒏

𝒌=𝟎

=𝟏

𝒏! ⋅ 𝒏!+

𝟏

(𝒏 − 𝟏)! ⋅ (𝒏 + 𝟏)!+ ⋯+

𝟏

𝟏! ⋅ (𝟐𝒏 − 𝟏)!+

𝟏

𝟎! ⋅ (𝟐𝒏)!

=𝟏

𝟎! ⋅ (𝟐𝒏)!+

𝟏

𝟏! ⋅ (𝟐𝒏 − 𝟏)!+ ⋯+

𝟏

𝒏! ⋅ 𝒏!= ∑

𝟏

(𝟐𝒏 − 𝒌) ⋅ 𝒌!

𝒏

𝒌=𝟎

∑𝟏

(𝟐𝒏− 𝒌)! ⋅ 𝒌!

𝟐𝒏

𝒌=𝒏+𝟏

=𝟏

(𝒏 − 𝟏)! ⋅ (𝒏 + 𝟏)!+⋯+

𝟏

𝟏! ⋅ (𝟐𝒏 − 𝟏)!+

𝟏

𝟎! ⋅ (𝟐𝒏)!+

𝟏

(𝒏!)𝟐−

𝟏

(𝒏!)𝟐

= 𝑰 −𝟏

(𝒏!)𝟐

𝟐𝑰 = ∑𝟏

(𝟐𝒏 − 𝒌)! ⋅ 𝒌!

𝟐𝒏

𝒌=𝟎

+𝟏

(𝒏!)𝟐=

𝟏

(𝒏!)𝟐+𝟒𝒏

(𝟐𝒏)!

𝛀𝐧 = √(𝟐𝒏)! ⋅ (𝟐∑𝟏

(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!

𝒏

𝒌=𝟎

−𝟒𝒏

(𝟐𝒏)!)

𝒏

= √(𝟐𝒏)! (𝟐𝑰 −𝟒𝒏

(𝟐𝒏)!)

𝒏

=

= √(𝟐𝒏)!

(𝒏!)𝟐𝒏

= √𝟐𝟐𝒏𝚪(𝒏 +

𝟏𝟐)

√𝝅𝚪(𝒏 + 𝟏)

𝒏

= 𝟒(𝟏

√𝝅⋅𝚪 (𝒏 +

𝟏𝟐)

𝚪(𝒏 + 𝟏))

𝟏𝒏

= 𝟒 ⋅ 𝒆

𝟏𝒏⋅𝐥𝐨𝐠(

𝟏

√𝝅⋅𝚪(𝒏+

𝟏𝟐)

𝚪(𝒏+𝟏))𝒏→∞→ 𝟒 ⋅ 𝒆𝟎

Therefore,


√(𝟐𝒏)! ⋅ (𝟐∑𝟏

(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!

𝒏

𝒌=𝟎

−𝟒𝒏

(𝟐𝒏)!)

𝒏

= 𝟒

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1529. For 𝒂, 𝒃, 𝒑, 𝒒 ∈ ℕ such that 𝒑(𝒒 − 𝒃) = 𝒂 + 𝟏. Find:

𝛀 = 𝐥𝐢𝐦𝒎→∞

𝟏

𝒎⋅ 𝐥𝐢𝐦𝒏→∞

∑∑𝒊𝒂𝒔𝒊𝒏𝒑 (𝒊𝒃

𝒏𝒒)

𝒏

𝒊=𝟏

𝟐𝒎

𝒑=𝟏

Proposed by Florică Anastase-Romania

Solution by Ruxandra Daniela Tonilă-Romania

𝒂𝒏 =∑𝒊𝒂𝒔𝒊𝒏𝒑 (𝒊𝒃

𝒏𝒒)

𝒏

𝒊=𝟏

=∑(𝒔𝒊𝒏(

𝒊𝒃

𝒏𝒒)

𝒊𝒃

𝒏𝒒

)

𝒑

𝒊𝒂+𝒃𝒑

𝒏𝒑𝒒

𝒏

𝒊=𝟏

=∑(𝒔𝒊𝒏(

𝒊𝒃

𝒏𝒒)

𝒊𝒃

𝒏𝒒

)

𝒑

𝒊𝒂+𝒃𝒑

𝒏𝒂+𝒃𝒑+𝟏

𝒏

𝒊=𝟏

⇔

∀𝒏 ∈ ℕ, ∃𝜻𝒏 > 0 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡: 1 − 𝜻𝒏 ≤ (𝒔𝒊𝒏(

𝒊𝒃

𝒏𝒒)

𝒊𝒃

𝒏𝒒

)

𝒑

≤ 𝟏 + 𝜻𝒏

( 𝟏 − 𝜻𝒏)∑𝒊𝒂+𝒃𝒑

𝒏𝒂+𝒃𝒑+𝟏≤ 𝒂𝒏 ≤

𝒏

𝒊=𝟏

( 𝟏 + 𝜻𝒏)∑𝒊𝒂+𝒃𝒑

𝒏𝒂+𝒃𝒑+𝟏

𝒏

𝒊=𝟏

𝒍𝒊𝒎𝒏→∞

∑𝒊𝒂+𝒃𝒑

𝒏𝒂+𝒃𝒑+𝟏= 𝒍𝒊𝒎𝒏→∞

𝟏

𝒏∑(

𝒊

𝒏)𝒂+𝒃𝒑

= ∫𝒙𝒂+𝒃𝒑𝒅𝒙 =𝟏

𝒂 + 𝒃𝒑 + 𝟏

𝟏

𝟎

𝒏

𝒊=𝟏

𝒏

𝒊=𝟏

Hence,

𝒍𝒊𝒎𝒏→∞

∑𝒊𝒂𝒔𝒊𝒏𝒑 (𝒊𝒃

𝒏𝒒) =

𝟏

𝒂 + 𝒃𝒑 + 𝟏

𝒏

𝒊=𝟏


𝒏𝒒)

𝒏

𝒊=𝟏

𝟐𝒎

𝒑=𝟏

= ∑𝟏

𝒂+ 𝒃𝒑 + 𝟏

𝟐𝒎

𝒑=𝟏

𝟎 ≤𝟏

𝒎⋅∑

𝟏

𝒂 + 𝒃𝒑 + 𝟏

𝟐𝒎

𝒑=𝟏

≤⏞𝑨𝑴−𝑮𝑴 𝟏

𝒎∑

𝟏

𝟑√𝒂𝒃𝒑𝟑

𝟐𝒎

𝒑=𝟏

=𝟏

𝟑√𝒂𝒃𝟑 ∑

𝟏

𝒎√𝒑𝟑

𝟐𝒎

𝒑=𝟏

=𝟏

𝟑√𝒂𝒃𝟑 ⋅

∑𝟏

√𝒑𝟑

𝟐𝒎

𝒑=𝟏

𝒎

𝐥𝐢𝐦𝒎→∞

𝟏

𝒎⋅∑

𝟏

𝒂 + 𝒃𝒑 + 𝟏

𝟐𝒎

𝒑=𝟏

= 𝐥𝐢𝐦𝒎→∞

𝟏

𝟑√𝒂𝒃𝟑 ⋅

∑𝟏

√𝒑𝟑

𝟐𝒎

𝒑=𝟏

𝒎=⏞

𝑳.𝑪−𝑺 𝟏

𝟑√𝒂𝒃𝟑 ⋅ 𝐥𝐢𝐦

𝒏→∞

𝟏

√𝟐𝒎𝟑

(𝒎 + 𝟏) −𝒎= 𝟎

Therefore,

www.ssmrmh.ro


𝛀 = 𝐥𝐢𝐦𝒎→∞

𝟏

𝒎⋅ 𝐥𝐢𝐦𝒏→∞


𝒏𝒒)

𝒏

𝒊=𝟏

𝟐𝒎

𝒑=𝟏

= 𝟎

1530. If (𝒂𝒏)𝒏≥𝟎, (𝒃𝒏)𝒏≥𝟎 are given by 𝒂𝟎 = 𝒃𝟎 = 𝟏,𝒂𝒏+𝟏 = 𝒂𝒏 + 𝒃𝒏,

𝒃𝒏+𝟏 = (𝒏𝟐 + 𝒏 + 𝟏)𝒂𝒏 + 𝒃𝒏, 𝒏 ≥ 𝟏. Find:


𝒏 ⋅ √∏(𝒂𝒌𝒃𝒌)

𝒏

𝒌=𝟏

𝒏

Proposed by Neculai Stanciu-Romania

Solution by George Florin Șerban-Romania

𝒂𝒏+𝟏 = 𝒂𝒏 + 𝒃𝒏 ⇒ 𝒃𝒏 = 𝒂𝒏+𝟏 − 𝒂𝒏 and 𝒃𝒏+𝟏 = (𝒏𝟐 + 𝒏+ 𝟏)𝒃𝒏, then

𝒃𝒏+𝟏 − 𝒃𝒏 = (𝒏𝟐 + 𝒏+ 𝟏)𝒂𝒏 ⇒ 𝒂𝒏+𝟐 − 𝒂𝒏+𝟏 − 𝒂𝒏+𝟏 + 𝒂𝒏 = (𝒏

𝟐 + 𝒏)𝒂𝒏 + 𝒂𝒏

𝒂𝒏+𝟐 − 𝟐𝒂𝒏+𝟏 = (𝒏𝟐 + 𝒏)𝒂𝒏 ⇒

𝒂𝒏+𝟐𝒂𝒏

− 𝟐𝒂𝒏+𝟏𝒂𝒏

= 𝒏𝟐 + 𝒏. 𝐋𝐞𝐭 𝒙𝒏 =𝒂𝒏+𝟏𝒂𝒏

⇒

𝒙𝒏+𝟏𝒙𝒏 − 𝟐𝒙𝒏 = 𝒏𝟐 + 𝒏. Applying mathematical induction to 𝑷(𝒏): 𝒙𝒏 = 𝒏+ 𝟏 from 𝒏 ≥

𝟎, we have: 𝑷(𝟎): 𝒙𝟎 =𝒂𝟏

𝒂𝟎= 𝟏 true.

Suppose that: 𝑷(𝒌): 𝒙𝒌 = 𝒌 + 𝟏 ⇒ 𝒙𝒌+𝟏 ⋅ (𝒌 + 𝟏) − 𝟐(𝒌 + 𝟏) = 𝒌𝟐 + 𝒌 ⇒

𝒙𝒌+𝟏 ⋅ (𝒌 + 𝟏) = 𝒌(𝒌 + 𝟏) + 𝟐(𝒌 + 𝟏) ⇒ 𝒙𝒌+𝟏 = 𝒌+ 𝟐. Hence,

𝒂𝒏+𝟏𝒂𝒏

= 𝒏+ 𝟏 ⇒∏𝒂𝒌+𝟏𝒂𝒌

𝒏

𝒌=𝟎

= (𝒏 + 𝟏)! ⇒ 𝒂𝒏+𝟏 = (𝒏 + 𝟏)! ⇒ 𝒂𝒏 = 𝒏!

∏𝒃𝒌𝒂𝒌

𝒏

𝒌=𝟏

=∏𝒌 ⋅ 𝒌!

𝒌!

𝒏

𝒌=𝟏

=∏𝒌

𝒏

𝒌=𝟏

= 𝒏!

Therefore,


𝒏 ⋅ √∏(𝒂𝒌𝒃𝒌)

𝒏

𝒌=𝟏

𝒏


𝒏

√𝒏!𝒏 = 𝐥𝐢𝐦

𝒏→∞√𝒏𝒏

𝒏!

𝒏

=𝑪−𝑫′𝑨


(𝒏 + 𝟏)𝒏+𝟏

(𝒏 + 𝟏)!⋅𝒏!

𝒏𝒏=


(𝟏 +𝟏

𝒏)𝒏

= 𝒆.

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1531. Find:

𝛀(𝒏) = 𝐥𝐢𝐦𝒙→𝟎

𝟐𝐭𝐚𝐧𝟐𝒙 ⋅ 𝟒𝐭𝐚𝐧𝟒𝒙 ⋅ … ⋅ (𝟐𝒏)𝐭𝐚𝐧(𝟐𝒏𝒙) − 𝟏

𝟑𝐭𝐚𝐧𝟑𝒙 ⋅ 𝟓𝐭𝐚𝐧𝟓𝒙 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝐭𝐚𝐧((𝟐𝒏+𝟏)𝒙) − 𝟏;𝒏 ∈ ℕ, 𝒏 ≥ 𝟏

Proposed by Mohammad Hamed Nasery-Afghanistan


Rewriting we have:

𝛀 = 𝐥𝐢𝐦𝒙→𝟎

𝒆𝐭𝐚𝐧 𝟐𝒙 𝐥𝐨𝐠 𝟐+𝐭𝐚𝐧 𝟒𝒙 𝐥𝐨𝐠 𝟒+⋯+𝐭𝐚𝐧(𝟐𝒏𝒙) 𝐥𝐨𝐠(𝟐𝒏) − 𝟏

𝒆𝐭𝐚𝐧 𝟑𝒙 𝐥𝐨𝐠 𝟑+𝐭𝐚𝐧 𝟓𝒙 𝐥𝐨𝐠 𝟓+⋯+𝐭𝐚𝐧(𝟐𝒏+𝟏)𝒙 𝐥𝐨𝐠(𝟐𝒏+𝟏) − 𝟏

𝒇(𝒙) = 𝒆𝐭𝐚𝐧 𝟐𝒙 𝐥𝐨𝐠 𝟐+𝐭𝐚𝐧 𝟒𝒙 𝐥𝐨𝐠 𝟒+⋯+𝐭𝐚𝐧(𝟐𝒏𝒙) 𝐥𝐨𝐠(𝟐𝒏),

𝒈(𝒙) = 𝒆𝐭𝐚𝐧 𝟑𝒙 𝐥𝐨𝐠 𝟑+𝐭𝐚𝐧 𝟓𝒙 𝐥𝐨𝐠 𝟓+⋯+𝐭𝐚𝐧(𝟐𝒏+𝟏)𝒙 𝐥𝐨𝐠(𝟐𝒏+𝟏)

⇒ 𝒇′(𝒙) = 𝒆𝐭𝐚𝐧 𝟐𝒙 𝐥𝐨𝐠 𝟐+𝐭𝐚𝐧 𝟒𝒙 𝐥𝐨𝐠 𝟒+⋯+𝐭𝐚𝐧(𝟐𝒏𝒙) 𝐥𝐨𝐠(𝟐𝒏) ⋅

⋅ (𝐥𝐨𝐠 𝟐 𝐬𝐞𝐜𝟐(𝟐𝒙) ⋅ 𝟐 + ⋯+ 𝐥𝐨𝐠(𝟐𝒏) 𝐬𝐞𝐜𝟐((𝟐𝒏 + 𝟏)𝒙) ⋅ 𝟐𝒏)

𝒈′(𝒙) = 𝒆𝐭𝐚𝐧 𝟑𝒙 𝐥𝐨𝐠 𝟑+𝐭𝐚𝐧 𝟓𝒙 𝐥𝐨𝐠 𝟓+⋯+𝐭𝐚𝐧(𝟐𝒏+𝟏)𝒙 𝐥𝐨𝐠(𝟐𝒏+𝟏) ⋅

⋅ (𝐥𝐨𝐠 𝟑 𝐬𝐞𝐜𝟐(𝟑𝒙) ⋅ 𝟑 +⋯+ 𝐥𝐨𝐠(𝟐𝒏 + 𝟏) 𝐬𝐞𝐜𝟐(𝟐𝒏 + 𝟏) 𝐬𝐞𝐜𝟐((𝟐𝒏 + 𝟏)𝒙) ⋅ (𝟐𝒏 + 𝟏))

That implies,

𝛀 = 𝐥𝐢𝐦𝒙→𝟎

𝒇′(𝒙)

𝒈′(𝒙)=

𝟐 𝐥𝐨𝐠 𝟐 + 𝟒 𝐥𝐨𝐠 𝟒 + ⋯+ 𝟐𝒏 𝐥𝐨𝐠𝟐𝒏

𝟑 𝐥𝐨𝐠 𝟑 + 𝟓 𝐥𝐨𝐠𝟓 + ⋯+ (𝟐𝒏 + 𝟏) 𝐥𝐨𝐠(𝟐𝒏 + 𝟏)=

=𝐥𝐨𝐠 𝟐𝟐 + 𝐥𝐨𝐠 𝟒𝟒 +⋯+ 𝐥𝐨𝐠(𝟐𝒏)𝟐𝒏

𝐥𝐨𝐠𝟑𝟑 + 𝐥𝐨𝐠 𝟓𝟓 +⋯+ 𝐥𝐨𝐠(𝟐𝒏 + 𝟏)𝟐𝒏+𝟏=

𝐥𝐨𝐠(𝟐𝟐 ⋅ 𝟒𝟒 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏)

𝐥𝐨𝐠(𝟑𝟑 ⋅ 𝟓𝟓 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝟐𝒏+𝟏)

Solution 2 by Serlea Kabay-Liberia

Recall, 𝐭𝐚𝐧(𝜶𝒙)~𝜶𝒙 and 𝒆𝜶𝒙 − 𝟏~𝜶𝒙;∀𝜶 ∈ ℝ.

𝛀(𝒏)~ 𝐥𝐢𝐦𝒙→𝟎

𝟐𝟐𝒙 ⋅ 𝟒𝟒𝒙 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏𝒙 − 𝟏

𝟑𝟑𝒙 ⋅ 𝟓𝟓𝒙 ⋅ … ⋅ (𝟐𝒏 + 𝟏)(𝟐𝒏+𝟏)𝒙 − 𝟏=

= 𝐥𝐢𝐦𝒙→𝟎

𝒆𝐥𝐨𝐠(𝟐𝟐𝒙⋅𝟒𝟒𝒙⋅…⋅(𝟐𝒏)𝟐𝒏𝒙) − 𝟏

𝒆𝐥𝐨𝐠(𝟑𝟑𝒙⋅𝟓𝟓𝒙⋅…⋅(𝟐𝒏+𝟏)(𝟐𝒏+𝟏)𝒙) − 𝟏

=


𝒆𝒙 𝐥𝐨𝐠 𝟐𝟐+𝒙 𝐥𝐨𝐠 𝟒𝟒+⋯+𝒙 𝐥𝐨𝐠(𝟐𝒏)𝟐𝒏 − 𝟏

𝒆𝒙 𝐥𝐨𝐠 𝟑𝟑+𝒙 𝐥𝐨𝐠 𝟓𝟓+⋯+𝒙 𝐥𝐨𝐠(𝟐𝒏+𝟏)𝟐𝒏+𝟏 − 𝟏

=

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𝒆𝒙 𝐥𝐨𝐠(∏ (𝟐𝒌)𝟐𝒌𝒏𝒌=𝟏 ) − 𝟏

𝒆𝒙 𝐥𝐨𝐠(∏ (𝟐𝒌+𝟏)𝟐𝒌+𝟏𝒏𝒌=𝟏 ) − 𝟏


𝒙 𝐥𝐨𝐠(∏ (𝟐𝒌)𝟐𝒌𝒏𝒌=𝟏 )

𝒙 𝐥𝐨𝐠(∏ (𝟐𝒌 + 𝟏)𝟐𝒌+𝟏𝒏𝒌=𝟏 )

=


𝐥𝐨𝐠(∏ (𝟐𝒌)𝟐𝒌𝒏𝒌=𝟏 )

𝐥𝐨𝐠(∏ (𝟐𝒌 + 𝟏)𝟐𝒌+𝟏𝒏𝒌=𝟏 )

=𝐥𝐨𝐠(𝟐𝟐 ⋅ 𝟒𝟒 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏)

𝐥𝐨𝐠(𝟑𝟑 ⋅ 𝟓𝟓 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝟐𝒏+𝟏)

Solution 3 by Obaidullah Jaihon-Afghanistan

𝟐𝐭𝐚𝐧 𝟐𝒙 ⋅ 𝟒𝐭𝐚𝐧 𝟒𝒙 ⋅ … ⋅ (𝟐𝒏)𝐭𝐚𝐧(𝟐𝒏𝒙) = 𝒕 ⇒

𝐥𝐨𝐠 𝒕 = 𝐥𝐨𝐠𝟐 𝐭𝐚𝐧 𝟐𝒙 + 𝐥𝐨𝐠 𝟒 𝐭𝐚𝐧 𝟒𝒙 +⋯+ 𝟐𝒏 𝐥𝐨𝐠(𝟐𝒏) 𝐭𝐚𝐧(𝟐𝒏𝒙)

(𝐥𝐨𝐠 𝒕)′ = 𝟐 𝐥𝐨𝐠𝟐 𝐬𝐞𝐜𝟐 𝟐𝒙 + 𝟒 𝐥𝐨𝐠𝟒 𝐬𝐞𝐜𝟐 𝟒𝒙 +⋯+ 𝟐𝒏 𝐥𝐨𝐠(𝟐𝒏) 𝐬𝐞𝐜𝟐(𝟐𝒏)

𝐥𝐢𝐦𝒙→𝟎(𝐥𝐨𝐠 𝒕)′ = 𝟐 𝐥𝐨𝐠𝟐 + 𝟒 𝐥𝐨𝐠𝟒 + ⋯+ 𝟐𝒏 𝐥𝐨𝐠(𝟐𝒏) = 𝐥𝐨𝐠(𝟐𝟐 ⋅ 𝟒𝟒 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏)

𝒑 = 𝟑𝐭𝐚𝐧 𝟑𝒙 ⋅ 𝟓𝐭𝐚𝐧 𝟓𝒙 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝐭𝐚𝐧(𝟐𝒏+𝟏)𝒙 − 𝟏

𝐥𝐨𝐠𝒑 = 𝐥𝐨𝐠𝟑 𝐭𝐚𝐧 𝟑𝒙 + 𝐥𝐨𝐠𝟓 𝐭𝐚𝐧 𝟓𝒙 +⋯+ 𝐥𝐨𝐠(𝟐𝒏+ 𝟏) 𝐭𝐚𝐧(𝟐𝒏+ 𝟏)

𝐥𝐢𝐦𝒙→𝟎(𝐥𝐨𝐠𝒑)′ = 𝐥𝐢𝐦

𝒙→𝟎(𝟑 𝐥𝐨𝐠𝟑 𝐬𝐞𝐜𝟐 𝟑𝒙 + 𝟓 𝐥𝐨𝐠𝟓 𝐬𝐞𝐜𝟐 𝟓𝒙 +⋯

+ (𝟐𝒏+ 𝟏) 𝐥𝐨𝐠(𝟐𝒏 + 𝟏) 𝐬𝐞𝐜𝟐(𝟐𝒏 + 𝟏)𝒙) =

= 𝟑 𝐥𝐨𝐠𝟑 + 𝟓 𝐥𝐨𝐠𝟓 + ⋯+ (𝟐𝒏+ 𝟏) 𝐥𝐨𝐠(𝟐𝒏 + 𝟏) = 𝐥𝐨𝐠(𝟑𝟑 ⋅ 𝟓𝟓 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝟐𝒏+𝟏)

𝛀 =𝐥𝐨𝐠 (𝟐𝟐 ⋅ 𝟒𝟒 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏)

𝐥𝐨𝐠 (𝟑𝟑 ⋅ 𝟓𝟓⋅ … ⋅ (𝟐𝒏+ 𝟏)𝟐𝒏+𝟏)

1532. Solve for integers:

𝟐𝒙𝟐 + 𝒙

𝒙𝟐 + 𝒙 + 𝟏+𝟏𝟖𝒙𝟐 + 𝟏𝟔𝒙 + 𝟑𝟎

𝒙𝟐 + 𝒙+ 𝟐+𝟖𝟒𝒙𝟐 + 𝟖𝟏𝒙 + 𝟐𝟒𝟎

𝒙𝟐 + 𝒙 + 𝟑+⋯+

𝒂𝒏 ⋅ 𝒙𝟐 + 𝒃𝒏 ⋅ 𝒙 + 𝒄𝒏𝒙𝟐 + 𝒙 + 𝒏

=𝟔𝒏𝟓 + 𝟏𝟓𝒏𝟒 + 𝟏𝟎𝒏𝟑 − 𝒏

𝟑𝟎;𝒏 ∈ ℕ∗ 𝐚𝐧𝐝 𝐟𝐢𝐧𝐝:

𝛀 = 𝐥𝐢𝐦𝐧→∞

(𝒂𝒏𝒃𝒏)

𝒄𝒏𝒏𝟐

Proposed by Costel Florea-Romania

Solution by George Florin Șerban-Romania

𝟔𝒏𝟓 + 𝟏𝟓𝒏𝟒 + 𝟏𝟎𝒏𝟑 − 𝒏 = 𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)(𝟑𝒏𝟐 + 𝟑𝒏 − 𝟏)

∑𝒌𝟒𝒏

𝒌=𝟏

=𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)(𝟑𝒏𝟐 + 𝟑𝒏 − 𝟏)

𝟑𝟎

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𝒂𝟏 = 𝟐, 𝒂𝟐 = 𝟏𝟖, 𝒂𝟑 = 𝟖𝟒, … , 𝒂𝒏 = 𝒏𝟒 + 𝒏

𝒃𝟏 = 𝟏, 𝒃𝟐 = 𝟏𝟔, 𝒃𝟑 = 𝟖𝟏,… , 𝒃𝒏 = 𝒏𝟒

𝒄𝟏 = 𝟎, 𝒄𝟐 = 𝟑𝟎, 𝒄𝟑 = 𝟐𝟒𝟎, … , 𝒄𝒏 = 𝒏𝟓 − 𝒏

(𝟐𝒙𝟐 + 𝒙

𝒙𝟐 + 𝒙 + 𝟏− 𝟏𝟒) + (

𝟏𝟖𝒙𝟐 + 𝟏𝟔𝒙 + 𝟑𝟎

𝒙𝟐 + 𝒙 + 𝟐− 𝟐𝟒) + (

𝟖𝟒𝒙𝟐 + 𝟖𝟏𝒙 + 𝟐𝟒𝟎

𝒙𝟐 + 𝒙 + 𝟑− 𝟑𝟒) +⋯

+ ((𝒏𝟒 + 𝒏)𝒙𝟐 + 𝒏𝟒𝒙 + 𝒏𝟓 − 𝒏)

𝒙𝟐 + 𝒙 + 𝒏− 𝒏𝟒) = 𝟎 ⇒

𝒙𝟐 − 𝟏

𝒙𝟐 + 𝒙 + 𝟏+𝟐𝒙𝟐 − 𝟐

𝒙𝟐 + 𝒙 + 𝟐+𝟑𝒙𝟐 − 𝟑

𝒙𝟐 + 𝒙 + 𝟑+⋯+

𝒏𝒙𝟐 − 𝒏

𝒙𝟐 + 𝒙 + 𝒏= 𝟎 ⟺

(𝒙𝟐 − 𝟏) (𝟏

𝒙𝟐 + 𝒙 + 𝟏+

𝟐

𝒙𝟐 + 𝒙 + 𝟐+⋯+

𝒏

𝒙𝟐 + 𝒙 + 𝒏) = 𝟎

Because 𝟏

𝒙𝟐+𝒙+𝟏+

𝟐

𝒙𝟐+𝒙+𝟐+⋯+

𝒏

𝒙𝟐+𝒙+𝒏≠ 𝟎 from 𝒙𝟐 + 𝒙 + 𝒏 > 0, ∀𝑛 ∈ ℕ

⇒ 𝒙𝟐 − 𝟏 = 𝟎 ⟺ 𝒙 ∈ {−𝟏, 𝟏}

𝛀 = 𝐥𝐢𝐦𝐧→∞

(𝒂𝒏𝒃𝒏)

𝒄𝒏𝒏𝟐= 𝐥𝐢𝐦𝒏→∞

(𝒏𝟒 + 𝒏

𝒏𝟒)

𝒏𝟓−𝒏

𝒏𝟐


[(𝟏 +𝟏

𝒏𝟑)𝒏𝟑

]

𝒏𝟒−𝟏

𝒏𝟒

= 𝒆𝐥𝐢𝐦𝒏→∞

𝒏𝟒−𝟏

𝒏𝟒 = 𝒆

1533. Let (𝒂𝒏)𝒏≥𝟏 −be a sequence of real numbers with 𝒂𝟎 = 𝟏 and

[(𝒂𝒏 − 𝒂𝒏−𝟏)(𝒏 + 𝟏)! 𝒏 − 𝒂𝒏𝒂𝒏−𝟏](𝒏 + 𝟏) = 𝒏𝟐𝒂𝒏𝒂𝒏−𝟏; 𝒏 ≥ 𝟎. Find:


( √𝒂𝒏+𝟏𝒏 + 𝟐

𝒏+𝟏)

𝒂

− (√𝒂𝒏𝒏 + 𝟏

𝒏)

𝒂

(√𝒂𝒏𝒏 + 𝟏

𝒏)

𝒂−𝟏


Solution by Mikael Bernardo-Mozambique

𝒂𝟎 = 𝟏; [(𝒂𝒏 − 𝒂𝒏−𝟏)(𝒏 + 𝟏)!𝒏 − 𝒂𝒏𝒂𝒏−𝟏](𝒏 + 𝟏) = 𝒏𝟐𝒂𝒏𝒂𝒏−𝟏; 𝒏 ≥ 𝟏 ⟺

(𝒂𝒏 − 𝒂𝒏−𝟏𝒏(𝒏 + 𝟏) ⋅ (𝒏 + 𝟏)! = 𝒂𝒏𝒂𝒏−𝟏(𝒏𝟐 + 𝒏 + 𝟏) ⟺

𝒂𝒏 − 𝒂𝒏−𝟏𝒂𝒏𝒂𝒏−𝟏

=(𝒏 + 𝟏)𝟐 − 𝒏

𝒏(𝒏 + 𝟏)(𝒏 + 𝟏)!⟺

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𝟏

𝒂𝒏−𝟏−𝟏

𝒂𝒏=

𝒏 + 𝟏

𝒏(𝒏 + 𝟏)!−

𝟏

(𝒏 + 𝟏)(𝒏 + 𝟏)!

𝟏

𝒂𝒏−

𝟏

𝒂𝒏−𝟏=

𝟏

(𝒏 + 𝟏) ⋅ (𝒏 + 𝟏)!−

𝟏

𝒏 ⋅ 𝒏!

𝟏

𝒂𝟏−𝟏

𝒂𝟎=

𝟏

𝟐 ⋅ 𝟐!−

𝟏

𝟏 ⋅ 𝟏!⇒ 𝒂𝒏 = (𝒏 + 𝟏) ⋅ (𝒏 + 𝟏)!

Hence,


( √𝒂𝒏+𝟏𝒏 + 𝟐

𝒏+𝟏)

𝒂

− (√𝒂𝒏𝒏 + 𝟏

𝒏)

𝒂

(√𝒂𝒏𝒏 + 𝟏

𝒏)

𝒂−𝟏 = 𝐥𝐢𝐦𝒏→∞

( √(𝒏 + 𝟐)!𝒏+𝟏

)𝒂

− (√(𝒏 + 𝟏)!𝒏

)𝒂

(√(𝒏 + 𝟏)!𝒏

)𝒂−𝟏 =


(√(𝒏 + 𝟏)!𝒏

)𝒂+𝟏−𝒂

⋅ ((√(𝒏 + 𝟐)!

𝒏+𝟏

√(𝒏 + 𝟏)!𝒏

)

𝒂

− 𝟏) =


(√(𝒏 + 𝟏)!𝒏

𝒏) ⋅ 𝒏 ⋅

(√(𝒏 + 𝟐)!

𝒏+𝟏

√(𝒏 + 𝟏)!𝒏

)

𝒂

− 𝟏

𝐥𝐨𝐠 ((√(𝒏 + 𝟐)!

𝒏+𝟏

√(𝒏 + 𝟏)!𝒏

)

𝒂

)

⋅ 𝐥𝐨𝐠((√(𝒏 + 𝟐)!

𝒏+𝟏

√(𝒏 + 𝟏)!𝒏

)

𝒂

)


(√(𝒏 + 𝟏)!𝒏

𝒏) =𝑪−𝑫


((𝒏 + 𝟐)!

(𝒏 + 𝟏)!⋅

𝒏𝒏

(𝒏 + 𝟏)𝒏+𝟏=𝟏

𝒆; (𝟏)


√(𝒏 + 𝟐)!𝒏+𝟏

√(𝒏 + 𝟏)!𝒏

= 𝟏 ⇒ 𝐥𝐢𝐦𝒏→∞

(√(𝒏 + 𝟐)!

𝒏+𝟏

√(𝒏 + 𝟏)!𝒏

)

𝒂

− 𝟏

𝐥𝐨𝐠((√(𝒏 + 𝟐)!

𝒏+𝟏

√(𝒏 + 𝟏)!𝒏

)

𝒂

)

= 𝟏; (𝟐)


𝐥𝐨𝐠 ((√(𝒏 + 𝟐)!

𝒏+𝟏

√(𝒏 + 𝟏)!𝒏

)

𝒏𝒂

) = 𝜶 ⋅ 𝐥𝐨𝐠 (𝐥𝐢𝐦𝒏→∞

(𝒏 + 𝟐)!

(𝒏 + 𝟏)!⋅

𝒏 + 𝟏

√(𝒏 + 𝟐)!𝒏+𝟏

⋅𝟏

𝒏 + 𝟏) = 𝒂 ⋅ 𝐥𝐨𝐠 𝒆

= 𝒂; (𝟑)

From (1),(2),(3) it follows that:


( √𝒂𝒏+𝟏𝒏 + 𝟐

𝒏+𝟏)

𝒂

− (√𝒂𝒏𝒏 + 𝟏

𝒏)

𝒂

(√𝒂𝒏𝒏 + 𝟏

𝒏)

𝒂−𝟏 =𝟏

𝒆⋅ 𝟏 ⋅ 𝒂 =

𝒂

𝒆

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1534.

(𝒂𝒏)𝒏≥𝟏, (𝒃𝒏)𝒏≥𝟏; 𝒂𝒏 = ∫ [𝒏𝟐

𝒙]

𝒏

𝟏

𝒅𝒙, 𝒃𝟏 > 1,

𝒃𝒏+𝟏 = 𝟏 + 𝐥𝐨𝐠(𝒃𝒏) , [∗] − 𝑮𝑰𝑭 . Find:


𝒂𝒏 ⋅ 𝐥𝐨𝐠 √𝒃𝒏𝒏

𝐥𝐨𝐠𝒏





𝐥𝐨𝐠𝒏= 𝐥𝐢𝐦𝒏→∞

𝒂𝒏 ⋅ 𝐥𝐨𝐠 𝒃𝒏𝒏 ⋅ 𝐥𝐨𝐠 𝒏


𝒂𝒏𝒏𝟐 ⋅ 𝐥𝐨𝐠 𝒏

⋅ 𝒏 𝐥𝐨𝐠 𝒃𝒏

= 𝛀𝟏 ⋅ 𝛀𝟐; (𝟏)

𝛀𝟏 = 𝐥𝐢𝐦𝒏→∞


𝐚𝐧𝐝 𝛀𝟐 = 𝐥𝐢𝐦𝒏→∞

𝒏 ⋅ 𝐥𝐨𝐠 𝒃𝒏

We have: 𝒏𝟐

𝒙− 𝟏 < [

𝒏𝟐

𝒙] ≤

𝒏𝟐

𝒙; ∀𝒙 ∈ ℝ, 𝒏 ∈ ℕ ⇒

∫ (𝒏𝟐

𝒙− 𝟏)

𝒏

𝟏

𝒅𝒙 < 𝒂𝒏 ≤ ∫𝒏𝟐

𝒙

𝒏

𝟏

𝒅𝒙 ⟺ 𝒏𝟐 𝐥𝐨𝐠𝒏 − (𝒏 − 𝟏) < 𝒂𝒏 ≤ 𝒏𝟐 𝐥𝐨𝐠 𝒏 ⟺

𝟏 −𝒏 − 𝟏

𝒏𝟐 𝐥𝐨𝐠 𝒏<

𝒂𝒏𝒏𝟐 𝐥𝐨𝐠 𝒏

≤ 𝟏


(𝟏 −𝒏 − 𝟏

𝒏𝟐 𝐥𝐨𝐠 𝒏) < 𝐥𝐢𝐦

𝒏→∞

𝒂𝒏𝒏𝟐 ⋅ 𝐥𝐨𝐠𝒏

≤ 𝟏 ⇒ 𝐥𝐢𝐦𝒏→∞


= 𝟏; (𝟐)

Now, 𝒃𝟏 > 1. Suppose that 𝒃𝒌 > 1;∀𝑘 ∈ ℕ and from 𝒃𝒌+𝟏 = 𝟏 + 𝐥𝐨𝐠 𝒃𝒌 we get

𝒃𝒌+𝟏 > 1;∀𝑘 ∈ ℕ. Thus, 𝒃𝒏 > 1;∀𝑛 ∈ ℕ.

𝒃𝒏+𝟏 = 𝟏 + 𝐥𝐨𝐠 𝒃𝒏 ⇒ 𝒃𝒏+𝟏 − 𝒃𝒏 = 𝟏 − 𝒃𝒏 + 𝐥𝐨𝐠 𝒃𝒏 ; (𝟑)

Let be the function 𝒇(𝒙) = 𝐥𝐨𝐠𝒙 − 𝒙 − 𝟏; (𝒙 > 1) with 𝒇′(𝒙) =𝟏

𝒙− 𝟏 < 0;∀𝑥 > 1

⇒ 𝒇−decreasing on (𝟏,∞) ⇒ 𝒇(𝒙) < 𝑓(𝟏) = 𝟎; ∀𝒙 > 1 ⇒

𝒇(𝒃𝒏) < 0 ⇒ 𝒃𝒏+𝟏 < 𝒃𝒏.

Since (𝒃𝒏)𝒏≥𝟏 −is decreasing and bounded, then (𝒃𝒏)𝒏≥𝟏 converges.

So, ∃𝒍 ∈ ℝ such that 𝐥𝐢𝐦𝒏→∞

𝒃𝒏 = 𝒍 ⇒ 𝒍 = 𝟏 + 𝐥𝐨𝐠 𝒍 ⇒ 𝒍 = 𝟏.

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𝛀𝟐 = 𝐥𝐢𝐦𝒏→∞

𝒏 ⋅ 𝐥𝐨𝐠 𝒃𝒏 = 𝐥𝐢𝐦𝒏→∞

𝐥𝐨𝐠 𝒃𝒏𝒏 = 𝐥𝐨𝐠 (𝐥𝐢𝐦

𝒏→∞𝒃𝒏𝒏) =

= 𝐥𝐨𝐠 (𝐥𝐢𝐦𝒏→∞

[(𝟏 + 𝒃𝒏 − 𝟏)𝟏

𝒃𝒏−𝟏]

𝒏(𝒃𝒏−𝟏)

) = 𝐥𝐨𝐠 (𝒆𝐥𝐢𝐦𝒏→∞

𝒏(𝒃𝒏−𝟏)) = 𝐥𝐢𝐦𝒏→∞

𝒏(𝒃𝒏 − 𝟏) =


𝒏

𝟏𝒃𝒏 − 𝟏

=𝑪−𝑺


𝒏 + 𝟏 − 𝒏

𝟏𝒃𝒏+𝟏 − 𝟏

−𝟏

𝒃𝒏 − 𝟏


(𝒃𝒏+𝟏 − 𝟏)(𝒃𝒏 − 𝟏)

𝒃𝒏 − 𝒃𝒏+𝟏=


(𝒃𝒏 − 𝟏)𝐥𝐨𝐠𝒃𝒏𝒃𝒏 − 𝐥𝐨𝐠 𝒃𝒏 − 𝟏


(𝒃𝒏 − 𝟏)𝟐

𝒃𝒏 − 𝐥𝐨𝐠 𝒃𝒏 − 𝟏=


𝟏

𝒃𝒏 − 𝐥𝐨𝐠𝒃𝒏 − 𝟏(𝒃𝒏 − 𝟏)𝟐


𝟏

𝟏𝒃𝒏 − 𝟏

−𝐥𝐨𝐠 𝒃𝒏(𝒃𝒏 − 𝟏)𝟐

=


𝟏

𝟏𝒃𝒏 − 𝟏

−𝐥𝐨𝐠(𝟏 + 𝒃𝒏 − 𝟏)

𝒃𝒏 − 𝟏⋅

𝟏𝒃𝒏 − 𝟏

= +∞; (𝟒)

From (1),(2),(3),(4) it follows that:



𝐥𝐨𝐠𝒏= ∞


𝒙 ∈ [𝟏, 𝒏] ⇒ 𝒏 ≤𝒏𝟐

𝒙≥ 𝒏𝟐

Let [𝒙𝟐

𝒏] = 𝒕, 𝒕 ∈ {𝒏, 𝒏 + 𝟏,… , 𝒏𝟐} ⇒

𝒏𝟐

𝒙− 𝟏 < 𝑡 ≤

𝒏𝟐

𝒙⇔

𝒏𝟐

𝒕+𝟏< 𝑥 ≤

𝒏𝟐

𝒕

𝒂𝒏 = ∑ ∫ 𝒕

𝒏𝟐

𝒕

𝒏𝟐

𝒕+𝟏

𝒅𝒙

𝒏𝟐−𝟏

𝒕=𝒏

= ∑ 𝒕(𝒏𝟐

𝒕−𝒏𝟐

𝒕 + 𝟏)

𝒏𝟐−𝟏

𝒕=𝒏

= 𝒏𝟐 (𝟏

𝒏 + 𝟏+

𝟏

𝒏 + 𝟐+⋯+

𝟏

𝒏𝟐)

Now, 𝒃𝟏 > 1,𝒃𝒏+𝟏 = 𝟏 + 𝐥𝐨𝐠(𝒃𝒏) ⇒ 𝒃𝒏 > 1,∀𝑛 ∈ ℕ (induction from 𝒏 ∈ ℕ) and

from 𝒃𝒏+𝟏 − 𝒃𝒏 = 𝟏 + 𝐥𝐨𝐠 𝒃𝒏 − 𝒃𝒏 ≤ 𝟎, because 𝐥𝐨𝐠(𝟏 + 𝒙) ≤ 𝒙, ∀𝒙 > −1 ⇒

(𝒃𝒏)𝒏≥𝟏−decreasing.

So, (𝒃𝒏)𝒏≥𝟏 −convergent sequence, then

∃ 𝒍 ∈ ℝ, 𝒍 > 0 such that 𝒍 = 𝐥𝐢𝐦𝒏→∞

𝒃𝒏 ⇒ 𝒍 = 𝟏 + 𝐥𝐨𝐠 𝒍 ⇒ 𝒍 = 𝟏.

Let 𝒄𝒏 = 𝟏 +𝟏

𝟐+𝟏

𝟑+⋯+

𝟏

𝒏− 𝐥𝐨𝐠 𝒏.

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𝐥𝐨𝐠 𝒏= 𝐥𝐢𝐦𝒏→∞

𝒂𝒏 ⋅ 𝐥𝐨𝐠 𝒃𝒏𝒏 ⋅ 𝐥𝐨𝐠𝒏


(𝒂𝒏

𝒏𝟐 ⋅ 𝐥𝐨𝐠 𝒏⋅ 𝒏 𝐥𝐨𝐠 𝒃𝒏) =


[(𝒄𝒏𝟐 − 𝒄𝒏𝐥𝐨𝐠𝒏

+ 𝟏) ⋅ 𝒏 𝐥𝐨𝐠 𝒃𝒏] = 𝐥𝐢𝐦𝒏→∞

(𝒏 𝐥𝐨𝐠 𝒃𝒏) =


𝒏(𝒃𝒏 − 𝟏) 𝐥𝐨𝐠(𝟏 + 𝒃𝒏 − 𝟏)𝟏

𝒃𝒏−𝟏 = 𝐥𝐢𝐦𝒏→∞

𝒏(𝒃𝒏 − 𝟏) = 𝐥𝐢𝐦𝒏→∞

𝒏 𝐥𝐨𝐠 𝒃𝒏−𝟏 =

= ⋯ = 𝐥𝐢𝐦𝒏→∞

𝒏 ⋅ 𝒃𝟏 = +∞

1535. Find:


(𝟏

𝒏(∑∑

𝒊𝟑 + 𝒋𝟑

𝒊𝟒 + 𝒋𝟒

𝒏

𝒋=𝟏

𝒏

𝒊=𝟏

−∑∑𝒌𝟑 − 𝒍𝟑

𝒌𝟒 − 𝒍𝟒

𝒏

𝒍=𝟏

𝒏

𝒌=𝟏

))

Proposed by Mikael Bernardo-Mozambique

Solution by Ty Halpen-Florida-USA


(𝟏

𝒏(∑∑

𝒊𝟑 + 𝒋𝟑

𝒊𝟒 + 𝒋𝟒

𝒏

𝒋=𝟏

𝒏

𝒊=𝟏

−∑∑𝒌𝟑 − 𝒍𝟑

𝒌𝟒 − 𝒍𝟒

𝒏

𝒍=𝟏

𝒏

𝒌=𝟏

))


(∑𝟏

𝒏∑𝟏

𝒏⋅(𝒊𝒏)

𝟑

+ (𝒋𝒏)

𝟑

(𝒊𝒏)𝟒

+ (𝒋𝒏)𝟒

𝒏

𝒋=𝟏

𝒏

𝒊=𝟏

−∑𝟏

𝒏∑𝟏

𝒏⋅(𝒌𝒏)

𝟑

− (𝒍𝒏)

𝟑

(𝒌𝒏)𝟒

− (𝒍𝒏)𝟒

𝒏

𝒍=𝟏

𝒏

𝒌=𝟏

) =

= ∫ ∫ (𝒙𝟑 + 𝒚𝟑

𝒙𝟒 + 𝒚𝟒+𝒙𝟑 − 𝒚𝟑

𝒙𝟒 − 𝒚𝟒)

𝟏

𝟎

𝒅𝒙𝟏

𝟎

𝒅𝒚 =

= ∫ ∫ (𝒙𝟑

𝒙𝟒 + 𝒚𝟒−

𝒙

𝟐(𝒙𝟐 + 𝒚𝟐)−

𝟏

𝟐(𝒙 + 𝒚))

𝟏

𝟎

𝒅𝒙𝟏

𝟎

𝒅𝒚 +∫ ∫ (𝒚𝟑

𝒙𝟒 + 𝒚𝟒−

𝒚

𝟐(𝒙𝟐 + 𝒚𝟐))𝒅𝒚

𝟏

𝟎

𝒅𝒙𝟏

𝟎

=

=𝟏

𝟐∫ (− 𝐥𝐨𝐠(𝟏 + 𝒕) + 𝐥𝐨𝐠 𝒕 − 𝐥𝐨𝐠(𝟏 + 𝒕𝟒) − 𝟒 𝐥𝐨𝐠 𝒕 − 𝐥𝐨𝐠(𝟏 + 𝒕𝟐) + 𝟐 𝐥𝐨𝐠 𝒕)𝒅𝒕𝟏

𝟎

=

=𝟏

𝟐∫ (− 𝐥𝐨𝐠 𝒕 − 𝐥𝐨𝐠(𝟏 + 𝒕) − 𝐥𝐨𝐠(𝟏 + 𝒕𝟐) + 𝐥𝐨𝐠(𝟏 + 𝒕𝟒))𝒅𝒕𝟏

𝟎

=

=𝑰𝑩𝑷 𝟏

𝟐(−𝟐 𝐥𝐨𝐠𝟐 − 𝟐∫ (

𝟏

𝟏 + 𝒕𝟐−

𝟐

𝟏 + 𝒕𝟒)

𝟏

𝟎

𝒅𝒕 =

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=𝟏

𝟐(−𝟐 𝐥𝐨𝐠 𝟐 −

𝝅

𝟐+𝟏

√𝟐∫ (

𝒕 + √𝟐

𝒕𝟐 + √𝟐𝒕 + 𝟏+

𝒕 − √𝟐

−𝒕𝟐 + √𝟐𝒕 − 𝟏)𝒅𝒕

𝟏

𝟎

=

=𝟏

𝟐(−𝟐 𝐥𝐨𝐠𝟐 −

𝝅

𝟐+√𝟐

𝟐(𝝅 + 𝟐 𝐥𝐨𝐠(𝟐 + √𝟐) − 𝐥𝐨𝐠 𝟐) =

=𝝅

𝟒(√𝟐 − 𝟏) −

𝟒 + √𝟐

𝟒⋅ 𝐥𝐨𝐠 𝟐 +

√𝟐

𝟐𝐥𝐨𝐠(𝟐 + √𝟐) ≅ 𝟎. 𝟐𝟓𝟓𝟒

1536. Find:


𝟏

𝒏√∑(−𝟏)𝒌 (

𝒏

𝒌) (𝒏 − 𝒌)𝒏

𝒏−𝟏

𝒌=𝟎

𝒏



∑(−𝟏)𝒌 (𝒏

𝒌) (𝒏 − 𝒌)𝒏

𝒏−𝟏

𝒌=𝟎

= ∑(−𝟏)𝒏(−𝟏)𝒌 (𝒏

𝒏 − 𝒌)𝒌𝒏

𝒏

𝒌=𝟏

=

= (−𝟏)𝒏∑(−𝟏)𝒌 (𝒏

𝒌)𝒌𝒏; (𝟏)

𝒏

𝒌=𝟏

We have:

∑(𝒏

𝒌) (−𝟏)𝒌𝒙𝒌

𝒏

𝒌=𝟎

= (𝟏 − 𝒙)𝒏

Differentiating w.r.t. 𝒙, we get:

∑(𝒏

𝒌) (−𝟏)𝒌𝒌𝒙𝒌−𝟏

𝒏

𝒌=𝟏

= (−𝟏)(𝟏 − 𝒙)𝒏−𝟏(𝒏) ⇒

∑(𝒏

𝒌) (−𝟏)𝒌𝒌𝒙𝒌

𝒏

𝒌=𝟏

= (−𝟏)(𝟏 − 𝒙)𝒏−𝟏(𝒏𝒙)

Differentiating w.r.t. 𝒙, we get:

∑(𝒏

𝒌)(−𝟏)𝒌𝒌𝟐𝒙𝒌−𝟏

𝒏

𝒌=𝟏

= (−𝟏)(𝟏 − 𝒙)𝒏−𝟏(𝒏) + 𝒏(𝒏 − 𝟏)(−𝟏)𝟐(𝟏 − 𝒙)𝒏−𝟐(𝒙)

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Multiply again by 𝒙, we get:

∑(𝒏

𝒌) (−𝟏)𝒌𝒌𝟐𝒙𝒌

𝒏

𝒌=𝟏

= (−𝟏)(𝟏 − 𝒙)𝒏−𝟏(𝒏𝒙) + 𝒏(𝒏 − 𝟏)(−𝟏)𝟐(𝟏 − 𝒙)𝒏−𝟐𝒙𝟐

Differentiating again, we get:

∑(−𝟏)𝒌 (𝒏

𝒌)𝒌𝟑𝒙𝒌−𝟏

𝒏

𝒌=𝟏

= (−𝟏)(𝟏 − 𝒙)∗𝒏−𝟏 + 𝒏(𝒏 − 𝟏)(−𝟏)𝟐(𝟐𝒙)(𝟏 − 𝒙)𝒏−𝟐 +

+𝒏(𝒏 − 𝟏)(𝒏− 𝟐)(−𝟏)𝟑(𝟏 − 𝒙)𝒏−𝟑𝒙𝟐

Continuing this way, we get:

∑(−𝟏)𝒌 (𝒏

𝒌)𝒌𝒏𝒙𝒌−𝟏

𝒏

𝒌=𝟏

= (𝟏 − 𝒙)𝒈(𝒙) + 𝒏! (−𝟏)𝒏𝒙𝒏

Putting 𝒙 = 𝟏, we get

∑(−𝟏)𝒌 (𝒏

𝒌)𝒌𝒏

𝒏

𝒌=𝟏

= 𝒏! (−𝟏)𝒏; (𝟐)

From (1),(2) it follows

∑(−𝟏)𝒏−𝒌 (𝒏

𝒌) (𝒏 − 𝒌)𝒏

𝒏

𝒌=𝟏

− (−𝟏)𝒏𝒏! (−𝟏)𝒏 = 𝒏!

Now,


𝟏

𝒏√∑(−𝟏)𝒌 (

𝒏

𝒌) (𝒏 − 𝒌)𝒏

𝒏−𝟏

𝒌=𝟎

𝒏


𝟏

𝒏√𝒏!𝒏

=𝑪−𝑫


(𝒏 + 𝟏)!

(𝒏 + 𝟏)𝒏+𝟏⋅𝒏𝒏

𝒏!=


𝟏

(𝟏 +𝟏𝒏)

𝒏 =𝟏

𝒆

Solution 2 by Felix Marin-Romania

∑(−𝟏)𝒌 (𝒏

𝒌) (𝒏 − 𝒌)𝒏

𝒏−𝟏

𝒌=𝟎

= ∑(−𝟏)𝒏−𝒌 (𝒏

𝒏 − 𝒌) [𝒏 − (𝒏 − 𝒌)]𝒏

𝒏

𝒌=𝟎

= (−𝟏)𝒏∑(−𝟏)𝒌 (𝒏

𝒌)𝒌𝒏

𝒏

𝒌=𝟎

= (−𝟏)𝒏∑(−𝟏)𝒌 (𝒏

𝒌) {𝒏! [𝒛𝒏]𝒆𝒌𝒛}

𝒏

𝒌=𝟎

= (−𝟏)𝒏𝒏! [𝒛𝒏]∑(𝒏

𝒌) (−𝒆𝒛)𝒌

𝒏

𝒌=𝟎

=

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= (−𝟏)𝒏𝒏! [𝒛𝒏](𝟏 − 𝒆𝒛)𝒏 = 𝒏! [𝒛𝒏](𝒆𝒛 − 𝟏)𝒏 = 𝒏! [𝒛𝒏] [𝒏!∑{𝒋

𝒏}𝒛𝒋

𝒋!

∞

𝒋=𝒏

]

{𝒋𝒏} −is the Stirling Number of the Second Kind and {𝒏

𝒏} = 𝟏.

∑(−𝟏)𝒌 (𝒏

𝒌) (𝒏 − 𝒌)𝒏

𝒏−𝟏

𝒌=𝟎

= 𝒏!

Therefore,


𝟏

𝒏√∑(−𝟏)𝒌 (

𝒏

𝒌) (𝒏 − 𝒌)𝒏

𝒏−𝟏

𝒌=𝟎

𝒏


𝟏

𝒏√𝒏!𝒏


√√𝟐𝝅𝒏𝒏+𝟏𝟐𝒆−𝒏

𝒏

𝒏=


(𝟐𝝅)𝟏𝟐𝒏𝒏 ⋅ 𝒏

𝟏𝟐𝒏𝒆−𝟏

𝒏=𝟏

𝒆≅ 𝟎. 𝟑𝟔𝟕𝟗

1537. 𝒙𝟎 = 𝟏, 𝒙𝟏 = √𝟐, 𝒙𝒏+𝟏 + 𝒙𝒏−𝟏 = √𝟐𝒙𝒏, 𝒏 ∈ ℕ∗. Find:

𝛀(𝒏) = ∑∑(𝒙𝟐𝒌+𝒊 + 𝒙𝟑𝒌+𝒊 + 𝒙𝟓𝒌+𝒊)

𝟖

𝒊=𝟏

𝒏

𝒌=𝟏


Solution 1 by Kamel Gandouli Rezgui-Tunisia

𝒙𝒏+𝟐 = √𝟐𝒙𝒏+𝟏 − 𝒙𝒏; 𝒂 = √𝟐, 𝒃 = −𝟏

𝑬: 𝒙𝟐 = √𝟐𝒙− 𝟏 characteristic equation⇒ 𝒙𝟐 − √𝟐𝒙+ 𝟏 = 𝟎,𝚫 = −𝟏, 𝒛𝟏,𝟐 =√𝟐+√𝟐𝒊

𝟐⇒

𝒙𝒏 = 𝝀𝐜𝐨𝐬𝒏𝝅

𝟒+ 𝝁𝐬𝐢𝐧

𝒏𝝅

𝟒, 𝒙𝟎 = 𝟏 ⇒ 𝝀 = 𝟏 and 𝒙𝟏 = √𝟐 ⇒ 𝝁 = 𝟏 ⇒

𝒙𝒏 = 𝐜𝐨𝐬𝒏𝝅

𝟒+ 𝐬𝐢𝐧

𝒏𝝅

𝟒= √𝟐 𝐜𝐨𝐬 (

(𝒏 − 𝟏)𝝅

𝟒)

𝒙𝟐𝒌+𝟏 = √𝟐𝐜𝐨𝐬 ((𝟐𝒌 + 𝒊 − 𝟏)𝝅

𝟒) = √𝟐𝐜𝐨𝐬 (

𝒌𝝅

𝟐+(𝒊 − 𝟏)𝝅

𝟒) ; (𝒊 = 𝟏, 𝟖̅̅ ̅̅ ̅) ⇒

∑𝒙𝟐𝒌+𝒊

𝟖

𝒊=𝟏

= √𝟐𝐜𝐨𝐬 (𝒌𝝅

𝟐) + √𝟐𝐜𝐨𝐬 (

𝒌𝝅

𝟐+𝝅

𝟒) + √𝟐𝐜𝐨𝐬 (

𝒌𝝅

𝟐+𝝅

𝟐) + 𝟒√𝟐𝐜𝐨𝐬 (

𝒌𝝅

𝟐+𝟑𝝅

𝟒)

+

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+√𝟐𝐜𝐨𝐬 (𝒌𝝅

𝟐+ 𝝅) + √𝟐𝐜𝐨𝐬 (

𝒌𝝅

𝟐+𝟓𝝅

𝟒) + √𝟐𝐜𝐨𝐬 (

𝒌𝝅

𝟐+𝟑𝝅

𝟐) + √𝟐 𝐜𝐨𝐬 (

𝒌𝝅

𝟐+𝟕𝝅

𝟒) = 𝟎

𝒙𝟑𝒌+𝒊 = √𝟐𝐜𝐨𝐬 ((𝟑𝒌 + 𝒊 − 𝟏)𝝅

𝟒) , 𝒙𝟓𝒌+𝒊 = √𝟐𝐜𝐨𝐬 (

(𝟓𝒌 + 𝒊 − 𝟏)𝝅

𝟒) ⇒

𝒙𝟑𝒌+𝒊 + 𝒙𝟓𝒌+𝒊 = 𝟐√𝟐𝐜𝐨𝐬 ((𝟖𝒌 + 𝟐𝒊 − 𝟐)𝝅

𝟖) 𝐜𝐨𝐬 (

𝒌𝝅

𝟒)

= 𝟐√𝟐𝐜𝐨𝐬 (𝒌𝝅 +(𝒊 − 𝟏)𝝅

𝟒) 𝐜𝐨𝐬 (

𝒌𝝅

𝟒)

∑(𝒙𝟓𝒌+𝒊 + 𝒙𝟑𝒌+𝒊)

𝟖

𝒊=𝟏

= 𝟐√𝟐∑𝐜𝐨𝐬(𝒌𝝅 +(𝒊 − 𝟏)𝝅

𝟒)𝐜𝐨𝐬 (

𝒌𝝅

𝟒)

𝟖

𝒊=𝟏

= 𝟎

Therefore,

𝛀(𝒏) =∑∑(𝒙𝟐𝒌+𝒊 + 𝒙𝟑𝒌+𝒊 + 𝒙𝟓𝒌+𝒊)

𝟖

𝒊=𝟏

𝒏

𝒌=𝟏

= 𝟎


Let generating function of (𝒙𝒏)𝒏≥𝟏 be

𝑨(𝒕) = 𝒙𝟎 + 𝒙𝟏𝒕 + 𝒙𝟐𝒕𝟐 + 𝒙𝟑𝒕

𝟑 +⋯

−√𝟐𝒕𝑨(𝒕) = −√𝟐𝒙𝟎𝒕 − √𝟐𝒙𝟏𝒕𝟐 − √𝟐𝒙𝟐𝒕

𝟑 −⋯

𝒕𝟐𝑨(𝒕) = 𝒙𝟎𝒕𝟐 + 𝒙𝟏𝒕

𝟑 +⋯

(𝟏 − √𝟐𝒕 + 𝒕𝟐)𝑨(𝒕) = 𝟏 ⇒ 𝑨(𝒕) =𝟏

𝟏 − √𝟐𝒕 + 𝒕𝟐=

𝟏

(𝟏 − 𝜶𝒕)(𝟏 − 𝜷𝒕), 𝐰𝐡𝐞𝐫𝐞

𝜶 =𝟏

𝟐(√𝟐 + √𝟐𝒊) =

𝟏

√𝟐(𝟏 + 𝒊) 𝐚𝐧𝐝 𝜷 =

𝟏

√𝟐(𝟏 − 𝒊).

𝑨(𝒕) =𝟏

(𝜶 − 𝜷)𝒕(

𝟏

𝟏 − 𝜶𝒕−

𝟏

𝟏 − 𝜷𝒕) =

𝟏

(𝜶 − 𝜷)𝒕[(𝟏 − 𝜶𝒕)−𝟏 − (𝟏 − 𝜷𝒕)−𝟏] =

=𝟏

𝜶− 𝜷(𝜶𝒏+𝟏 − 𝜷𝒏+𝟏) =

=𝟏

√𝟐𝒊[𝐜𝐨𝐬

(𝒏 + 𝟏)𝝅

𝟒+ 𝒊 𝐬𝐢𝐧

(𝒏 + 𝟏)𝝅

𝟒− 𝐜𝐨𝐬

(𝒏 + 𝟏)𝝅

𝟒+ 𝒊 𝐬𝐢𝐧

(𝒏 + 𝟏)𝝅

𝟒]

Thus, 𝒙𝒏 = √𝟐𝐬𝐢𝐧(𝒏+𝟏)𝝅

𝟒. Now,

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∑𝒙𝟐𝒌+𝒊

𝟖

𝒊=𝟏

= √𝟐∑𝐬𝐢𝐧 ((𝟐𝒌 + 𝒊 + 𝟏)𝝅

𝟒)

𝟖

𝒊=𝟏

=

= √𝟐 [𝟐 𝐬𝐢𝐧 ((𝟐𝒌 + 𝟏𝟏)𝝅

𝟖)(𝐜𝐨𝐬

𝟕𝝅

𝟖+ 𝐜𝐨𝐬

𝟓𝝅

𝟖+ 𝐜𝐨𝐬

𝟑𝝅

𝟖+ 𝐜𝐨𝐬

𝝅

𝟖)] =

= √𝟐 [𝟐 𝐬𝐢𝐧 ((𝟐𝒌 + 𝟏𝟏)𝝅

𝟖)(𝐜𝐨𝐬 (𝝅 −

𝝅

𝟖) + 𝐜𝐨𝐬 (𝝅 −

𝟑𝝅

𝟖) + 𝐜𝐨𝐬

𝟑𝝅

𝟖+ 𝐜𝐨𝐬

𝝅

𝟖)] = 𝟎

Similarly,

∑𝒙𝟑𝒌+𝒊

𝟖

𝒊=𝟏

= 𝟎 𝐚𝐧𝐝 ∑𝒙𝟓𝒌+𝒊

𝟖

𝒊=𝟏

= 𝟎

Therefore,

𝛀(𝒏) =∑∑(𝒙𝟐𝒌+𝒊 + 𝒙𝟑𝒌+𝒊 + 𝒙𝟓𝒌+𝒊)

𝟖

𝒊=𝟏

𝒏

𝒌=𝟏

= 𝟎

1538. For 𝒂, 𝒃 > 𝟎 prove that:

∫𝒙𝟐 − 𝒂

𝒙𝟐 + 𝒃

∞

−∞

𝐬𝐢𝐧 (𝒙

√𝒃𝐥𝐨𝐠 (

𝒂 + 𝒃

𝒂))𝒅𝒙

𝒙= 𝟎


Solution by Kartick Chandra Betal-India

∫𝒙𝟐 − 𝒂

𝒙𝟐 + 𝒃

∞

−∞

𝐬𝐢𝐧 (𝒙


𝒂 + 𝒃

𝒂))𝒅𝒙

𝒙= 𝟐∫

𝒙𝟐 − 𝒂

𝒙𝟐 + 𝒃

∞

𝟎

𝐬𝐢𝐧 (𝒙


𝒂 + 𝒃

𝒂))𝒅𝒙

𝒙=

= 𝟐∫𝟏

𝒙⋅ 𝐬𝐢𝐧 (

𝒙


𝒂 + 𝒃

𝒂))

∞

𝟎

𝒅𝒙 − 𝟐(𝒂 + 𝒃)∫𝟏

𝒙(𝒙𝟐 + 𝒃)⋅𝒙

√𝒃𝐥𝐨𝐠 (𝟏 +

𝒃

𝒂)

∞

𝟎

𝒅𝒙 =

= 𝟐 ⋅𝝅

𝟐− 𝟐(𝒂 + 𝒃)∫

𝟏

𝒙𝟐 + 𝒃∫ 𝐜𝐨𝐬(𝒙𝒚)

𝟏

√𝒃𝐥𝐨𝐠(𝟏+

𝒃𝒂)

𝟎

𝒅𝒚∞

𝟎

𝒅𝒙 =

= 𝝅 − 𝟐(𝒂 + 𝒃)∫ ∫𝐜𝐨𝐬(√𝒃𝒙𝒚)

√𝒃(𝟏 + 𝒙𝟐)

∞

𝟎

𝒅𝒙

𝟏


𝒃𝒂)

𝟎

𝒅𝒚 =

= 𝝅 − 𝟐(𝒂 + 𝒃)∫𝝅

𝟐√𝒃⋅ 𝒆−√𝒃𝒚

𝟏


𝒃𝒂)

𝟎

𝒅𝒚 =

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= 𝝅−𝝅(𝒂 + 𝒃)

√𝒃⋅𝟏

√𝒃[𝟏 − 𝒆− 𝐥𝐨𝐠(𝟏+

𝒃𝒂)] = 𝝅 −

𝝅(𝒂 + 𝒃)

𝒃(𝟏 −

𝒂

𝒂 + 𝒃) = 𝟎

1539. Prove that:

𝑭𝟏(𝒌, 𝒌; 𝒌 + 𝟏; 𝟏 − 𝒎)𝟐 =𝚪(𝒌 + 𝟏)

𝚪(𝒌)∫

𝒅𝒙

(𝟏 + 𝒙)(𝒎 + 𝒙)𝒌

∞

𝟎

where 𝑭𝟏(. , . ; . ; . ) −𝟐 𝒉𝒚𝒑𝒆𝒓𝒈𝒆𝒐𝒎𝒆𝒕𝒓𝒊𝒄 function, 𝒎 ∈ ℝ+, 𝒌 ∈ ℕ.

Proposed by Simon Peter-Madagascar

Solution by Syed Shahabudeen-Kerala-India

𝑭𝟏(𝒌,𝒌; 𝒌 + 𝟏; 𝟏 −𝒎)𝟐 =

= (𝒎)−𝒌 𝑭𝟏 (𝒌, 𝟏; 𝒌 + 𝟏;𝒎 − 𝟏

𝒎)𝟐 (𝒂𝒑𝒑𝒍𝒚 𝑷𝒇𝒂𝒇𝒇 𝑻𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏)

=𝚪(𝒌 + 𝟏)

𝒎𝒌𝚪(𝒌)∫

(𝟏 − 𝒕)𝒌−𝟏

(𝟏 − (𝒎− 𝟏𝒎 ) 𝒕)

𝒌

𝟏

𝟎

𝒅𝒕; (𝒂𝒑𝒑𝒍𝒚 𝑬𝒖𝒍𝒆𝒓 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍)

=𝚪(𝒌 + 𝟏)

𝚪(𝒌)∫

(𝟏 − 𝒕)𝒌−𝟏

(𝒎 − (𝒎− 𝟏)𝒕)𝒌

𝟏

𝟎

𝒅𝒕 =𝚪(𝒌 + 𝟏)

𝚪(𝒌)∫

𝟏

(𝟏 − 𝒕) (𝒎+𝒕

𝟏 − 𝒕)𝒌𝒅𝒕

𝟏

𝟎

=𝒙=

𝒕𝟏−𝒕

=𝚪(𝒌 + 𝟏)

𝚪(𝒌)∫

𝒙 + 𝟏

(𝒎+ 𝒙)𝒌𝒅𝒙

(𝒙 + 𝟏)𝟐

∞

𝟎

=𝚪(𝒌 + 𝟏)

𝚪(𝒌)∫

𝒅𝒙

(𝒙 + 𝟏)(𝒎+ 𝒙)𝒌

∞

𝟎

1540. Prove that:

∏𝒏+ 𝐜𝐨𝐬 (

𝒏𝝅𝟑)

𝒏 + 𝐜𝐨𝐭 (𝝅𝟑) 𝐬𝐢𝐧 (

𝒏𝝅𝟑)

∞

𝒏=𝟏

=𝝅√𝟐 ⋅ 𝚪 (

𝟏𝟐) 𝚪 (

𝟓𝟏𝟐)

𝚪𝟐 (𝟏𝟒) 𝚪 (

𝟏𝟑) 𝚪 (

𝟕𝟔) 𝚪 (

𝟏𝟏𝟏𝟐)


Solution by Amrit Awasthi-India

𝐜𝐨𝐬 (𝒏𝝅

𝟑) and 𝐬𝐢𝐧 (

𝒏𝝅

𝟑) have same sighn, ∀𝒏 = 𝟔𝒌 + 𝟏 or 𝒏 = 𝟔𝒌 + 𝟓.

Therefore, rewriting the product we get:

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𝑷 =∏𝟔𝒌+ 𝟐 −

𝟏𝟐

𝟔𝒌 + 𝟐 +𝟏𝟐

∞

𝒌=𝟎

⋅𝟔𝒌 + 𝟑 − 𝟏

𝟔𝒌 + 𝟑⋅𝟔𝒌 + 𝟔 + 𝟏

𝟔𝒌 + 𝟔=

=𝟏𝟏 ⋅ 𝟕

𝟓 ⋅ 𝟑 ⋅ 𝟗𝐥𝐢𝐦𝒏→∞

∏

𝚪(𝒏+ 𝟏 +𝟏𝟒)

𝚪 (𝟏 +𝟏𝟒)

𝚪 (𝒏 + 𝟏 +𝟏𝟑)

𝚪 (𝟏 +𝟏𝟑)

𝚪(𝒏 + 𝟏 +𝟏𝟏𝟏𝟐)

𝚪(𝟏 +𝟏𝟏𝟏𝟐)

𝚪(𝒏 + 𝟏 +𝟕𝟔)

𝚪(𝟏 +𝟕𝟔)

𝚪 (𝒏 + 𝟏 +𝟓𝟏𝟐)

𝚪 (𝟏 +𝟓𝟏𝟐)

𝚪 (𝒏 + 𝟏 +𝟏𝟐)

𝚪 (𝟏 +𝟏𝟐)

𝚪(𝒏 + 𝟏 +𝟑𝟒)

𝚪(𝟏 +𝟑𝟒)

𝚪(𝒏 + 𝟏)𝚪(𝟐)

∞

𝒌=𝟏

=

=𝟏𝟏 ⋅ 𝟕

𝟓 ⋅ 𝟑 ⋅ 𝟗⋅

𝚪 (𝟏 +𝟓𝟏𝟐)𝚪 (𝟏 +

𝟏𝟐) 𝚪(𝟏 +

𝟑𝟒)

𝚪 (𝟏 +𝟏𝟒) 𝚪(𝟏 +

𝟏𝟑) 𝚪 (𝟏 +

𝟏𝟏𝟐)𝚪 (𝟏 +

𝟕𝟔)=

=𝟏𝟏 ⋅ 𝟕

𝟓 ⋅ 𝟑 ⋅ 𝟗⋅

𝟓𝟏𝟐 ⋅

𝟏𝟐 ⋅𝟑𝟒 ⋅ 𝚪 (

𝟓𝟏𝟐)𝚪 (

𝟏𝟐)𝚪 (

𝟑𝟒)

𝟏𝟒 ⋅𝟏𝟑 ⋅𝟏𝟏𝟏𝟐 ⋅

𝟕𝟔 ⋅ 𝚪 (

𝟏𝟒)𝚪 (

𝟏𝟑)𝚪 (

𝟏𝟏𝟐)𝚪 (

𝟕𝟔)=

=𝚪(𝟓𝟏𝟐)𝚪 (

𝟏𝟐)𝚪 (

𝟑𝟒)

𝚪(𝟏𝟒)𝚪 (

𝟏𝟑)𝚪 (

𝟏𝟏𝟐)𝚪 (

𝟕𝟔)=𝚪 (𝟓𝟏𝟐)𝚪(

𝟏𝟐) 𝚪(𝟏 −

𝟏𝟒)

𝚪(𝟏𝟒) 𝚪(

𝟏𝟑) 𝚪(

𝟏𝟏𝟐)𝚪 (

𝟕𝟔)=

𝝅√𝟐 ⋅ 𝚪 (𝟏𝟐)𝚪 (

𝟓𝟏𝟐)

𝚪𝟐 (𝟏𝟒)𝚪 (

𝟏𝟑)𝚪 (

𝟕𝟔)𝚪 (

𝟏𝟏𝟏𝟐)

Therefore,

∏𝒏+ 𝐜𝐨𝐬 (

𝒏𝝅𝟑 )

𝒏 + 𝐜𝐨𝐭 (𝝅𝟑) 𝐬𝐢𝐧 (

𝒏𝝅𝟑 )

∞

𝒏=𝟏

=𝝅√𝟐 ⋅ 𝚪 (

𝟏𝟐)𝚪 (

𝟓𝟏𝟐)

𝚪𝟐 (𝟏𝟒)𝚪 (

𝟏𝟑)𝚪 (

𝟕𝟔)𝚪 (

𝟏𝟏𝟏𝟐)

1541. If all the derivatives of 𝒇(𝒙) are defined at 𝒙 = 𝟏, then prove that:

𝒇(𝒆−𝒙) = ∑(−𝒙)𝒌

𝒌![𝑩𝒌(𝑫)𝒇(𝒙)]|𝒙=𝟏

∞

𝒌=𝟎

where, 𝑫 ≔𝒅

𝒅𝒙 and 𝑩𝒌(𝒙) is the Bell polynomial.



𝐋𝐞𝐭 𝒇(𝒙) = ∑𝒂𝒏𝒙𝒏

∞

𝒏=𝟎

, 𝐭𝐡𝐞𝐧

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𝒇(𝒆−𝒙) = ∑𝒂𝒏𝒆−𝒏𝒙

∞

𝒏=𝟎

= ∑𝒂𝒏

∞

𝒏=𝟎

∑(−𝒏𝒙)𝒌

𝒌!

∞

𝒌=𝟎

Hence,

𝒇(𝒆−𝒙) = ∑(−𝒙)𝒌

𝒌!

∞

𝒌=𝟎

∑𝒏𝒌𝒂𝒏

∞

𝒏=𝟎

𝐍𝐨𝐰, 𝐥𝐞𝐭: 𝑭𝒎(𝒙) = ∑𝒏𝒎𝒂𝒏𝒙𝒏

∞

𝒏=𝟎

= ∑𝒄(𝒎, 𝒏)𝒇(𝒏)(𝒙)𝒙𝒏𝒎

𝒏=𝟎

𝒙𝑭𝒎′ (𝒙) = 𝑭𝒎+𝟏(𝒙) ⇒ 𝒄(𝒎,𝒏) = 𝒄(𝒎− 𝟏, 𝒏 − 𝟏) + 𝒏𝒄(𝒎− 𝟏, 𝒏), where

𝟎 ≤ 𝒏 ≤ 𝒎, 𝒄(𝒎, 𝟎) = 𝜹𝒎𝟎 and 𝒄(𝒏, 𝒏) = 𝟏, where 𝜹𝒎𝒏 is the Kronecker delta. It is known from the definition of Bell polynomials that,

𝑩𝒎(𝒙) = ∑𝑺(𝒎,𝒌)𝒙𝒌𝒎

𝒌=𝟎

; 𝑺(𝒎, 𝒌) − (𝑺𝒕𝒊𝒓𝒍𝒊𝒏𝒈 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒆𝒄𝒐𝒏𝒅 𝒌𝒊𝒏𝒅)

𝑺(𝒎, 𝒌) = 𝑺(𝒎− 𝟏,𝒌 − 𝟏) + 𝒌𝑺(𝒎− 𝟏, 𝒌), where 𝟎 ≤ 𝒌 ≤ 𝒎, using this property and knowing the fact that 𝑺(𝒏,𝒏) = 𝟏 and 𝑺(𝒏, 𝟎) = 𝜹𝒏𝟎, we can show that:

𝒙𝑩𝒎′ (𝒙𝑫) = 𝑩𝒎+𝟏(𝒙𝑫) since 𝒄(𝒎,𝒏) and 𝑺(𝒎, 𝒌) satisfies the same recurrence relation

with same boundary/initial conditions, we conclude that: 𝑺(𝒎, 𝒌) = 𝒄(𝒎, 𝒌) ⇒ 𝑭𝒎(𝒙) = 𝑩𝒎(𝒙𝑫)𝒇(𝒙) ⇒ 𝑭𝒎(𝟏) = [𝑩𝒌(𝑫)𝒇(𝒙)]|𝒙=𝟏

1542. If 𝚽𝒏 = ∑ ∑ (𝒏𝒊) (𝒏

𝒋)𝒏

𝒋=𝟎 𝐜𝐨𝐬 (𝟐𝝅(𝒋−𝒊)

𝟕)𝒏

𝒊=𝟎 −

𝟐∑ (𝒏𝒊) (𝒏

𝒋)𝟎≤𝒊<𝑗≤𝑛 𝐜𝐨𝐬 (

𝟐𝝅(𝒊−𝒋)

𝟕). Define 𝑴 = {√𝚽𝒏

𝐜𝐨𝐬(𝒏𝝅)𝒏

𝐬𝐢𝐧 (𝒏𝝅

𝟒) |𝒏 ∈ ℕ}.

Find 𝑴′ −derived set.

Proposed by Surjeet Singhania-India


𝚽𝒏 =∑∑(𝒏

𝒊)(𝒏

𝒋)

𝒏

𝒋=𝟎

𝐜𝐨𝐬 (𝟐𝝅(𝒋 − 𝒊)

𝟕)

𝒏

𝒊=𝟎

− 𝟐 ∑ (𝒏

𝒊) (𝒏

𝒋)

𝟎≤𝒊<𝑗≤𝑛

𝐜𝐨𝐬 (𝟐𝝅(𝒊 − 𝒋)

𝟕)

Evaluate these finite series one by one

∑∑(𝒏

𝒋) (𝒏

𝒌)𝐜𝐨𝐬 (

𝟐𝝅(𝒌 − 𝒋)

𝟕)

𝒏

𝒌=𝟎

𝒏

𝒋=𝟎

= 𝓡(∑(𝒏

𝒌) 𝒆−

𝟐𝒊𝝅𝒌𝟕

𝐧

𝐤=𝟎

∑(𝒏

𝒋)

𝒏

𝒋=𝟎

𝒆𝟐𝝅𝒊𝒋𝟕 ) =

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= (𝟏 + 𝐞𝐱𝐩 (𝟐𝝅

𝟕))𝒏

(𝟏 + 𝐞𝐱𝐩 (−𝟐𝝅

𝟕))𝒏

= 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏 (𝝅

𝟕)

∑ (𝒏

𝒌)(𝒏

𝒋) 𝐜𝐨𝐬 (

𝟐𝝅(𝒋 − 𝒌)

𝟕)

𝟎≤𝒌<𝑗≤𝑛

= 𝟐𝟐𝒏−𝟏 𝐜𝐨𝐬𝟐𝒏 (𝝅

𝟕) −

𝟏

𝟐(𝟐𝒏

𝒏)

Hence our 𝚽𝒏 = (𝟐𝒏𝒏). Denote 𝑿𝒏 = √𝚽𝒏

𝐜𝐨𝐬(𝒏𝝅)𝒏


𝟒).

For finding derived set we need to find possible convergent sequences

𝑿𝟒𝒏 = 𝟎,𝑿𝟒𝒏+𝟏, 𝑿𝟒𝒏+𝟑 → ±√𝟐

𝟖 and 𝑿𝟒𝒏+𝟐 → ±𝟒.

Hence, 𝑴′ = {𝟎, ±𝟒, ±√𝟐

𝟖}

1543. If 𝒂, 𝒃, 𝒏, 𝒌 ∈ ℕ and 𝑺(𝒂, 𝒃, 𝒏) = {𝒌|𝒌 ≡ 𝒂(𝒎𝒐𝒅 𝒃), 𝒌|𝒏} then prove

that:

∑𝒙𝒂𝒌

𝟏 − 𝒙𝒃𝒌

∞

𝒌=𝟏

= ∑|𝑺(𝒂, 𝒃, 𝒏)|𝒙𝒏∞

𝒏=𝟏



Observe that if |𝒙| < 1, then

𝒙𝒂

𝟏 − 𝒙𝒃= 𝒙𝒂(𝟏 + 𝒙𝒃 + 𝒙𝟐𝒃 + 𝒙𝟑𝒃 + 𝒙𝟒𝒃 +⋯)

𝒙𝟐𝒂

𝟏 − 𝒙𝟐𝒃= 𝒙𝟐𝒂(𝟏 + 𝒙𝟐𝒃 + 𝒙𝟒𝒃 + 𝒙𝟔𝒃 + 𝒙𝟖𝒃 +⋯)

𝒙𝟑𝒂

𝟏 − 𝒙𝟑𝒃= 𝒙𝟑𝒂(𝟏 + 𝒙𝟑𝒃 + 𝒙𝟔𝒃 + 𝒙𝟗𝒃 + 𝒙𝟏𝟐𝒃 +⋯)

Adding them, we have:

∑𝒙𝒂𝒌

𝟏 − 𝒙𝒃𝒌

∞

𝒌=𝟏

= ∑𝒂𝒏𝒙𝒏

∞

𝒏=𝟏

Where 𝒂𝒏 is the number of solutions of 𝒂𝒑 + 𝒃𝒑𝒒 = 𝒏, where 𝒑 ∈ ℕ and 𝒒 ∈ ℕ + {𝟎} for

the some given values of 𝒂 and 𝒃, thus 𝒂𝒏 is the number of divisors of 𝒏 of the form 𝒃𝒒 +

𝒂.

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Using the definition of 𝑺(𝒂, 𝒃, 𝒏) we can show that 𝒂𝒏 = |𝑺(𝒂, 𝒃, 𝒏)| and this completes

the proof.

1544. Find:

𝛀 = ∫√𝒙 𝐥𝐨𝐠 𝒙

𝒙𝟑 + 𝒙√𝒙 + 𝟏𝒅𝒙

∞

𝟏




𝒙𝟑 + 𝒙√𝒙+ 𝟏𝒅𝒙

∞

𝟏

=𝒙=𝒙𝟐

𝟒∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟔 + 𝒙𝟑 + 𝟏

∞

𝟏

𝒅𝒙 =𝒙=𝟏𝒙− 𝟒∫

𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟔 + 𝒙𝟑 + 𝟏𝒅𝒙

𝟏

𝟎

=

= 𝟒∫𝒙𝟐(𝒙𝟑 − 𝟏) 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟗

𝟏

𝟎

𝒅𝒙 = 𝟒∫(𝒙𝟓 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟗

𝟏

𝟎

𝒅𝒙 =

= 𝟒∑∫ (𝒙𝟗𝒏+𝟓 − 𝒙𝟗𝒏+𝟐) 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

=

∞

𝒏=𝟎

𝟒∑(𝟏

(𝟗𝒏 + 𝟑)𝟐−

𝟏

(𝟗𝒏 + 𝟔)𝟐)

∞

𝒏=𝟎

=

=𝟒

𝟖𝟏∑(

𝟏

(𝒏+𝟏𝟑)𝟐 −

𝟏

(𝒏 +𝟐𝟑)𝟐)

∞

𝒏=𝟎

Therefore,


𝒙𝟑 + 𝒙√𝒙+ 𝟏𝒅𝒙

∞

𝟏

=𝟒

𝟖𝟏[𝝍(𝟏) (

𝟏

𝟑) − 𝝍(𝟏) (

𝟐

𝟑)]

Solution 2 by Ajetunmobi Abdulqoyyum-Nigeria


𝒙𝟑 + 𝒙√𝒙 + 𝟏𝒅𝒙

∞

𝟏

=𝒙=𝒕𝟐

𝟒∫𝒕𝟐 𝐥𝐨𝐠 𝒕

𝒕𝟔 + 𝒕𝟑 + 𝟏𝒅𝒕

∞

𝟏

𝛀 = ∫𝒕𝟐 𝐥𝐨𝐠 𝒕

𝒕𝟔 + 𝒕𝟑 + 𝟏𝒅𝒕

∞

𝟏

=𝒕=𝟏𝒕− 𝟒∫

𝒕𝟐 𝐥𝐨𝐠 𝒕

𝒕𝟔 + 𝒕𝟑 + 𝟏𝒅𝒕

𝟏

𝟎

𝛀 = −𝟒∫(𝟏 − 𝒕𝟑)𝒕𝟐 𝐥𝐨𝐠 𝒕

𝟏 − 𝒕𝟗𝒅𝒕

𝟏

𝟎

= −∫𝒕𝟐 𝐥𝐨𝐠 𝒕

𝟏 − 𝒕𝟗𝒅𝒕

𝟏

𝟎

+ 𝟒∫𝒕𝟓 𝐥𝐨𝐠 𝒕

𝟏 − 𝒕𝟗𝒅𝒕

𝟏

𝟎

=𝒕𝟗=𝒙

=𝟒

𝟖𝟏(−∫

𝒕𝟏𝟑−𝟏 𝐥𝐨𝐠 𝒕


𝟏

𝟎

+∫𝒕𝟐𝟑−𝟏 𝐥𝐨𝐠 𝒕


𝟏

𝟎

) =

Therefore,

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𝒙𝟑 + 𝒙√𝒙+ 𝟏𝒅𝒙

∞

𝟏

=𝟒

𝟖𝟏[𝝍(𝟏) (

𝟏

𝟑) − 𝝍(𝟏) (

𝟐

𝟑)]


𝝍(𝒎) = −∫𝒕𝒛−𝟏

𝟏 − 𝒕𝐥𝐨𝐠𝒎 𝒕 𝒅𝒕

𝟏

𝟎


𝒙𝟑 + 𝒙√𝒙+ 𝟏𝒅𝒙

∞

𝟏

=𝒙=√𝒙

𝟒∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟔 + 𝒙𝟑 + 𝟏𝒅𝒙

∞

𝟎

=𝒙=𝟏𝒙

= −𝟒∫𝒙𝟐 𝐥𝐨𝐠 𝒙

𝒙𝟔 + 𝒙𝟑 + 𝟏𝒅𝒙

𝟏

𝟎

= 𝟒{∫𝒙𝟓 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟗𝒅𝒙

𝟏

𝟎

−∫𝒙𝟐 𝐥𝐨𝐠𝒙

𝟏 − 𝒙𝟗𝒅𝒙

𝟏

𝟎

}

𝑨 = ∫𝒙𝟓 𝐥𝐨𝐠 𝒙

𝟏 − 𝒙𝟗𝒅𝒙

𝟏

𝟎

=𝟏

𝟖𝟏∫𝒙−𝟏𝟑 𝐥𝐨𝐠 𝒙


𝟏

𝟎

= −𝟏

𝟖𝟏𝝍(𝟏) (

𝟐

𝟑)

𝑩 = ∫𝒙𝟐 𝐥𝐨𝐠𝒙

𝟏 − 𝒙𝟗𝒅𝒙

𝟏

𝟎

=𝟏

𝟖𝟏∫𝒙−𝟐𝟑 𝐥𝐨𝐠 𝒙


𝟏

𝟎

= −𝟏

𝟖𝟏𝝍(𝟏) (

𝟏

𝟑)

Therefore,

𝛀 = 𝟒(𝑨 − 𝑩) =𝟒

𝟖𝟏[𝝍(𝟏) (

𝟏

𝟑) − 𝝍(𝟏) (

𝟐

𝟑)]

Solution 4 by Probal Chakraborty-Kolkata-India

𝝍(𝒎) = −∫𝒕𝒛−𝟏

𝟏 − 𝒕𝐥𝐨𝐠𝒎 𝒕 𝒅𝒕

𝟏

𝟎


𝒙𝟑 + 𝒙√𝒙 + 𝟏𝒅𝒙

∞

𝟏

= ∫√𝒙 𝐥𝐨𝐠 𝒙

(𝒙𝟑𝟐)𝟐

+ 𝒙𝟑𝟐 + 𝟏

𝒅𝒙∞

𝟎

=𝒙𝟑𝟐=𝒕 𝟐

𝟑∫

𝐥𝐨𝐠 (𝒕𝟐𝟑)

𝒕𝟐 + 𝒕 + 𝟏𝒅𝒕

𝟏

𝟎

=

=𝟒

𝟗∫

𝐥𝐨𝐠 𝒕

𝒕𝟐 + 𝒕 + 𝟏𝒅𝒕

𝟏

𝟎

=𝟒

𝟗∫

𝟏 − 𝒕

𝟏 − 𝒕𝟑𝐥𝐨𝐠 𝒕𝒅𝒕

𝟏

𝟎

=𝟒

𝟗∫

𝐥𝐨𝐠 𝒕

𝟏 − 𝒕𝟑𝒅𝒕

𝟏

𝟎

−𝟒

𝟗∫𝒕 𝐥𝐨𝐠 𝒕

𝟏 − 𝒕𝟑𝒅𝒕

𝟏

𝟎

=𝒕=𝒚

𝟏𝟑

=𝟒

𝟖𝟏∫𝒚−𝟐𝟑 𝐥𝐨𝐠 𝒚

𝟏 − 𝒚𝒅𝒚

𝟏

𝟎

=𝟒

𝟖𝟏∫𝒚−𝟏𝟑 𝐥𝐨𝐠 𝒚

𝟏 − 𝒚𝒅𝒚

𝟏

𝟎

=𝟒

𝟖𝟏[𝝍(𝟏) (

𝟏

𝟑) − 𝝍(𝟏) (

𝟐

𝟑)]

1545. Find:

𝛀(𝟏, 𝟐) = ∫𝐥𝐨𝐠 𝒙

(𝟏 + 𝐥𝐨𝐠 𝒙)𝟐𝒅𝒙 ;𝛀(𝟐, 𝟑) = ∫

𝐥𝐨𝐠𝟐 𝒙

(𝟏 + 𝐥𝐨𝐠 𝒙)𝟑𝒅𝒙

𝛀(𝒎,𝒏) = ∫𝐥𝐨𝐠𝒎 𝒙

(𝟏 + 𝐥𝐨𝐠 𝒙)𝒏𝒅𝒙 ,𝒎, 𝒏 ∈ ℕ

Proposed by Durmuş Ogmen-Turkyie

www.ssmrmh.ro


Solution by Mikael Bernardo-Mozambique

𝛀(𝟏, 𝟐) = ∫𝐥𝐨𝐠𝒙

(𝟏 + 𝐥𝐨𝐠𝒙)𝟐𝒅𝒙 =

𝐥𝐨𝐠 𝒙=𝒖∫

𝒖𝒆𝒖

(𝟏 + 𝒖)𝟐𝒅𝒖 = ∫

𝒆𝒖

𝟏 + 𝒖𝒅𝒖 −∫

𝒆𝒖

(𝟏 + 𝒖)𝟐𝒅𝒖 =

𝑰𝑩𝑷

= ∫𝒆𝒖

𝟏 + 𝒖𝒅𝒖 + ∫

𝒆𝒖

𝟏 + 𝒖𝒅𝒖 − ∫

𝒆𝒖

𝟏 + 𝒖𝒅𝒖 =

𝒆𝒖

𝟏 + 𝒖+ 𝑪 =

𝒙

𝟏 + 𝐥𝐨𝐠 𝒙+ 𝑪

𝛀(𝟐, 𝟑) = ∫𝐥𝐨𝐠𝟐 𝒙

(𝟏 + 𝐥𝐨𝐠 𝒙)𝟑𝒅𝒙 =

𝐥𝐨𝐠 𝒙=𝒖∫

𝒖𝟐𝒆𝒖

(𝟏 + 𝒖)𝟑𝒅𝒖 = ∫

𝒖𝒆𝒖

(𝟏 + 𝒖)𝟐𝒅𝒖 −∫

𝒖𝒆𝒖

(𝟏 + 𝒖)𝟑𝒅𝒖

𝛀(𝟐, 𝟑) = 𝛀(𝟏, 𝟐) − ∫𝒖𝒆𝒖

(𝟏 + 𝒖)𝟐𝒅𝒖 + ∫

𝒖𝒆𝒖

(𝟏 + 𝒖)𝟑𝒅𝒖 =

𝑰𝑩𝑷

=𝒙

𝟏 + 𝐥𝐨𝐠𝒙− ∫

𝒖𝒆𝒖

(𝟏 + 𝒖)𝟐𝒅𝒖 −

𝒆𝒖

(𝟏 + 𝒖)𝟐+∫

𝒆𝒖

(𝟏 + 𝒖)𝟐𝒅𝒖 =

=𝒙

𝟏 + 𝐥𝐨𝐠𝒙−

𝒆𝒖

(𝟏 + 𝒖)𝟐=

𝒙

𝟏 + 𝐥𝐨𝐠 𝒙−

𝒙

(𝟏 + 𝐥𝐨𝐠 𝒙)𝟐+ 𝑪

𝛀(𝒎,𝒏) = ∫𝐥𝐨𝐠𝒎 𝒙

(𝟏 + 𝐥𝐨𝐠 𝒙)𝒏𝒅𝒙 ,𝒎, 𝒏 ∈ ℕ

∵𝟏

(𝟏 + 𝒖)𝒏=∑(−𝟏)𝒌−(𝒏−𝟏) ⋅

(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!

(𝒌 − (𝒏 − 𝟏))!⋅ 𝒖𝒌−(𝒏−𝟏)

∞

𝒌=𝟎

, ∀𝒏 ≥ 𝟐

𝐏𝐮𝐭 𝒖 = 𝐥𝐨𝐠𝒙 ⇒𝟏

(𝟏 + 𝐥𝐨𝐠 𝒙)𝒏=∑(−𝟏)𝒌−(𝒏−𝟏) ⋅

(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!

(𝒌 − (𝒏 − 𝟏))!⋅ (𝐥𝐨𝐠𝒙)𝒌−(𝒏−𝟏)

∞

𝒌=𝟎

𝛀(𝒎, 𝒏) = ∑(−𝟏)𝒌−(𝒏−𝟏) ⋅(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!

(𝒌 − (𝒏 − 𝟏))!⋅ ∫(𝐥𝐨𝐠 𝒙)𝒎+𝒌−(𝒏−𝟏) 𝒅𝒙

∞

𝒌=𝟎

=

= ∑(−𝟏)𝒌−(𝒏−𝟏) ⋅(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!

(𝒌 − (𝒏 − 𝟏))!⋅𝝏𝒎+𝒌−(𝒏−𝟏)

𝝏𝒂𝒎+𝒌−(𝒏−𝟏)|𝒂=𝟎

∞

𝒌=𝟎

∫𝒙𝒂 𝒅𝒙 =

=∑(−𝟏)𝒌−(𝒏−𝟏) ⋅(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!

(𝒌 − (𝒏 − 𝟏))!⋅𝝏𝒎+𝒌−(𝒏−𝟏)

𝝏𝒂𝒎+𝒌−(𝒏−𝟏)|𝒂=𝟎

⋅𝒙𝒂+𝟏

𝒂 + 𝟏+ 𝑪

∞

𝒌=𝟎

1546. Find:

𝛀 = ∫ 𝐬𝐢𝐧−𝟏 (𝟏

𝟏 + 𝒙𝟐)𝒅𝒙

∞

𝟎

Proposed by Ajetunmobi Abdulqoyyum-Nigeria

www.ssmrmh.ro



𝛀 = ∫ 𝐬𝐢𝐧−𝟏 (𝟏

𝟏 + 𝒙𝟐)𝒅𝒙

∞

𝟎

=𝑰𝑩𝑷𝒙 ⋅ 𝐬𝐢𝐧−𝟏 (

𝟏

𝟏 + 𝒙𝟐) + 𝟐∫

𝒙

(𝟏 + 𝒙𝟐)√𝒙𝟐 + 𝟐𝒅𝒙

∞

𝟎

=𝒖=√𝒙𝟐+𝟐

= 𝒙 ⋅ 𝐬𝐢𝐧−𝟏 (𝟏

𝟏 + 𝒙𝟐) + 𝟐∫

𝟏

𝒖𝟐 − 𝟏𝒅𝒖

∞

𝟏

=

= 𝒙 ⋅ 𝐬𝐢𝐧−𝟏 (𝟏

𝟏 + 𝒙𝟐) + 𝟐∫ (

𝟏

𝟐(𝒖 − 𝟏)−

𝟏

𝟐(𝒖 + 𝟏))𝒅𝒖

∞

𝟏

=

= [𝒙 ⋅ 𝐬𝐢𝐧−𝟏 (𝟏

𝟏 + 𝒙𝟐) + 𝐥𝐨𝐠 (

√𝒙𝟐 + 𝟐 − 𝟏

√𝒙𝟐 + 𝟐 + 𝟏)]𝟎

∞

= − 𝐥𝐨𝐠 (√𝟐− 𝟏

√𝟐+ 𝟏)

𝛀 = ∫ 𝐬𝐢𝐧−𝟏 (𝟏

𝟏 + 𝒙𝟐)𝒅𝒙

∞

𝟎

= 𝐥𝐨𝐠(√𝟐 + 𝟏) − 𝐥𝐨𝐠(√𝟐 − 𝟏)

Solution 2 by Abdul Mukhtar-Nigeria

𝛀 = ∫ 𝐬𝐢𝐧−𝟏 (𝟏

𝟏 + 𝒙𝟐)𝒅𝒙

∞

𝟎

= ∫ 𝐜𝐬𝐜−𝟏(𝟏 + 𝒙𝟐) 𝒅𝒙∞

𝟎

=𝑰𝑩𝑷

= [𝒙 ⋅ 𝐜𝐬𝐜−𝟏(𝟏 + 𝒙𝟐)]𝟎∞ + 𝟐∫

𝒙

(𝟏 + 𝒙𝟐)√𝒙𝟐 + 𝟐𝒅𝒙

∞

𝟎

𝛀 = 𝟐∫𝒙

(𝟏 + 𝒙𝟐)√𝒙𝟐 + 𝟐𝒅𝒙

∞

𝟎

=𝒚=√𝒙𝟐+𝟐

∫𝟐𝒚

(𝒚𝟐 − 𝟏)𝒚𝒅𝒚

∞

√𝟐

=𝟏

𝟐∫

𝟐(𝒚 + 𝟏) − 𝟐(𝒚 − 𝟏)

𝒚𝟐 − 𝟏𝒅𝒚

∞

√𝟐

= [𝐥𝐨𝐠 |𝒚 − 𝟏

𝒚 + 𝟏|]√𝟐

∞

= 𝐥𝐨𝐠√𝟐 + 𝟏

√𝟐 − 𝟏

1547. Prove that:

∫ 𝒆−𝒙∏(𝟏 − 𝒆−𝟔𝒙𝒌)(𝟏 + 𝒆−𝟔𝒙𝒌+𝒙)(𝟏 + 𝒆−𝟔𝒙𝒌+𝟓𝒙)𝒅𝒙

∞

𝒌=𝟏

∞

𝟎

=𝝅

√𝟐⋅

𝐬𝐢𝐧𝐡 (𝟐𝝅√𝟐𝟑 )

𝐜𝐨𝐬𝐡(𝟐𝝅√𝟐𝟑 ) − 𝐜𝐨𝐬 (

𝟐𝝅𝟑 )

Proposed by Syed Shahabudeen-India

www.ssmrmh.ro


Solution by Kaushik Mahanta-Assam-India

Recall definition of Jacobi’s triple product identity:

∑ 𝒒𝒌𝟐𝒛𝒌

∞

𝒌=−∞

=∏(𝟏 − 𝒒𝟐𝒌)(𝟏 − 𝒛 − 𝟏 𝒒𝟐𝒌−𝟏)(𝟏 + 𝒛𝒒𝟐𝒌−𝟏)

∞

𝒌=𝟏

Comparing, we get:

𝒒𝟐𝒌 = 𝒆−𝟔𝒙𝒌, 𝒒 = 𝒆−𝟑𝒙; 𝒛𝒒𝟐𝒌−𝟏 = 𝒆𝒙𝒆−𝟔𝒙𝒌 ⇒𝒛 ⋅ 𝒆−𝟔𝒙𝒌

𝒆−𝟑𝒙= 𝒆−𝟔𝒙𝒌 ⋅ 𝒆𝒙 ⇒ 𝒛 = 𝒆−𝟐𝒙

𝑰 = ∫ 𝒆−𝒙 ∑ (𝒆−𝟑𝒙)𝒌𝟐(𝒆−𝟐𝒙)𝒌𝒅𝒙

∞

𝒌=−∞

∞

𝟎

= ∑ ∫ 𝒆−𝒙(𝟏+𝟐𝒌+𝟑𝒌𝟐)𝒅𝒙

∞

𝟎

∞

𝒌=−∞

=∑𝟏

𝟑𝒌𝟐 + 𝟐𝒌+ 𝟏

∞

𝒌=𝟏

∑𝟏

𝒂𝒌𝟐 + 𝒃𝒌 + 𝒄=

∞

𝒌=−∞

𝟐𝝅

√𝚫⋅

𝐬𝐢𝐧 (𝝅√𝚫𝒂 )

𝐜𝐨𝐬 (𝝅√𝚫𝒂 ) − 𝒄𝒐𝒔 (

𝝅𝒃𝒂 )

For 𝒂 = 𝟑, 𝒃 = 𝟐, 𝒄 = 𝟏,𝚫 = −𝟖 ⇒

∑𝟏

𝟑𝒌𝟐 + 𝟐𝒌 + 𝟏

∞

𝒌=𝟏

=𝟐𝝅

𝒊𝟐√𝟐⋅

𝐬𝐢𝐧 (𝒊𝟐√𝟐𝝅𝟑 )

𝐜𝐨𝐬 (𝒊𝟐√𝟐𝝅𝟑 ) − 𝐜𝐨𝐬 (

𝟐𝒊𝟑 )

=𝝅

√𝟐⋅

𝐬𝐢𝐧𝐡 (𝟐𝝅√𝟐𝟑 )

𝐜𝐨𝐬𝐡(𝟐𝝅√𝟐𝟑 ) − 𝐜𝐨𝐬 (

𝟐𝝅𝟑 )


𝛀 = ∫ ∫𝒙

√𝒙𝟐 + 𝒚𝟐

√𝟐−𝒙𝟐

𝒙

𝒅𝒙𝟏

𝟎

𝒅𝒚

Proposed by Durmuş Ogmen-Turkiye


Let 𝒙 = 𝒓 𝐬𝐢𝐧𝜽 , 𝒚 = 𝒓 𝐬𝐢𝐧𝜽, then 𝒙𝟐 + 𝒚𝟐 = 𝒓𝟐, 𝒅𝒙𝒅𝒚 = 𝒓 𝒅𝒓𝒅𝜽 and

𝑹 = {(𝒙, 𝒚)|𝒙 ≤ 𝒚 ≤ √𝟐 − 𝒙𝟐, 𝟎 ≤ 𝒙 ≤ 𝟏} = {(𝒓, 𝜽)|𝟎 ≤ 𝒓 ≤ √𝟐,𝝅𝟒 ≤ 𝜽 ≤

𝝅𝟐}

𝛀 = ∫ ∫𝒙

√𝒙𝟐 + 𝒚𝟐

√𝟐−𝒙𝟐

𝒙

𝒅𝒙𝟏

𝟎

𝒅𝒚 = ∫ ∫𝒓𝐜𝐨𝐬𝜽

𝒓⋅ 𝒓 𝒅𝒓𝒅𝜽

√𝟐

𝟎

𝝅𝟐

𝝅𝟒

=

www.ssmrmh.ro


= (∫ 𝐜𝐨𝐬𝜽𝒅𝜽

𝝅𝟐

𝝅𝟒

)(∫ 𝒓√𝟐

𝟎

𝒅𝒓) = 𝐬𝐢𝐧𝜽|𝝅𝟒

𝝅𝟐 ⋅𝟏

𝟐𝒓𝟐|

𝟎

√𝟐

=𝟐 − √𝟐

𝟐


𝛀 = ∫ ∫𝒙

√𝒙𝟐 + 𝒚𝟐

√𝟐−𝒙𝟐

𝒙

𝒅𝒙𝟏

𝟎

𝒅𝒚

= ∫ ∫𝒙

√𝒙𝟐 + 𝒚𝟐𝒅𝒙𝒅𝒚

√𝟐

𝟎

𝟏

𝟎

+∫ ∫𝒙

√𝒙𝟐 + 𝒚𝟐

√𝟐−𝒚𝟐

𝟎

𝒅𝒙√𝟐

𝟏

𝒅𝒚

= ∫ (√𝒙𝟐 + 𝒚𝟐)𝟏

𝟎

|𝟎

𝒚

𝒅𝒚 +∫ (√𝒙𝟐 + 𝒚𝟐)√𝟐

𝟏

|

𝟎

√𝟐−𝒚𝟐

𝒅𝒚 =

= ∫ (√𝟐− 𝟏)𝒚𝟏

𝟎

𝒅𝒚 +∫ (√𝟐 − 𝒚)√𝟐

𝟏

𝒅𝒚 = (√𝟐 − 𝟏)𝒚𝟐

𝟐|𝟎

𝟏

+ (√𝟐𝒚 −𝒚𝟐

𝟐)|𝟏

√𝟐

=𝟐 − √𝟐

𝟐

Solution 3 by Katrick Chandra Betal-India

𝛀 = ∫ ∫𝒙

√𝒙𝟐 + 𝒚𝟐

√𝟐−𝒙𝟐

𝒙

𝒅𝒙𝟏

𝟎

𝒅𝒚 = ∫ 𝒙 [𝐥𝐨𝐠 (𝒚 + √𝒙𝟐 + 𝒚𝟐)]𝒙

√𝟐−𝒙𝟐𝟏

𝟎

𝒅𝒙 =

= ∫ 𝒙 𝐥𝐨𝐠(√𝟐 − 𝒙𝟐 + √𝟐

𝒙 + √𝟐𝒙)𝒅𝒙 =

𝟏

𝟎

= ∫ 𝒙 𝐥𝐨𝐠 (√𝟐 + √𝟐 − 𝒙𝟐)𝒅𝒙𝟏

𝟎

− 𝐥𝐨𝐠(𝟏 + √𝟐 )∫ 𝒙𝟏

𝟎

𝒅𝒙 = ∫ 𝒙 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

=

=𝟏

𝟐∫ 𝐥𝐨𝐠(√𝟐 + √𝟐 − 𝒙)𝒅𝒙𝟏

𝟎

−𝟏

𝟐𝐥𝐨𝐠(𝟏 + √𝟐) − [

𝒙𝟐

𝟐𝐥𝐨𝐠𝒙 −

𝒙𝟐

𝟒]𝟎

𝟏

=

=𝟏

𝟐∫ 𝐥𝐨𝐠(√𝟐 + √𝒙 + 𝟏)𝒅𝒙𝟏

𝟎

−𝟏

𝟐𝐥𝐨𝐠(𝟏 + √𝟐) +

𝟏

𝟒=

=𝟏

𝟒−𝟏

𝟐𝐥𝐨𝐠(𝟏 + √𝟐) +

𝟏

𝟐[𝒙 𝐥𝐨𝐠(√𝟐 + √𝟏 + 𝒙)]

𝟎

𝟏−𝟏

𝟐∫

𝒙

√𝟐+ √𝟏 + 𝒙⋅

𝒅𝒙

𝟐√𝟏 + 𝒙

𝟏

𝟎

=

=𝟏

𝟒−𝟏

𝟐𝐥𝐨𝐠(𝟏 + √𝟐) +

𝟏

𝟐𝐥𝐨𝐠(𝟐√𝟐) −

𝟏

𝟒∫(𝒙 − 𝟏)(√𝒙 − √𝟐)

(𝒙 − 𝟐)√𝒙

𝟏

𝟎

𝒅𝒙 =

=𝟏

𝟒−𝟏

𝟐𝐥𝐨𝐠 (

𝟏 + √𝟐

𝟐√𝟐) −

𝟏

𝟐∫ (𝒙 − √𝟐)𝒅𝒙√𝟐

𝟏

−𝟏

𝟐[𝐥𝐨𝐠(𝒙 + √𝟐)]

𝟏

√𝟐=

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=𝟏

𝟒−𝟏

𝟐𝐥𝐨𝐠 (

𝟏 + √𝟐

𝟐√𝟐) −

𝟏

𝟐[𝒙𝟐

𝟐− √𝟐𝒙]

𝟏

√𝟐

− 𝐥𝐨𝐠 (𝟐√𝟐

𝟏+ √𝟐) =

=𝟏

𝟒−𝟏

𝟒+𝟐 − √𝟐

𝟐= 𝟏 −

𝟏

√𝟐


𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐

(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙𝒙 + 𝟑𝒙𝒆𝒙 + 𝒙)𝒅𝒙


Solution 1 by Hussain Reza Zadah-Afghanistan

𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐

(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙𝒙 + 𝟑𝒙𝒆𝒙 + 𝒙)𝒅𝒙 =

= ∫(𝟑𝒙𝒆𝒙 + 𝟐)𝒅𝒙

𝒙(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏)= ∫

𝟑𝒆𝒙 +𝟐𝒙

(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙)𝟐 − 𝟏𝒅𝒙 =

𝒖=𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙

= ∫𝒅𝒖

𝒖𝟐 − 𝟏=𝟏

𝟐𝐥𝐨𝐠 |

𝒖 − 𝟏

𝒖 + 𝟏| + 𝑪 =

𝟏

𝟐𝐥𝐨𝐠 |

𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏

𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏| + 𝑪


𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐


= ∫𝟑𝒙𝒆𝒙 + 𝟐

𝒙(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =

= ∫𝟑𝒆𝒙 +

𝟐𝒙

(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =

𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒕

=𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒕

∫𝟏

(𝒕 − 𝟏)(𝒕 + 𝟏)𝒅𝒕 =

𝟏

𝟐∫𝒅𝒕

𝒕 − 𝟏−𝟏

𝟐∫𝒅𝒕

𝒕 + 𝟏=

=𝟏

𝟐𝐥𝐨𝐠 |

𝒕 − 𝟏

𝒕 + 𝟏| + 𝑪 =

𝟏

𝟐𝐥𝐨𝐠 |

𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏

𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏| + 𝑪

Solution 3 by Timson Azeez Folorunsho-Nigeria

𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐


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= ∫𝟑𝒙𝒆𝒙 + 𝟐


= ∫𝟑𝒆𝒙 +

𝟐𝒙

(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =

𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒕

=𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒕

∫𝟏

(𝒕 − 𝟏)(𝒕 + 𝟏)𝒅𝒕 =

𝟏

𝟐∫𝒅𝒕

𝒕 − 𝟏−𝟏

𝟐∫𝒅𝒕

𝒕 + 𝟏=

=𝟏

𝟐𝐥𝐨𝐠 |

𝒕 − 𝟏

𝒕 + 𝟏| + 𝑪 =

𝟏

𝟐𝐥𝐨𝐠 |

𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏

𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏| + 𝑪


𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐


= ∫𝟑𝒙𝒆𝒙 + 𝟐


= ∫𝟑𝒆𝒙 +

𝟐𝒙

(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =

𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒚

=𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒚

∫𝟏

(𝒚 − 𝟏)(𝒚 + 𝟏)𝒅𝒕 =

𝟏

𝟐∫

𝒅𝒕

𝒚 − 𝟏−𝟏

𝟐∫

𝒅𝒕

𝒚 + 𝟏=

=𝟏

𝟐𝐥𝐨𝐠 |

𝒚 − 𝟏

𝒚 + 𝟏| + 𝑪 =

𝟏

𝟐𝐥𝐨𝐠 |

𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏

𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 + 𝟏| + 𝑪

1550. Prove that:

∫𝒅𝒙

𝟏 + √𝟏 + 𝐬𝐢𝐧𝟐 𝒙

𝝅𝟐

𝟎

=𝝅

𝟐⋅𝚪𝟒 (

𝟏𝟒) − 𝟖𝝅𝟐

(𝟐𝝅)𝟑𝟐𝚪𝟐 (

𝟏𝟒)=𝝅

𝟐(𝑮 −

𝟏

𝝅𝑮)

where 𝑮 =𝚪𝟐(

𝟏

𝟒)

(𝟐𝝅)𝟑𝟐

denotes Gauss Constant.


Solution by Kartick Chandra Betal-India

∫𝒅𝒙

𝟏 + √𝟏 + 𝐬𝐢𝐧𝟐 𝒙

𝝅𝟐

𝟎

= ∫√𝟏+ 𝐬𝐢𝐧𝟐 𝒙 − 𝟏

𝐬𝐢𝐧𝟐 𝒙𝒅𝒙

𝝅𝟐

𝟎

=

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= −(√𝟏 + 𝐬𝐢𝐧𝟐 𝒙 − 𝟏) 𝐜𝐨𝐭 𝒙|𝟎

𝝅𝟐+∫

𝟐 𝐬𝐢𝐧𝒙 𝐜𝐨𝐬 𝒙

𝟐√𝟏 + 𝐬𝐢𝐧𝟐 𝒙𝐜𝐨𝐭 𝒙

𝝅𝟐

𝟎

𝒅𝒙 =


[(√𝟏 + 𝐬𝐢𝐧𝟐 𝒙 − 𝟏)𝐜𝐨𝐭 𝒙] + ∫𝐜𝐨𝐬𝟐 𝒙

√𝟏+ 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙

𝝅𝟐

𝟎

=


𝐬𝐢𝐧𝒙 𝐜𝐨𝐬 𝒙

𝟏 + √𝟏 + 𝐬𝐢𝐧𝟐 𝒙+ ∫

𝟏 − 𝒙

√(𝟏 + 𝒙𝟐)(𝟏 − 𝒙𝟐)

𝟏

𝟎

𝒅𝒙 =

= −∫𝒙𝟐

√𝟏− 𝒙𝟒

𝟏

𝟎

𝒅𝒙 + ∫𝒅𝒙

√𝟏 − 𝒙𝟒

𝟏

𝟎

= −𝟏

𝟒∫𝒙𝟏𝟐+𝟏𝟒−𝟏

√𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 +𝟏

𝟒∫

𝒙𝟏𝟒−𝟏

√𝟏 − 𝒙

𝟏

𝟎

𝒅𝒙 =

= −𝟏

𝟒∫ 𝒙

𝟑𝟒−𝟏(𝟏 − 𝒙)

𝟏𝟐−𝟏

𝟏

𝟎

𝒅𝒙 +𝟏

𝟒∫ 𝒙

𝟏𝟒−𝟏(𝟏 − 𝒙)

𝟏𝟐−𝟏

𝟏

𝟎

𝒅𝒙 =

= −𝟏

𝟒⋅𝚪 (𝟑𝟒)𝚪(

𝟏𝟐)

𝚪 (𝟓𝟒)

+𝟏

𝟒⋅𝚪 (𝟏𝟒)𝚪 (

𝟏𝟐)

𝚪 (𝟑𝟒)

=

= −√𝝅

𝟒⋅√𝟐𝝅

𝟏𝟒𝚪𝟐 (

𝟏𝟒)+√𝝅

𝟒⋅𝚪𝟐 (

𝟏𝟒)

𝝅√𝟐=𝚪𝟒 (

𝟏𝟒) − 𝟖𝝅

𝟐

𝟒√𝟐𝝅 ⋅ 𝚪𝟐 (𝟏𝟒)=𝝅

𝟐⋅𝚪𝟒 (

𝟏𝟒) − 𝟖𝝅

𝟐

(𝟐𝝅)𝟑𝟐 ⋅ 𝚪𝟐 (

𝟏𝟒)=

=𝝅

𝟐⋅

{

𝚪𝟐 (𝟏𝟒)

(𝟐𝝅)𝟑𝟐

−𝟏

𝚪𝟐 (𝟏𝟒)

(𝟐𝝅)𝟑𝟐

⋅\𝒑𝒊}

=𝝅

𝟐(𝑮 −

𝟏

𝝅𝑮)

1551. Find:

𝛀 = ∫ 𝒙𝟐 ⋅ 𝐬𝐢𝐧−𝟏 𝒙 ⋅ 𝐬𝐢𝐧−𝟏(𝟒𝒙𝟑 − 𝟑𝒙)𝒅𝒙𝟏

𝟎

Proposed by Ty Halpen-Florida-USA

Solution by Asmat Qatea-Afghanistan

𝛀 = ∫ 𝒙𝟐 ⋅ 𝐬𝐢𝐧−𝟏 𝒙 ⋅ 𝐬𝐢𝐧−𝟏(𝟒𝒙𝟑 − 𝟑𝒙)𝒅𝒙𝟏

𝟎

=𝒙=𝐬𝐢𝐧𝒖

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= −∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ 𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧(𝟑𝒖)) 𝐜𝐨𝐬𝒖𝒅𝒖

𝝅𝟐

𝟎

=

= −∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ 𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧(𝟑𝒖)) 𝐜𝐨𝐬𝒖𝒅𝒖

𝝅𝟔

𝟎

−∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ 𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧(𝟑𝒖)) 𝐜𝐨𝐬𝒖𝒅𝒖

𝝅𝟐

𝝅𝟔

= −∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ (𝟑𝒖) 𝐜𝐨𝐬𝒖𝒅𝒖

𝝅𝟔

𝟎

–∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ (−𝟑𝒖 + 𝝅) 𝐜𝐨𝐬𝒖𝒅𝒖

𝝅𝟐

𝝅𝟔

=

= −𝟑∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖𝟐 ⋅ 𝐜𝐨𝐬𝒖𝒅𝒖

𝝅𝟔

𝟎⏟ 𝑰𝟏

+ 𝟑∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖𝟐 ⋅ 𝐜𝐨𝐬𝒖𝒅𝒖

𝝅𝟐

𝝅𝟔⏟

𝑰𝟐

− 𝝅∫ 𝒖 ⋅ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝐜𝐨𝐬 𝒖

𝝅𝟐

𝝅𝟔

𝒅𝒖⏟

𝑰𝟑

∵ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝐜𝐨𝐬 𝒖 =𝟏

𝟒(𝐜𝐨𝐬𝒖 − 𝐜𝐨𝐬(𝟑𝒖)) ⇒ ∫𝐬𝐢𝐧𝟐 𝒖 𝐜𝐨𝐬𝒖𝒅𝒖

=𝟏

𝟒∫(𝐜𝐨𝐬𝒖 − 𝐜𝐨𝐬(𝟑𝒖))𝒅𝒖

𝑰𝟏 = −𝟑

𝟒[𝒖𝟐 (𝐬𝐢𝐧𝒖 −

𝐬𝐢𝐧(𝟑𝒖)

𝟑) − 𝟐𝒖(−𝐜𝐨𝐬 𝒖 +

𝐜𝐨𝐬(𝟑𝒖)

𝟗) + 𝟐(−𝐬𝐢𝐧 𝒖 +


𝟐𝟕)]𝟎

𝝅𝟔

=

= −𝝅𝟐

𝟐𝟖𝟖−√𝟑𝝅

𝟖+𝟐𝟓

𝟑𝟔

𝑰𝟐 = −𝟑

𝟒[𝒖𝟐 (𝐬𝐢𝐧𝒖 −


𝟑) − 𝟐𝒖(−𝐜𝐨𝐬 𝒖 +


𝟗) + 𝟐(−𝐬𝐢𝐧 𝒖 +


𝟐𝟕)]𝝅𝟔

𝝅𝟐

=𝟕𝟏𝝅𝟐

𝟐𝟖𝟖−√𝟑𝝅

𝟖−𝟑𝟏

𝟑𝟔

𝑰𝟏 + 𝑰𝟐 =𝟕𝟎𝝅𝟐

𝟐𝟖𝟖−√𝟑𝝅

𝟒−𝟏

𝟔

𝑰𝟑 = −𝝅∫ 𝒖 ⋅ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝐜𝐨𝐬𝒖

𝝅𝟐

𝝅𝟔

𝒅𝒖 = −𝝅

𝟒∫ 𝒖(𝐜𝐨𝐬 𝒖 − 𝐜𝐨𝐬(𝟑𝒖))

𝝅𝟐

𝝅𝟔

𝒅𝒖 =

= −𝝅

𝟒[𝒖 (𝐬𝐢𝐧𝒖 −


𝟑)—𝐜𝐨𝐬𝒖 +


𝟗]𝝅𝟔

𝝅𝟐

= −𝟐𝟑𝝅𝟐

𝟏𝟒𝟒+√𝟑𝝅

𝟖

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𝛀 = 𝑰𝟏 + 𝑰𝟐 + 𝑰𝟑 =𝝅𝟐

𝟏𝟐−√𝟑𝝅

𝟖−𝟏

𝟔

1552. Prove that:

∫ 𝐥𝐨𝐠(𝒙𝟐 + 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)) 𝒅𝒙

𝝅𝟐

𝟎

= 𝝅 𝐥𝐨𝐠(𝐥𝐨𝐠 𝟐)


Solution by Luca Paes Barreto-Pernambuco-Brazil

It is well-known that:

∫𝐜𝐨𝐬 (𝒔 𝐭𝐚𝐧−𝟏 (

𝒙− 𝐥𝐨𝐠 𝐜𝐨𝐬 𝒙))

(𝒙𝟐 + 𝐥𝐨𝐠𝟐 𝐜𝐨𝐬 𝒙)𝒔𝟐

𝝅𝟐

𝟎

𝒅𝒙 =𝝅

𝟐⋅𝟏

𝐥𝐨𝐠𝒔 𝟐

Differentiable both sides w.r.t. 𝒔, we have:

𝝏𝒔(𝐜𝐨𝐬 (𝒔 𝐭𝐚𝐧−𝟏 (

𝒙− 𝐥𝐨𝐠 𝐜𝐨𝐬 𝒙))

(𝒙𝟐 + 𝐥𝐨𝐠𝟐 𝐜𝐨𝐬 𝒙)𝒔𝟐

)|

𝒔=𝟎

= −𝟏

𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝐥𝐨𝐠𝟐 𝐜𝐨𝐬 𝒙)

𝝏𝒔 (𝝅

𝟐⋅𝟏

𝐥𝐨𝐠𝒔 𝟐)|𝒔=𝟎

= −𝝅

𝟐⋅ 𝐥𝐨𝐠(𝐥𝐨𝐠𝟐) ⇒

−𝟏

𝟐∫ 𝐥𝐨𝐠(𝒙𝟐 + 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)) 𝒅𝒙

𝝅𝟐

𝟎

= −𝝅

𝟐⋅ 𝐥𝐨𝐠(𝐥𝐨𝐠𝟐)

Therefore,

∫ 𝐥𝐨𝐠(𝒙𝟐 + 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)) 𝒅𝒙

𝝅𝟐

𝟎

= 𝝅 𝐥𝐨𝐠(𝐥𝐨𝐠𝟐)

1553. Find:

𝛀 = ∫𝑳𝒊𝟑(𝒙)

𝟏 − 𝒙𝐥𝐨𝐠 𝒙

𝟏

𝟎

𝒅𝒙



We know that:

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𝑳𝒊𝟑(𝒙)

𝟏 − 𝒙= ∑𝑯𝒏

(𝟑)𝒙𝒏∞

𝒏=𝟏

⇒ 𝛀 = ∫𝑳𝒊𝟑(𝒙)


𝟏

𝟎

𝒅𝒙 = ∑𝑯𝒏(𝟑)∫ 𝒙𝒏 𝐥𝐨𝐠(𝒙)𝒅𝒙

𝟏

𝟎

∞

𝒏=𝟏

∵ ∫ 𝒙𝒏 𝐥𝐨𝐠𝒂(𝒙) 𝒅𝒙𝟏

𝟎

=(−𝟏)𝒂𝒂!

(𝒏 + 𝟏)𝒂+𝟏; 𝒇𝒐𝒓 𝒂 = 𝟏 ⇒

∫ 𝒙𝒏 𝐥𝐨𝐠(𝒙)𝒅𝒙𝟏

𝟎

=(−𝟏)𝟏𝟏!

(𝒏 + 𝟏)𝟏+𝟏= −

𝟏

(𝒏 + 𝟏)𝟐⇒∑𝑯𝒏

(𝟑)∫ 𝒙𝒏 𝐥𝐨𝐠(𝒙)𝒅𝒙𝟏

𝟎

∞

𝒏=𝟏

=

= ∑𝑯𝒏(𝟑) (−𝟏)

(𝒏 + 𝟏)𝟐

∞

𝒏=𝟏

= −∑𝑯𝒏(𝟑)

(𝒏 + 𝟏)𝟐

∞

𝒏=𝟏

= −∑𝑯𝒏−𝟏(𝟑)

𝒏𝟐

∞

𝒏=𝟏

We know that: 𝑯𝒏−𝟏(𝟑) = 𝑯𝒏

(𝟑) −𝟏

𝒏𝟑.

−∑𝑯𝒏−𝟏(𝟑)

𝒏𝟐

∞

𝒏=𝟏

= −∑𝟏

𝒏𝟐(𝑯𝒏

(𝟑)−𝟏

𝒏𝟑)

∞

𝒏=𝟏

= −∑𝑯𝒏(𝟑)

𝒏𝟐

∞

𝒏=𝟏

+∑𝟏

𝒏𝟓

∞

𝒏=𝟏

=

= −𝟏𝟏

𝟐𝜻(𝟓) + 𝟐𝜻(𝟐)𝜻(𝟑) + 𝜻(𝟓)


𝛀 = ∫𝑳𝒊𝟑(𝒙)

𝟏 − 𝒙𝐥𝐨𝐠𝒙

𝟏

𝟎

𝒅𝒙 =𝝏

𝝏𝒂∫𝒙𝒂𝑳𝒊𝟑(𝒙)


𝟏

𝟎

=𝝏

𝝏𝒂∫ 𝑳𝒊𝟑(𝒙)∑𝒙𝒌

∞

𝒌=𝟎

𝒅𝒙𝟏

𝟎

=

=𝝏

𝝏𝒂∑∫ 𝒙𝒂+𝒌𝑳𝒊𝟑(𝒙)

𝟏

𝟎

𝒅𝒙

∞

𝒌=𝟎

=

= 𝐥𝐢𝐦𝒂→𝟎

𝝏

𝝏𝒂∑(

𝜻(𝟑)

𝒂 + 𝒌 + 𝟏−

𝜻(𝟐)

(𝒂 + 𝒌 + 𝟏)𝟐+

𝑯𝒂+𝒌+𝟏(𝒂 + 𝒌 + 𝟏)𝟑

)

∞

𝒌=𝟎

=

= 𝐥𝐢𝐦𝒂→𝟎

∑ (−𝜻(𝟑)

(𝒂 +𝒎)𝟐+

𝟐𝜻(𝟐)

(𝒂 +𝒎)𝟑+𝝏

𝝏𝒂

𝑯𝒂+𝒎(𝒂 +𝒎)𝟑

)

∞

𝒎=𝟎

=

= ∑ (−𝜻(𝟑)

𝒎𝟐+𝟐𝜻(𝟐)

𝒎𝟑)

∞

𝒎=𝟎

+ ∑ (𝝏

𝝏𝒂


)

∞

𝒎=𝟏

=

∵𝝏

𝝏𝒂

𝑯𝒂+𝒎(𝒂 + 𝒎)𝟑

= −𝟑𝑯𝒂+𝒎(𝒂 +𝒎)𝟒

+𝟏

(𝒂 +𝒎)𝟑(𝜻(𝟐) − 𝑯𝒂+𝒎

𝟐 )

⇒ 𝐥𝐢𝐦𝒂→𝟎

𝝏

𝝏𝒂


= −𝟑𝑯𝒂+𝒎(𝒂 +𝒎)𝟒

+𝟏

(𝒂 +𝒎)𝟑(𝜻(𝟐) − 𝑯𝒂+𝒎

𝟐 )

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⇒ 𝛀 = ∑ (−𝜻(𝟑)

𝒎𝟐+𝟐𝜻(𝟐)

𝒎𝟑)

∞

𝒎=𝟏

+ ∑ (−𝟑𝑯𝒂+𝒎(𝒂 +𝒎)𝟒

+𝟏

(𝒂 +𝒎)𝟑(𝜻(𝟐) − 𝑯𝒂+𝒎

𝟐 ))

∞

𝒎=𝟏

=

= 𝜻(𝟐)𝜻(𝟑) + ∑ (−𝟑𝑯𝒂+𝒎(𝒂 +𝒎)𝟒

+𝟏

(𝒂 +𝒎)𝟑(𝜻(𝟐) − 𝑯𝒂+𝒎

𝟐 ))

∞

𝒎=𝟏

∵ ∑𝑯𝒎𝒎𝟒

∞

𝒎=𝟏

= 𝟑𝜻(𝟓) − 𝜻(𝟐)𝜻(𝟑) 𝒂𝒏𝒅 ∑𝑯𝒎𝟐

𝒎𝟑

∞

𝒎=𝟏

= 𝟑𝜻(𝟐)𝜻(𝟑) −𝟗

𝟐𝜻(𝟓)

Therefore,

𝛀 = 𝟐𝜻(𝟐)𝜻(𝟑) + (−𝟗𝜻(𝟓) + 𝟒𝜻(𝟐)𝜻(𝟑) − 𝟑𝜻(𝟐)𝜻(𝟑) +𝟗

𝟐𝜻(𝟓)) = 𝟐𝜻(𝟐)𝜻(𝟑) −

𝟗

𝟐𝜻(𝟓)

𝛀 = ∫𝑳𝒊𝟑(𝒙)


𝟏

𝟎

𝒅𝒙 = 𝟐𝜻(𝟐)𝜻(𝟑) −𝟗

𝟐𝜻(𝟓)𝜻(𝟑)

1554. Find:

𝛀 = ∫ ∫ ∫𝒚 𝐥𝐨𝐠 𝒙

(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)(𝟏 + 𝒛𝟐)

∞

𝟏

∞

𝟏

𝒅𝒙∞

𝟎

𝒅𝒚 𝒅𝒛

Proposed by Probal Chakraborty-India


𝛀 = ∫ ∫ ∫𝒚 𝐥𝐨𝐠𝒙

(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)(𝟏 + 𝒛𝟐)

∞

𝟏

∞

𝟏

𝒅𝒙∞

𝟎

𝒅𝒚 𝒅𝒛 =𝝅

𝟐∫ ∫

𝒚 𝐥𝐨𝐠 𝒙

(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)

∞

𝟏

𝒅𝒙∞

𝟏

𝒅𝒚

=𝝅

𝟐∫ ∫

−𝒚 𝐥𝐨𝐠 𝒙

(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)

𝟏

𝟎

𝒅𝒙𝟏

𝟎

𝒅𝒚 = −𝝅

𝟐∫

𝟏

𝒚(𝟏 + 𝒚𝟐)

𝟏

𝟎

∫𝐥𝐨𝐠 𝒙

(𝟏 +𝒙𝒚)𝟐 𝒅𝒙

𝟏

𝟎

𝒅𝒚 =𝒕=𝒙𝒚

= −𝝅

𝟐∫

𝟏

𝟏 + 𝒚𝟐∫𝐥𝐨𝐠 𝒕 + 𝐥𝐨𝐠 𝒚

(𝟏 + 𝒕)𝟐𝒅𝒕

𝟏𝒚

𝟎

𝒅𝒚𝟏

𝟎

=

= −𝝅

𝟐∫

𝟏

𝟏 + 𝒚𝟐

𝟏

𝟎

∫𝐥𝐨𝐠 𝒕

(𝟏 + 𝒕)𝟐𝒅𝒕

𝟏𝒚

𝟎

𝒅𝒚 −𝝅

𝟐∫

𝐥𝐨𝐠 𝒚

(𝟏 + 𝒚𝟐)

𝟏

𝟎

∫𝟏

𝟏 + 𝒕𝟐

𝟏𝒚

𝟎

𝒅𝒕𝒅𝒚 =

=𝝅

𝟐∫

𝟏

𝟏 + 𝒚𝟐[𝐥𝐨𝐠(𝒕 + 𝟏) −

𝒕 𝐥𝐨𝐠 𝒕

𝒕 + 𝟏]𝟎

𝟏𝒚

𝒅𝒚𝟏

𝟎

+𝝅

𝟐∫

𝐥𝐨𝐠𝒚

𝟏 + 𝒚𝟐[

𝟏

(𝒚 + 𝟏)]𝟎

𝟏𝒚𝒅𝒚

𝟏

𝟎

=

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=𝝅

𝟐∫𝐥𝐨𝐠(𝟏 + 𝒚) − 𝐥𝐨𝐠 𝒚 +

𝐥𝐨𝐠 𝒚𝒚 + 𝟏

𝟏 + 𝒚𝟐

𝟏

𝟎

𝒅𝒚 −𝝅

𝟐∫

𝐥𝐨𝐠 𝒚

(𝟏 + 𝒚)(𝟏 + 𝒚𝟐)𝒅𝒚

𝟏

𝟎

=

=𝝅

𝟐∫𝐥𝐨𝐠(𝟏 + 𝒚)

𝟏 + 𝒚𝟐𝒅𝒚

𝟏

𝟎

−𝝅

𝟐∫

𝐥𝐨𝐠𝒚

𝟏 + 𝒚𝟐𝒅𝒚

𝟏

𝟎

=𝒚=𝐭𝐚𝐧 𝜽

=𝝅

𝟐∫ 𝐥𝐨𝐠(𝟏 + 𝐭𝐚𝐧𝜽)𝒅𝜽

𝝅𝟒

𝟎

−𝝅

𝟐∫ 𝐥𝐨𝐠(𝐭𝐚𝐧𝜽)𝒅𝜽

𝝅𝟒

𝟎⏟ −𝑮

∫ 𝐥𝐨𝐠(𝟏 + 𝐭𝐚𝐧 𝜽) 𝒅𝜽

𝝅𝟒

𝟎

= ∫ 𝐥𝐨𝐠 (√𝟐𝐜𝐨𝐬 (𝝅

𝟒− 𝜽))𝒅𝜽

𝝅𝟒

𝟎

−∫ 𝐥𝐨𝐠(𝐜𝐨𝐬 𝜽)𝒅𝜽

𝝅𝟒

𝟎

=

= ∫ 𝐥𝐨𝐠(√𝟐 𝐜𝐨𝐬 𝜽)𝒅𝜽 −

𝝅𝟒

𝟎

∫ 𝐥𝐨𝐠(𝐜𝐨𝐬 𝜽)𝒅𝜽

𝝅𝟒

𝟎

=𝝅

𝟖𝐥𝐨𝐠 𝟐

Therefore,

𝛀 = ∫ ∫ ∫𝒚 𝐥𝐨𝐠 𝒙

(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)(𝟏 + 𝒛𝟐)

∞

𝟏

∞

𝟏

𝒅𝒙∞

𝟎

𝒅𝒚 𝒅𝒛 =𝝅

𝟐(𝑮 +

𝝅

𝟖𝐥𝐨𝐠 𝟐)

1555. If 𝐬𝐞𝐜𝝅

𝟕< 𝑎 ≤ 𝑏 then find:

𝛀(𝒂, 𝒃) = ∫ (𝐭𝐚𝐧−𝟏(𝒙

𝐬𝐞𝐜𝝅𝟕− 𝒙 𝐭𝐚𝐧

𝝅𝟕

) − 𝐭𝐚𝐧−𝟏 (𝒙 𝐬𝐞𝐜𝝅

𝟕− 𝐭𝐚𝐧

𝝅

𝟕))𝒅𝒙

𝒃

𝒂


Solution 1 by Mohamed Rostami-Afghanistan

𝛀(𝒂, 𝒃) = ∫ (𝐭𝐚𝐧−𝟏 (𝒙

𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧

𝝅𝟕

) − 𝐭𝐚𝐧−𝟏 (𝒙 𝐬𝐞𝐜𝝅


𝝅

𝟕))𝒅𝒙

𝒃

𝒂

=

= ∫ (𝐭𝐚𝐧−𝟏(𝒙 𝐜𝐨𝐬

𝝅𝟕

𝟏 − 𝒙𝐬𝐢𝐧𝝅𝟕

) − 𝐭𝐚𝐧−𝟏 (𝒙 − 𝐬𝐢𝐧

𝝅𝟕

𝐜𝐨𝐬𝝅𝟕

))𝒅𝒙𝒃

𝒂

𝐭𝐚𝐧−𝟏(𝒙 𝐜𝐨𝐬

𝝅𝟕


) = 𝜶 ⇒ 𝐭𝐚𝐧𝜶 =𝒙 𝐜𝐨𝐬

𝝅𝟕

𝟏 − 𝒙 𝐬𝐢𝐧𝝅𝟕

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𝐭𝐚𝐧−𝟏(𝒙 − 𝐬𝐢𝐧

𝝅𝟕


) = 𝜷 ⇒ 𝐭𝐚𝐧𝜷 =𝒙 − 𝐬𝐢𝐧

𝝅𝟕


𝜶 − 𝜷 = 𝜸 ⇒𝐭𝐚𝐧𝜶 − 𝐭𝐚𝐧𝜷

𝟏 + 𝐭𝐚𝐧𝜶 𝐭𝐚𝐧𝜷= 𝐭𝐚𝐧 𝜸 ⇒

𝒙 𝐜𝐨𝐬𝝅𝟕


−𝒙 − 𝐬𝐢𝐧

𝝅𝟕


𝟏 +𝒙 𝐜𝐨𝐬

𝝅𝟕


⋅𝒙 − 𝐬𝐢𝐧

𝝅𝟕


= 𝐭𝐚𝐧𝜸 ⇒

𝐭𝐚𝐧 𝜸 =

𝒙(𝐜𝐨𝐬𝟐𝝅𝟕 − 𝐬𝐢𝐧

𝟐𝝅𝟕) − 𝒙 + 𝒙

𝟐 𝐬𝐢𝐧𝝅𝟕 + 𝐬𝐢𝐧

𝝅𝟕

𝐜𝐨𝐬𝝅𝟕 (𝟏 − 𝒙 𝐬𝐢𝐧

𝝅𝟕)

𝟏 − 𝒙𝐬𝐢𝐧𝝅𝟕 + 𝒙

𝟐 − 𝒙𝐬𝐢𝐧𝝅𝟕


=

=

𝟏


[−𝒙 (𝟏 − 𝐜𝐨𝐬𝟐𝝅𝟕 ) + 𝒙

𝟐 𝐬𝐢𝐧𝝅𝟕 + 𝐬𝐢𝐧

𝝅𝟕]

𝟏 − 𝟐𝒙 𝐬𝐢𝐧𝝅𝟕 + 𝒙

𝟐

𝐭𝐚𝐧𝜸 =𝐭𝐚𝐧

𝝅𝟕 (−𝟐𝒙 𝐬𝐢𝐧

𝝅𝟕 + 𝒙

𝟐 + 𝟏)

𝟏 − 𝟐𝒙 𝐬𝐢𝐧𝝅𝟕 + 𝒙

𝟐= 𝐭𝐚𝐧

𝝅

𝟕⇒ 𝜸 = 𝜶− 𝜷 =

𝝅

𝟕

Therefore,

𝛀(𝒂, 𝒃) = ∫ (𝜶 − 𝜷)𝒃

𝒂

𝒅𝒙 = ∫𝝅

𝟕

𝒃

𝒂

𝒅𝒙 =𝝅

𝟕(𝒃 − 𝒂).


∵ 𝐭𝐚𝐧−𝟏 𝒙 ± 𝐭𝐚𝐧−𝟏 𝒚 = 𝐭𝐚𝐧−𝟏 (𝒙 ± 𝒚

𝟏 ∓ 𝒙𝒚) ⇒

𝐭𝐚𝐧−𝟏 (𝒙


𝝅𝟕

) − 𝐭𝐚𝐧−𝟏 (𝒙𝐬𝐞𝐜𝝅


𝝅

𝟕) = 𝐭𝐚𝐧−𝟏 (

𝒙


𝝅𝟕

) =

=𝝅

𝟐− 𝐭𝐚𝐧−𝟏 (


𝝅𝟕

𝒙)

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𝐭𝐚𝐧−𝟏 (𝒙 𝐬𝐞𝐜𝝅


𝝅

𝟕) + 𝐭𝐚𝐧−𝟏 (


𝝅𝟕

𝒙) =

= 𝐭𝐚𝐧−𝟏

(


𝝅𝟕

𝒙 + 𝒙𝐬𝐞𝐜𝝅𝟕 − 𝐭𝐚𝐧

𝝅𝟕

𝟏 − (𝒙 𝐬𝐞𝐜𝝅𝟕 − 𝐭𝐚𝐧

𝝅𝟕) ⋅


𝝅𝟕

𝒙 )

=

= 𝐭𝐚𝐧−𝟏 (𝐬𝐞𝐜

𝝅𝟕 − 𝒙 𝐭𝐚𝐧

𝝅𝟕 + 𝒙

𝟐 𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧

𝝅𝟕

𝒙 − 𝒙𝐬𝐞𝐜𝟐𝝅𝟕 + 𝐬𝐞𝐜

𝝅𝟕 𝐭𝐚𝐧

𝝅𝟕 + 𝒙

𝟐 𝐬𝐞𝐜𝝅𝟕 𝐭𝐚𝐧

𝝅𝟕 − 𝒙 𝐭𝐚𝐧

𝟐 𝝅𝟕

) =

= 𝐭𝐚𝐧−𝟏 (𝐬𝐞𝐜

𝝅𝟕 − 𝟐𝒙 𝐭𝐚𝐧

𝝅𝟕 + 𝒙

𝟐 𝐬𝐞𝐜𝝅𝟕

𝐬𝐞𝐜𝝅𝟕 𝐭𝐚𝐧

𝝅𝟕 + 𝒙

𝟐 𝐬𝐞𝐜𝝅𝟕 𝐭𝐚𝐧

𝝅𝟕 − 𝟐𝒙 𝐭𝐚𝐧

𝟐 𝝅𝟕

) =(∗)

𝐬𝐞𝐜𝝅

𝟕𝐭𝐚𝐧

𝝅

𝟕+ 𝒙𝟐 𝐬𝐞𝐜

𝝅

𝟕𝐭𝐚𝐧

𝝅

𝟕− 𝟐𝒙 𝐭𝐚𝐧𝟐

𝝅

𝟕= 𝐭𝐚𝐧

𝝅

𝟕(𝐬𝐞𝐜

𝝅

𝟕− 𝟐𝒙 𝐭𝐚𝐧

𝝅

𝟕+ 𝒙𝟐 𝐬𝐞𝐜

𝝅

𝟕)

=(∗)𝐭𝐚𝐧−𝟏(

𝟏

𝐭𝐚𝐧𝝅𝟕

) =𝝅

𝟐−𝝅

𝟕

Therefore,

𝛀(𝒂, 𝒃) =𝝅

𝟕(𝒃 − 𝒂).

1556. If 𝟏 < 𝑎 ≤ 𝑏 then find:

𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛

𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛

𝒃

𝒂

𝒃

𝒂

𝒃

𝒂


Solution 1 by Serlea Kabay-Liberia

𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛


𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

=

= ∫ ∫ ∫ (𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 − 𝝅)𝒃

𝒂

𝒅𝒙𝒅𝒚𝒅𝒛𝒃

𝒂

𝒃

𝒂

=(∗)

(∵ ∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙 = 𝒙 𝐭𝐚𝐧−𝟏 𝒙 − 𝐥𝐨𝐠(√𝟏 + 𝒙𝟐) )

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=(∗)∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝒅𝒚𝒅𝒛

𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

+∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 𝒚𝒅𝒙𝒅𝒚𝒅𝒛𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

+ ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 𝒛𝒅𝒙𝒅𝒚𝒅𝒛𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

−𝝅(𝒃 − 𝒂)𝟐 =

= (𝒃 − 𝒂)𝟐 (∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝒃

𝒂

+∫ 𝐭𝐚𝐧−𝟏 𝒚𝒅𝒚𝒃

𝒂

+∫ 𝐭𝐚𝐧−𝟏 𝒛𝒃

𝒂

𝒅𝒛) − 𝝅(𝒃 − 𝒂)𝟐 =

= 𝟑(𝒃 − 𝒂)𝟐 (𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏

𝟐(𝒂𝟐 + 𝟏

𝒃𝟐 + 𝟏)) − 𝝅(𝒃 − 𝒂)𝟐

Solution 2 by Remus Florin Stanca-Romania

𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛

𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙) = 𝐭𝐚𝐧−𝟏 (

𝒚 + 𝒛 − 𝒙(𝟏 − 𝒚𝒛)

𝟏 − 𝒙(𝒚 + 𝒛) − 𝒚𝒛) =

= 𝐭𝐚𝐧−𝟏 (

𝒚 + 𝒛𝟏 − 𝒚𝒛 + 𝒙

𝟏 + 𝒙𝒚 + 𝒛𝟏 − 𝒚𝒛

) = 𝐭𝐚𝐧−𝟏 (𝒚 + 𝒛

𝟏 − 𝒚𝒛) + 𝐭𝐚𝐧−𝟏 𝒛 − 𝝅 =

= 𝐭𝐚𝐧−𝟏 (𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒚) + 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒛)

𝟏 − 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒚) ⋅ 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒛)) + 𝐭𝐚𝐧−𝟏 𝒙 − 𝝅 =

= 𝐭𝐚𝐧−𝟏(𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒚) + 𝐭𝐚𝐧−𝟏 𝒛) + 𝐭𝐚𝐧−𝟏 𝒙 − 𝝅 =

= 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 − 𝝅

∫ 𝐭𝐚𝐧−𝟏 𝒙𝒃

𝒂

𝒅𝒙 = [𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 −𝟏

𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝟏)]

𝒂

𝒃

=

= 𝒃𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 −𝟏

𝟐𝐥𝐨𝐠(𝒃𝟐 + 𝟏) +

𝟏

𝟐𝐥𝐨𝐠(𝒂𝟐 + 𝟏)

𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛


𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

=

= ∫ ∫ (𝒃 𝐭𝐚𝐧−𝟏 𝒃 −𝟏

𝟐𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +

𝟏

𝟐𝐥𝐨𝐠(𝒂𝟐 + 𝟏) + (𝒃 − 𝒂) 𝐭𝐚𝐧−𝟏 𝒚

𝒃

𝒂

𝒃

𝒂

+ (𝒃 − 𝒂) 𝐭𝐚𝐧−𝟏 𝒛)𝒅𝒚𝒅𝒛 − 𝝅(𝒃 − 𝒂)𝟐 =

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= ∫ (𝒃(𝒃 − 𝒂) 𝐭𝐚𝐧−𝟏 𝒃 −𝟏

𝟐(𝒃 − 𝒂) 𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝒂 𝐭𝐚𝐧−𝟏 𝒂(𝒃 − 𝒂)

𝒃

𝒂

+𝟏

𝟐𝐥𝐨𝐠(𝒂𝟐 + 𝟏)(𝒃 − 𝒂)

+ (𝒃 − 𝒂) (𝒃 𝐭𝐚𝐧−𝟏 𝒃 −𝟏

𝟐𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +

𝟏

𝟐𝐥𝐨𝐠(𝒂𝟐 + 𝟏))

+ (𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒛)𝒅𝒛 − 𝝅(𝒃 − 𝒂)𝟐 =

= 𝟑𝒃(𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒃 −𝟑

𝟐(𝒃 − 𝒂)𝟐 𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝟑𝒂(𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒂

+𝟑

𝟐(𝒃 − 𝒂)𝟐 𝐥𝐨𝐠(𝒂𝟐 + 𝟏) − 𝝅(𝒃 − 𝒂)𝟐

Therefore,

𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛


𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

=

= 𝟑𝒃(𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒃 −𝟑

𝟐(𝒃 − 𝒂)𝟐 𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝟑𝒂(𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒂

+𝟑

𝟐(𝒃 − 𝒂)𝟐 𝐥𝐨𝐠(𝒂𝟐 + 𝟏) − 𝝅(𝒃 − 𝒂)𝟐


𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 =

{

𝐭𝐚𝐧−𝟏 (

𝒙 + 𝒚

𝟏 − 𝒙𝒚) , 𝒙𝒚 < 1

𝝅 + 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚

𝟏 − 𝒙𝒚) , 𝒙 > 0, 𝑦 > 0(𝒙𝒚 > 1)

−𝝅 + 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚

𝟏− 𝒙𝒚) , 𝒙 < 0, 𝑦 < 0(𝑥𝑦 > 1)

𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 = 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚

𝟏− 𝒙𝒚) ⇒ 𝒙 ∈ (𝟏,∞), 𝒚 ∈ (𝟏,∞)

𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 = 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚

𝟏 − 𝒙𝒚) + 𝐭𝐚𝐧−𝟏 𝒛

{𝒙𝒚 > 1 ⇒

𝒙 + 𝒚

𝟏 − 𝒙𝒚< 0

𝒛 ∈ (𝟏,∞)

𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 = 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛

𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)

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∫ ∫ ∫ (𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛)𝒃

𝒂


𝒂

𝒃

𝒂

=

= 𝝅(𝒃 − 𝒂)𝟐 +∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛


𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

𝟑∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 𝒙𝒃

𝒂


𝒂

𝒃

𝒂

= 𝝅(𝒃 − 𝒂)𝟐 +∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛


𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

𝟑(𝒃 − 𝒂)𝟐 [𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 −𝟏

𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝟏)]

𝒂

𝒃

=

= 𝝅(𝒃 − 𝒂)𝟑 +∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛


𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

Therefore,

𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛


𝒃

𝒂

𝒃

𝒂

𝒃

𝒂

=

= 𝟑(𝒃 − 𝒂)𝟐 (𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏

𝟐(𝒂𝟐 + 𝟏

𝒃𝟐 + 𝟏)) − 𝝅(𝒃 − 𝒂)𝟑

1557. Prove that:

∫ 𝒆−𝒙𝟐𝑯𝟐𝒏(𝜶𝒙)

∞

−∞

𝒅𝒙 = √𝝅(𝟐𝒏)!

𝒏!(𝜶𝟐 − 𝟏)𝒏, 𝒏 > 0

where, 𝑯𝟐𝒏 −Hermite polynomials.

Proposed by Tobi Joshua-Nigeria

Solution by Serlea Kabay-Liberia

Using Hermite polynomial as a special case of the Laguerre polynomial,

𝑯𝟐𝒏(𝜶𝒙) = (−𝟒)𝒏𝒏! 𝑳𝒏

−𝟏𝟐(𝒂𝟐𝒙𝟐) ⇒ ∫ 𝒆−𝒙

𝟐𝑯𝟐𝒏(𝜶𝒙)

∞

−∞

𝒅𝒙

= (−𝟒)𝒏𝒏!∫ 𝒆−𝒙𝟐𝑳𝒏−𝟏𝟐(𝒂𝟐𝒙𝟐)

∞

−∞

𝒅𝒙

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𝑰 = ∫ 𝒆−𝒙𝟐𝑯𝟐𝒏(𝜶𝒙)

∞

−∞

𝒅𝒙 = (−𝟒)𝒏𝒏! ∫ 𝒆−𝒙𝟐𝑳𝒏−𝟏𝟐(𝒂𝟐𝒙𝟐)

∞

−∞

𝒅𝒙

𝑳𝒏(𝒂)(𝒙) = ∑(−𝟏)𝒌 (

𝒏 + 𝒂

𝒏 − 𝒌)𝒙𝟐𝒌

𝒌!

𝒏

𝒌=𝟎

⇒ 𝑳𝒏(−𝟏𝟐)(𝒂𝟐𝒙𝟐) = ∑(−𝟏)𝒌(

𝒏 −𝟏𝟐

𝒏 − 𝒌)𝒂𝟐𝒌𝒙𝟐𝒌

𝒌!

𝒏

𝒌=𝟎

∵ 𝑰 = (−𝟒)𝒏𝒏!∫ 𝒆−𝒙𝟐∑(−𝟏)𝒌(

𝒏 −𝟏𝟐


𝒌!

𝒏

𝒌=𝟎

𝒅𝒙 =∞

−∞

= 𝟐(−𝟒)𝒏𝒏!∫ 𝒆−𝒙𝟐∑(−𝟏)𝒌 (

𝒏 −𝟏𝟐


𝒌!

𝒏

𝒌=𝟎

𝒅𝒙∞

𝟎

Using Dominated convergence theorem:

𝑰 = 𝟐𝟐𝒏+𝟏(−𝟏)𝒏∑(−𝟏)𝒌(𝒏 −

𝟏𝟐

𝒏 − 𝒌)𝒂𝟐𝒌

𝒌!

𝒏

𝒌=𝟎

∫ 𝒆−𝒙𝟐𝒙𝟐𝒌

∞

𝟎

𝒅𝒙 =

= 𝟐𝟐𝒏(−𝟏)𝒏∑(−𝟏)𝒌(𝒏 −

𝟏𝟐

𝒏 − 𝒌)𝒂𝟐𝒌𝚪 (𝒌 +

𝟏𝟐)

𝒌!

𝒏

𝒌=𝟎

(𝒏 −

𝟏𝟐

𝒏 − 𝒌) =

𝚪 (𝒏 +𝟏𝟐)

𝚪(𝒏 + 𝟏 − 𝒌)𝚪(𝒌 +𝟏𝟐)=

(𝟐𝒏)!√𝝅

𝟒𝒏𝒏! 𝚪(𝒏 + 𝟏 − 𝒌)𝚪(𝒌 +𝟏𝟐)

𝑰 = (𝟐𝒏)! √𝝅(−𝟏)𝒏∑(−𝟏)𝒌𝒏

𝒌=𝟎

𝒂𝟐𝒌𝚪(𝒌 +𝟏𝟐)

𝒌! 𝚪(𝒏 + 𝟏 − 𝒌)𝚪(𝒌 +𝟏𝟐)=

= (𝟐𝒏)!√𝝅(−𝟏)𝒏∑(−𝟏)𝒌𝒂𝟐𝒌

𝒌! (𝒏 − 𝒌)!

𝒏

𝒌=𝟎

⇒ 𝑰 =(𝟐𝒏)! √𝝅(−𝟏)𝒏

𝒏!∑(−𝟏)𝒌

𝒏! 𝒂𝟐𝒌

𝒌! (𝒏 − 𝒌)!𝒂𝟐𝒌

𝒏

𝒌=𝟎

=

=(𝟐𝒏)! √𝝅(−𝟏)𝒏

𝒏!∑(

𝒏

𝒌) (−𝒂𝟐)𝒌

𝒏

𝒌=𝟎

𝑰 =(𝟐𝒏)! √𝝅(−𝟏)𝒏

𝒏!(𝟏 − 𝒂𝟐)𝒏 =

√𝝅 ⋅ (𝟐𝒏)! ⋅ (𝒂𝟐 − 𝟏)𝒏

𝒏!

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Therefore,

∫ 𝒆−𝒙𝟐𝑯𝟐𝒏(𝜶𝒙)

∞

−∞

𝒅𝒙 = √𝝅(𝟐𝒏)!

𝒏!(𝜶𝟐 − 𝟏)𝒏, 𝒏 > 0

𝑳𝒏𝒂(⋅) −Laguerre polynomial.

1558. Prove that:

𝛀 = ∫𝐭𝐚𝐧−𝟏(𝒂𝒙)

𝐬𝐢𝐧𝐡(𝒃𝒙)

∞

𝟎

𝒅𝒙 =𝝅

𝒃(𝐥𝐨𝐠(

𝚪 (𝒃𝟐𝝅𝒂

)

𝚪 (𝟏𝟐+

𝒃𝟐𝝅𝒂

)) +

𝟏

𝟐𝐥𝐨𝐠 (

𝒃

𝟐𝝅𝒂))

Proposed by Ose Favour-Nigeria

Solution 1 by Felix Marin-Romania

∫𝐭𝐚𝐧−𝟏(𝒂𝒙)

𝐬𝐢𝐧𝐡(𝒃𝒙)𝒅𝒙

∞

𝟎

= ∫𝒂𝒙


∞

𝟎

∫𝒅𝒚

𝒚𝟐 + (𝒂𝒙)𝟐

∞

𝟏

𝒅𝒙 =

𝒔𝒈𝒏 (𝒃)∫𝒂𝒙

𝒚𝟐 + (𝒂𝒙)𝟐𝟏

𝐬𝐢𝐧(|𝒃|𝒙)

∞

𝟏

𝒅𝒙𝒅𝒚 =

|𝒃|𝒙→𝝅𝒙

𝒗=𝝅𝒂|𝒃| 𝝅

𝒃∫ ∫

𝒗𝒙

𝒚𝟐 + (𝒗𝒙)𝟐𝟏

𝐬𝐢𝐧𝐡(𝝅𝒙)𝒅𝒙𝒅𝒚

∞

𝟎

∞

𝟏

=

=𝝅

𝒃∫ [𝒊∫

(𝒚 + 𝒗𝒙𝒊)−𝟏 − (𝒚 − 𝒗𝒙𝒊)−𝟏

𝟐 𝐬𝐢𝐧𝐡(𝝅𝒙)𝒅𝒙

∞

𝟎

] 𝒅𝒚∞

𝟏

=

=𝝅

𝒃∫ [∑(−𝟏)𝒏(𝒚 + 𝟐𝒏𝒗)−𝟏 −

𝟏

𝟐(𝒚 + 𝒗𝒙)−𝟏|𝒙=𝟎

∞

𝒏=𝟎

] 𝒅𝒚∞

𝟏

=

=𝝅

𝒃∫ {∑[

𝟏

𝒚 + 𝟐𝒏𝒗−

𝟏

𝒚 + (𝟐𝒏 + 𝟏)𝒗] −

𝟏

𝟐𝒚

∞

𝒏=𝟎

}∞

𝟏

𝒅𝒚 =

=𝝅

𝒃∫ {

𝟏

𝟐𝒗∑[

𝟏

𝒏 +𝒚𝟐𝒗

−𝟏

𝒏 +𝟏𝟐 +

𝒚𝟐𝒗

]

∞

𝒏=𝟎

−𝟏

𝟐𝒚}𝒅𝒚

∞

𝟏

=

=𝝅

𝒃∫ {

𝟏

𝟐𝒗[𝚿(

𝟏

𝟐+𝒚

𝟐𝒗) −𝚿(

𝒚

𝟐𝒗)] −

𝟏

𝟐𝒚} 𝒅𝒚

∞

𝟏

=

=𝝅

𝒃[𝐥𝐨𝐠(

𝚪(𝟏𝟐+

𝒚[𝟐𝒗]

)

𝚪 (𝒚[𝟐𝒗]

)) −

𝟏

𝟐𝐥𝐨𝐠𝒚]

𝟏

∞

=𝝅

𝒃[𝟏

𝟐𝐥𝐨𝐠 (

𝟏

𝟐𝒗) − 𝐥𝐨𝐠(

𝚪 ([𝟏 + 𝒗][𝟐𝒗]

)

𝚪 (𝟏[𝟐𝒗]

))]

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Therefore,



∞

𝟎

𝒅𝒙 =𝝅

𝒃(𝐥𝐨𝐠(

𝚪 (𝒃𝟐𝝅𝒂)

𝚪 (𝟏𝟐 +

𝒃𝟐𝝅𝒂)

) +𝟏

𝟐𝐥𝐨𝐠 (

𝒃

𝟐𝝅𝒂))


𝐋𝐞𝐭 𝑰(𝒂, 𝒃) = ∫𝟏

𝐬𝐢𝐧𝐡(𝒃𝒙)𝐭𝐚𝐧−𝟏 (

𝒙

𝒂)𝒅𝒙

∞

𝟎

Using Feynmann parametrization technique for integrating 𝐬𝐢𝐧𝒙

𝒙 , we have:

𝑰(𝒂, 𝒃) = ∫𝟏


∞

𝟎

∫𝐬𝐢𝐧 𝒕𝒙

𝒕

∞

𝟎

𝒆−𝒂𝒕𝒅𝒕𝒅𝒙 =

= ∫𝒆−𝒂𝒕

𝒕

∞

𝟎



∞

𝟎

𝒅𝒙𝒅𝒕 = ∫ 𝒆−𝒕𝒙 𝐬𝐢𝐧 𝒕𝒙𝒅𝒙∞

𝟎

=𝒕

𝒕𝟐 + 𝒏𝟐; (∵

𝟏

𝐬𝐢𝐧𝐡 𝒙=

𝟐𝒆−𝒙

𝟏 − 𝒆−𝟐𝒙)


𝐬𝐢𝐧𝐡(𝒃𝒙)𝒅𝒙

∞

𝟎

= 𝟐∑∫ 𝒆−𝒙(𝟐𝒃𝒏+𝒏) 𝐬𝐢𝐧 𝒕𝒙𝒅𝒙∞

𝟎

∞

𝒏=𝟎

= 𝟐𝒕∑𝟏

𝒕𝟐 + (𝟐𝒃𝒏 + 𝒃)𝟐

∞

𝒏=𝟎

Again, using Weierstrass product of 𝐜𝐨𝐬𝐡 𝒙:

𝐜𝐨𝐬𝐡 (𝝅𝒙

𝒃) =∏(𝟏+

𝒙𝟐

(𝟐𝒌𝒏+ 𝒌)𝟐)

𝒌≥𝟎

Taking logarithmic differention w.r.t. 𝒙, we get:

𝝅

𝟐𝒃𝐭𝐚𝐧𝐡 (

𝝅𝒙

𝟐𝒃) = 𝟐𝒙∑

𝟏

𝒙𝟐 + (𝟐𝒌𝒏 + 𝒌)𝒌≥𝟎

𝑰(𝒂, 𝒃) =𝝅

𝟐𝒃∫

𝒆−𝒂𝒕

𝒕∫

𝐬𝐢𝐧 𝒕𝒙

𝐬𝐢𝐧𝐡(𝒃𝒙)𝒅𝒙𝒅𝒕

∞

𝟎

∞

𝟎

=𝝅

𝟐𝒃∫

𝒆−𝒂𝒕

𝒕𝐭𝐚𝐧𝐡 (

𝝅𝒕

𝟐𝒃)

∞

𝟎

𝒅𝒕

Now, using Leibniz rule w.r.t. 𝒂, we get

𝑰(𝒂′, 𝒃) = −𝝅

𝟐𝒃∫ 𝒆−𝒂𝒕 𝐭𝐚𝐧𝐡 (

𝝅𝒕

𝟐𝒃)

∞

𝟎

𝒅𝒕 = −𝝅

𝟐𝒃∫

𝟏 − 𝒆𝝅𝒕𝒃

𝟏 + 𝒆𝝅𝒕𝒃

𝒆−𝒂𝒕∞

𝟎

𝒅𝒕 =

= −𝝅

𝟐𝒃∫ [−𝟏 + 𝟐∑(−𝟏)𝒏𝒆

𝝅𝒕𝒏𝒃

𝒏≥𝟎

] 𝒆−𝒂𝒕∞

𝟎

𝒅𝒕 =

=𝝅

𝟐𝒃− 𝝅∑(−𝟏)𝒏

𝟏

𝒏𝝅 + 𝒂𝒏≥𝟎

=𝝅

𝟐𝒃𝒂−𝟏

𝝅[𝝍(

𝒂 + 𝝅

𝟐𝝅) − 𝝍(

𝒂

𝟐𝝅)]

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Therefore,



∞

𝟎

𝒅𝒙 =𝝅

𝒃(𝐥𝐨𝐠(


𝚪 (𝟏𝟐 +

𝒃𝟐𝝅𝒂)

) +𝟏

𝟐𝐥𝐨𝐠 (

𝒃

𝟐𝝅𝒂))

Solution 3 by Syed Shahabudeen-India



∞

𝟎

𝒅𝒙 = ∫𝟏

𝐬𝐢𝐧𝐡(𝒃𝒙)𝑳 {𝐬𝐢𝐧(𝒙𝒕)

𝒕}𝒔=𝟏𝒂

∞

𝟎

𝒅𝒙 =

= ∫𝒆−𝒕𝒂

𝒕

∞

𝟎

∫𝐬𝐢𝐧(𝒙𝒕)𝒆𝒃𝒙

𝒆𝟐𝒃𝒙 − 𝟏

∞

𝟎

𝒅𝒙𝒅𝒕 = ∫𝒆−𝒕𝒂

𝒕

∞

𝟎

∫ 𝐬𝐢𝐧(𝒙𝒕)𝒆−𝒃𝒙∞

𝟎

∑𝒆−𝟐𝒃𝒙𝒌∞

𝒌=𝟎

𝒅𝒙𝒅𝒕 =


𝒕∑∫ 𝒆(−𝟐𝒃𝒌+𝒃)𝒙 𝐬𝐢𝐧(𝒙𝒕)

∞

𝟎

∞

𝒌=𝟎

𝒅𝒙𝒅𝒕∞

𝟎


𝒕∑𝑳{𝐬𝐢𝐧(𝒙𝒕)}𝒔=(𝟐𝒃𝒌+𝒃)

∞

𝒌=𝟎

∞

𝟎

𝒅𝒕 =


𝒕∑

𝒕

(𝟐𝒃𝒌 + 𝒃)𝟐 + 𝒕𝟐

∞

𝒌=𝟎

𝒅𝒕∞

𝟎

= ∫ 𝒆−𝒕𝒂∑

𝟏

(𝟐𝒃𝒌 + 𝒃)𝟐 + 𝒕𝟐

∞

𝒌=𝟎

𝒅𝒕∞

𝟎

∑𝟏

(𝟐𝒃𝒌 + 𝒃)𝟐 + 𝒕𝟐

∞

𝒌=𝟎

=𝟏

𝒃𝟐∑

𝟏

(𝟐𝒌 + 𝟏)𝟐 +𝒕𝟐

𝒃𝟐

∞

𝒌=𝟎

=𝝅

𝟐𝒃𝒕𝐭𝐚𝐧𝐡 (

𝝅𝒕

𝟐𝒃)

𝛀 =𝝅

𝟐𝒃∫ 𝒆−

𝒕𝒂𝐭𝐚𝐧𝐡 (

𝝅𝒕𝟐𝒃)

𝒕

∞

𝟎

𝒅𝒕 = (𝝅

𝟐𝒃)𝟐

∫ 𝒆−𝒕𝒂𝐭𝐚𝐧𝐡 (

𝝅𝒕𝟐𝒃)

𝝅𝒕𝟐𝒃

𝒅𝒕∞

𝟎

=𝒙=𝝅𝒕𝟐𝒃

=𝝅

𝟐𝒃∫ 𝒆−

𝟐𝒃𝝅𝒂𝒙

∞

𝟎

𝐭𝐚𝐧𝐡𝒙

𝒙𝒅𝒙 =

𝝅

𝟐𝒃𝑳 {𝐭𝐚𝐧𝐡𝒙

𝒙}𝒔=𝟐𝒃𝝅𝒂

It is well-know that:

𝑳 {𝐭𝐚𝐧𝐡 𝒙

𝒙}𝒔=𝟐𝒃𝝅𝒂

= 𝟐 𝐥𝐨𝐠(√𝒔𝚪 (

𝒔𝟒)

𝟐𝚪(𝒔 + 𝟐𝟒 )

) ⇒

𝛀 =𝝅

𝒃𝐥𝐨𝐠

(

√𝟐𝒃𝝅𝒂𝚪 (

𝒃𝟐𝝅𝒂)

𝟐𝚪(

𝒃𝝅𝒂 + 𝟏

𝟐)

)

=𝝅

𝒃(𝐥𝐨𝐠(

𝚪 (𝒃𝟐𝝅𝒂

)

𝚪(𝟏𝟐 +

𝒃𝟐𝝅𝒂)

) +𝟏

𝟐𝐥𝐨𝐠 (

𝒃

𝟐𝝅𝒂)

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Therefore,



∞

𝟎

𝒅𝒙 =𝝅

𝒃(𝐥𝐨𝐠(


𝚪 (𝟏𝟐 +

𝒃𝟐𝝅𝒂)

) +𝟏

𝟐𝐥𝐨𝐠 (

𝒃

𝟐𝝅𝒂))

1559. If 𝟏

√𝟑𝟏< 𝑎 ≤ 𝑏 then find:

𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟑𝟎𝒙𝟑 − 𝟏𝟎𝒙

𝟑𝟏𝒙𝟐 − 𝟏)

𝒃

𝒂

𝒅𝒙



𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟑𝟎𝒙𝟑 − 𝟏𝟎𝒙

𝟑𝟏𝒙𝟐 − 𝟏)

𝒃

𝒂

𝒅𝒙 = ∫ 𝐭𝐚𝐧−𝟏 (𝟏𝟎𝒙 − 𝟑𝟎𝒙𝟑

𝟏 − 𝟑𝟏𝒙𝟐)

𝒃

𝒂

𝒅𝒙

𝐭𝐚𝐧−𝟏 (𝟏𝟎𝒙 − 𝟑𝟎𝒙𝟑

𝟏 − 𝟑𝟏𝒙𝟐) = 𝐭𝐚𝐧−𝟏 (

𝟓𝒙 + 𝟓𝒙(𝟏 − 𝟔𝒙𝟐)

𝟏 − 𝟑𝟏𝒙𝟐) = 𝐭𝐚𝐧−𝟏 (

𝟓𝒙𝟏− 𝟔𝒙𝟐

+ 𝟓𝒙

𝟏 −𝟐𝟓𝒙𝟐

𝟏 − 𝟔𝒙𝟐

)

𝑭𝒐𝒓 𝒙 ≥𝟏

√𝟑𝟏⇒

𝟐𝟓𝒙𝟐

𝟏 − 𝟔𝒙𝟐≥ 𝟏 ⇒ 𝐭𝐚𝐧−𝟏 (

𝟏𝟎𝒙 − 𝟑𝟎𝒙𝟑

𝟏 − 𝟑𝟏𝒙𝟐) =

= 𝐭𝐚𝐧−𝟏 (𝟓𝒙

𝟏 − 𝟔𝒙𝟐) + 𝐭𝐚𝐧−𝟏(𝟓𝒙) − 𝝅

𝐭𝐚𝐧−𝟏 (𝟓𝒙

𝟏 − 𝟔𝒙𝟐) = 𝐭𝐚𝐧−𝟏 (

𝟐𝒙 + 𝟑𝒙

𝟏 − 𝟔𝒙𝟐) = 𝐭𝐚𝐧−𝟏(𝟐) + 𝐭𝐚𝐧−𝟏(𝟑𝒙)

𝐭𝐚𝐧−𝟏 (𝟏𝟎𝒙 − 𝟑𝟎𝒙𝟑

𝟏 − 𝟑𝟏𝒙𝟐) = 𝐭𝐚𝐧−𝟏(𝟐𝒙) + 𝐭𝐚𝐧−𝟏(𝟑𝒙) + 𝐭𝐚𝐧−𝟏(𝟓𝒙) − 𝝅

𝛀(𝒂, 𝒃) = ∫ (−𝝅 + 𝐭𝐚𝐧−(𝟐𝒙) + 𝐭𝐚𝐧−𝟏(𝟑𝒙) + 𝐭𝐚𝐧−𝟏(𝟓𝒙))𝒃

𝒂

𝒅𝒙 =

= [−𝝅𝒙 + 𝒙 𝐭𝐚𝐧−𝟏(𝟐𝒙) + 𝒙 𝐭𝐚𝐧−𝟏(𝟑𝒙) + 𝒙 𝐭𝐚𝐧−𝟏(𝟓𝒙) −𝟏

𝟒𝐥𝐨𝐠(𝟒𝒙𝟐 + 𝟏) −

𝟏

𝟔(𝟗𝒙𝟐 + 𝟏)

−𝟏

𝟏𝟎(𝟐𝟓𝒙𝟐 + 𝟏)]

𝒂

𝒃

=

= [𝒙 𝐭𝐚𝐧−𝟏 (𝟑𝟎𝒙𝟑 − 𝟏𝟎𝒙

𝟑𝟏𝒙𝟐 − 𝟏) −

𝟏

𝟒𝐥𝐨𝐠(𝟒𝒙𝟐 + 𝟏) −

𝟏

𝟔(𝟗𝒙𝟐 + 𝟏) −

𝟏

𝟏𝟎(𝟐𝟓𝒙𝟐 + 𝟏)]

𝒂

𝒃

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Therefore,

𝛀(𝒂, 𝒃) = 𝒃 𝐭𝐚𝐧−𝟏 (𝟑𝟎𝒃𝟑 − 𝟏𝟎𝒃

𝟑𝟏𝒃𝟐 − 𝟏) − 𝒂 𝐭𝐚𝐧−𝟏 (

𝟑𝟎𝒂𝟑 − 𝟏𝟎𝒂

𝟑𝟏𝒂𝟐 − 𝟏) −

𝟏

𝟒𝐥𝐨𝐠(

𝟒𝒃𝟐 + 𝟏

𝟒𝒂𝟐 + 𝟏) −

−𝟏

𝟔𝐥𝐨𝐠(

𝟗𝒃𝟐 + 𝟏

𝟗𝒂𝟐 + 𝟏) −

𝟏

𝟏𝟎𝐥𝐨𝐠 (

𝟐𝟓𝒃𝟐 + 𝟏

𝟐𝟓𝒂𝟐 + 𝟏)

1560. If 𝟎 < 𝑎 ≤ 𝑏 <𝝅

𝟔 then find:

𝛀(𝒂, 𝒃) = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐

𝐜𝐨𝐬𝟐 𝒙 − 𝟑 𝐬𝐢𝐧𝟐 𝒙

𝒃

𝒂

𝒅𝒙



𝛀(𝒂, 𝒃) = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐

𝐜𝐨𝐬𝟐 𝒙 − 𝟑𝐬𝐢𝐧𝟐 𝒙

𝒃

𝒂

𝒅𝒙 = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐

𝐜𝐨𝐬𝟐 𝒙 (𝟏 + 𝟑 𝐭𝐚𝐧𝟐 𝒙)

𝒃

𝒂

𝒅𝒙 ⇒𝐭𝐚𝐧 𝒙=𝒕

𝛀 = ∫(𝟏 + 𝒕𝟐)𝟐

𝟏 − 𝟑𝒕𝟐𝒅𝒕 = ∫

𝟏 + 𝟐𝒕𝟐 + 𝒕𝟒

𝟏 − 𝟑𝒕𝟐𝒅𝒕 = ∫(−

𝟏

𝟑𝒕𝟐 −

𝟕

𝟗+

𝟏𝟔𝟗

𝟏 − 𝟑𝒕𝟐)𝒅𝒕 =

= −𝟏

𝟗𝒕𝟑 −

𝟕

𝟗𝒕 +𝟏𝟔

𝟐𝟕∫

𝟏

𝟏𝟑− 𝒕𝟐

𝒅𝒕 = −𝟏

𝟗𝒕𝟑 −

𝟕

𝟗𝒕 +

𝟏𝟔

𝟐𝟕⋅√𝟑

𝟐𝐥𝐨𝐠(

𝟏

√𝟑+ 𝒕

𝟏

√𝟑− 𝒕)+ 𝑪 =

= −𝟏

𝟗𝒕𝟑 −

𝟕

𝟗𝒕 +

𝟏𝟔√𝟑

𝟐𝟕⋅𝟏

𝟐𝐥𝐨𝐠 (

𝟏 + √𝟑𝒕

𝟏 − √𝟑𝒕) + 𝑪 =

= −𝟏

𝟗𝒕𝟑 −

𝟕

𝟗𝒕 +

𝟏𝟔√𝟑

𝟐𝟕𝐭𝐚𝐧𝐡−𝟏(√𝟑𝒕) + 𝑪

𝛀(𝒂, 𝒃) = [−𝟏

𝟗𝐭𝐚𝐧𝟑 𝒙 −

𝟕

𝟗𝐭𝐚𝐧𝒙 +

𝟏𝟔√𝟑

𝟐𝟕𝐭𝐚𝐧𝐡−𝟏(√𝟑 𝐭𝐚𝐧 𝒙)]

𝒂

𝒃

𝛀(𝒂, 𝒃) = −𝟏

𝟗𝐭𝐚𝐧𝟑 𝒃 −

𝟕

𝟗𝐭𝐚𝐧 𝒃 +

𝟏𝟔√𝟑

𝟐𝟕𝐭𝐚𝐧𝐡−𝟏(√𝟑 𝐭𝐚𝐧 𝒃) +

𝟏

𝟗𝐭𝐚𝐧𝟑 𝒂 +

𝟕

𝟗𝐭𝐚𝐧𝒂 −

−𝟏𝟔√𝟑

𝟐𝟕𝐭𝐚𝐧𝐡−𝟏(√𝟑 𝐭𝐚𝐧 𝒂)

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𝛀(𝒂, 𝒃) = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐


𝒃

𝒂

𝒅𝒙 = ∫𝟏

𝐜𝐨𝐬𝟐 𝒙⋅(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐

𝟏 − 𝟑 𝐭𝐚𝐧𝟐 𝒙

𝒃

𝒂

𝒅𝒙 =𝐭𝐚𝐧 𝒙=𝒕

= ∫(𝟏 + 𝒕𝟐)𝟐

𝟏 − 𝟑𝒕𝟐

𝐭𝐚𝐧 𝒃

𝐭𝐚𝐧 𝒂

𝒅𝒕 = ∫𝟏 + 𝒕𝟒 + 𝟐𝒕𝟐

𝟏 − 𝟑𝒕𝟐

𝐭𝐚𝐧 𝒃

𝐭𝐚𝐧 𝒂

𝒅𝒕 =𝟏

𝟑∫

𝟑𝒕𝟒 + 𝟔𝒕𝟐 + 𝟑

𝟏 − 𝟑𝒕𝟐

𝐭𝐚𝐧 𝒃

𝐭𝐚𝐧 𝒂

𝒅𝒕 =

=𝟏

𝟑∫

𝟑𝒕𝟒 − 𝒕𝟐 + 𝟕𝒕𝟐 + 𝟑

𝟏 − 𝟑𝒕𝟐

𝐭𝐚𝐧 𝒃

𝐭𝐚𝐧 𝒂

𝒅𝒕 =𝟏

𝟑∫ (−𝒕𝟐 +

𝟕𝒕𝟐 + 𝟑

𝟏 − 𝟑𝒕𝟐)𝒅𝒕

𝐭𝐚𝐧 𝒃

𝐭𝐚𝐧 𝒂

=

=𝟏

𝟑∫ (−𝒕𝟐 +

𝟏

𝟑(−𝟕 +

𝟏𝟔

𝟏 − 𝟑𝒕𝟐))

𝐭𝐚𝐧 𝒃

𝐭𝐚𝐧 𝒂

𝒅𝒕 = [−𝒕𝟑

𝟗−𝟕𝒕

𝟗−𝟏𝟔

𝟐𝟕⋅√𝟑

𝟐𝐥𝐨𝐠 |

𝒕 −𝟏

√𝟑

𝒕 +𝟏

√𝟑

|]

𝐭𝐚𝐧 𝒂

𝐭𝐚𝐧 𝒃

𝛀(𝒂, 𝒃) = −𝐭𝐚𝐧𝟑 𝒃

𝟗−𝟕 𝐭𝐚𝐧 𝒃

𝟗−𝟖√𝟑

𝟐𝟕𝐥𝐨𝐠 |

√𝟑 𝐭𝐚𝐧 𝒃 − 𝟏

√𝟑 𝐭𝐚𝐧 𝒃 + 𝟏| +𝐭𝐚𝐧𝟑 𝒂

𝟗

+𝟕 𝐭𝐚𝐧𝒂

𝟗𝐥𝐨𝐠 |

√𝟑 𝐭𝐚𝐧 𝒂 − 𝟏

√𝟑 𝐭𝐚𝐧 𝒂 + 𝟏|

Solution 3 by Ghuiam Shah Naseri-Afghanistan

𝛀(𝒂, 𝒃) = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐


𝒃

𝒂

𝒅𝒙 = ∫𝟏

𝐜𝐨𝐬𝟐 𝒙⋅(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐

𝟏 − 𝟑 𝐭𝐚𝐧𝟐 𝒙

𝒃

𝒂

𝒅𝒙 ⇒𝐭𝐚𝐧 𝒙=𝒖

𝛀 = ∫(𝟏 + 𝒖𝟐)𝟐

𝟏 − 𝟑𝒖𝟐𝒅𝒖 = ∫

𝒖𝟒 + 𝟐𝒖𝟐 + 𝟏

−𝟑𝒖𝟐 + 𝟏𝒅𝒖

= ∫(𝟖√𝟑

𝟐𝟕⋅

𝟏

𝒖 +√𝟑𝟑

−𝟖√𝟑

𝟐𝟕⋅

𝟏

𝒖 −√𝟑𝟑

−𝟏

𝟑𝒖𝟐 −

𝟕

𝟗)𝒅𝒖 =

= ∫𝟖√𝟑

𝟐𝟕⋅

𝟏

𝒖 +√𝟑𝟑

𝒅𝒖 −∫𝟖√𝟑

𝟐𝟕⋅

𝟏

𝒖 −√𝟑𝟑

𝒅𝒖 −∫𝟏

𝟑𝒖𝟐𝒅𝒖 − ∫

𝟕

𝟗𝒅𝒖 =

=𝟖√𝟑

𝟐𝟕𝐥𝐨𝐠 (𝒖 +

√𝟑

𝟑) −

𝟖√𝟑

𝟑𝐥𝐨𝐠 (𝒖 −

√𝟑

𝟑) −

𝟏

𝟑𝒖𝟑 −

𝟕

𝟗𝒖

𝛀(𝒂, 𝒃) = [𝟖√𝟑

𝟐𝟕𝐥𝐨𝐠(𝐭𝐚𝐧 𝒙 +

√𝟑

𝟑) −

𝟖√𝟑

𝟑𝐥𝐨𝐠 (𝐭𝐚𝐧𝒙 −

√𝟑

𝟑) −

𝟏

𝟑𝐭𝐚𝐧𝟑 𝒙 −

𝟕

𝟗𝐭𝐚𝐧𝒙]

𝒂

𝒃

=

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=𝟖√𝟑

𝟐𝟕𝐥𝐨𝐠 (𝐭𝐚𝐧𝒃 +

√𝟑

𝟑) −

𝟖√𝟑

𝟑𝐥𝐨𝐠 (𝐭𝐚𝐧𝒃 −

√𝟑

𝟑) −

𝟏

𝟑𝐭𝐚𝐧𝟑 𝒃 −

𝟕

𝟗𝐭𝐚𝐧 𝒃 −

−(𝟖√𝟑

𝟐𝟕𝐥𝐨𝐠 (𝐭𝐚𝐧 𝒂 +

√𝟑

𝟑) −

𝟖√𝟑

𝟑𝐥𝐨𝐠 (𝐭𝐚𝐧𝒂 −

√𝟑

𝟑) −

𝟏

𝟑𝐭𝐚𝐧𝟑 𝒂 −

𝟕

𝟗𝐭𝐚𝐧 𝒂) =

=𝟖√𝟑

𝟐𝟕𝐥𝐨𝐠(

𝐭𝐚𝐧𝒃 +√𝟑𝟑

𝐭𝐚𝐧𝒃 −√𝟑𝟑

) −𝟏

𝟑𝐭𝐚𝐧𝟑 𝒃 −

𝟕

𝟗𝐭𝐚𝐧 𝒃 −

𝟖√𝟑

𝟐𝟕𝐥𝐨𝐠(

𝐭𝐚𝐧𝒂 +√𝟑𝟑

𝐭𝐚𝐧𝒂 −√𝟑𝟑

)+𝟏

𝟑𝐭𝐚𝐧𝟑 𝒂

+𝟕

𝟗𝐭𝐚𝐧𝒂


𝛀 = ∫ 𝒙−𝟐 ⋅ 𝒆−𝟒𝒙 ⋅ 𝐬𝐢𝐧𝟐(𝟐𝒙)∞

𝟎

𝒅𝒙


Solution 1 by Ty Halpen-Florida-USA

We will parametrize the integral:

𝑰(𝒂) = ∫ 𝒙−𝟐 ⋅ 𝒆−𝒂𝒙 ⋅ 𝐬𝐢𝐧𝟐(𝟐𝒙)∞

𝟎

𝒅𝒙

𝝏𝟐𝑰(𝒂)

𝝏𝒂𝟐= ∫ 𝒆−𝟐𝒂 𝐬𝐢𝐧𝟐(𝟐𝒙)𝒅𝒙

∞

𝟎

=

= [𝒆−𝒂𝒙(𝒂𝟐 𝐜𝐨𝐬(𝟒𝒙) − 𝒂𝟐 − 𝟒𝒂 𝐬𝐢𝐧(𝟒𝒙) − 𝟏𝟔)

𝟐𝒂(𝒂𝟐 + 𝟏𝟔)]𝒙=𝟎

𝒙=∞

=𝟖

𝒂(𝒂𝟐 + 𝟏𝟔)

𝝏𝑰(𝒂)

𝝏𝒂= ∫

𝟖

𝒂(𝒂𝟐 + 𝟏𝟔)𝒅𝒂 =

𝟏

𝟐𝐥𝐨𝐠𝒂 −

𝟏

𝟒𝐥𝐨𝐠(𝒂𝟐 + 𝟏𝟔) + 𝑪𝟏

𝑰(𝒂) = ∫(𝟏

𝟐𝐥𝐨𝐠𝒂 −

𝟏

𝟒𝐥𝐨𝐠(𝒂𝟐 + 𝟏𝟔) + 𝑪𝟏)𝒅𝒂 =

𝑰𝑩𝑷

= −𝒂

𝟒𝐥𝐨𝐠(𝒂𝟐 + 𝟏𝟔) +

𝒂

𝟐𝐥𝐨𝐠 𝒂 − 𝟐 𝐭𝐚𝐧−𝟏 (

𝒂

𝟒) + 𝒂𝑪𝟏 + 𝑪𝟐

Now, notice that 𝐥𝐢𝐦𝒂→∞

𝑰(𝒂) = 𝟎 and 𝑰(𝒂) = 𝟎 from the famous Dirichlet integral:

𝑰(𝟎) = 𝝅 = 𝑪𝟐

𝐥𝐢𝐦𝒂→∞

𝑰(𝟎) = 𝟎 = −𝟐 (𝝅

𝟐) + 𝒂𝑪𝟏 + 𝝅 ⇒ 𝑪𝟏 = 𝟎

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Then,

𝑰(𝟒) =𝝅

𝟐− 𝐥𝐨𝐠 𝟐


𝛀 = ∫ 𝒙−𝟐 ⋅ 𝒆−𝟒𝒙 ⋅ 𝐬𝐢𝐧𝟐(𝟐𝒙)∞

𝟎

𝒅𝒙 = ∫𝒆−𝟒𝒙 𝐬𝐢𝐧𝟐(𝟐𝒙)

𝒙𝟐𝒅𝒙

∞

𝟎

=

= ∫ 𝒆−𝟒𝒙 𝐬𝐢𝐧𝟐(𝟐𝒙)(∫ 𝒚𝒆−𝒙𝒚𝒅𝒚∞

𝟎

)𝒅𝒙∞

𝟎

= ∫ ∫ 𝒚 𝐬𝐢𝐧𝟐(𝟐𝒙)𝒆−(𝒚+𝟒)𝒙∞

𝟎

𝒅𝒙𝒅𝒚∞

𝟎

=

=𝟏

𝟐∫ ∫ 𝒚(𝟏 − 𝐜𝐨𝐬(𝟒𝒙))𝒆−(𝒚+𝟒)𝒙

∞

𝟎

∞

𝟎

𝒅𝒙𝒅𝒕 =𝟏

𝟐∫ (

𝒚

𝒚 + 𝟒−

𝒚(𝒚 + 𝟒)

(𝒚 + 𝟏)𝟐 + 𝟏𝟔)

∞

𝟎

𝒅𝒚 =

= ∫ (𝟐𝒚 + 𝟖

𝒚𝟐 + 𝟖𝒚 + 𝟑𝟐−

𝟐

𝒚 + 𝟒+

𝟖

(𝒚 + 𝟒)𝟐 + 𝟏𝟔)

∞

𝟎

𝒅𝒚 =

= [𝐥𝐨𝐠 (𝒚𝟐 + 𝟖𝒚 + 𝟑𝟐

𝒚𝟐 + 𝟖𝒚 + 𝟏𝟔) + 𝟐 𝐭𝐚𝐧−𝟏 (

𝒚 + 𝟒

𝟒)]𝟎

∞

=𝝅

𝟐− 𝐥𝐨𝐠 𝟐


𝛀 = ∫ 𝒙−𝟐 ⋅ 𝒆−𝟒𝒙 ⋅ 𝐬𝐢𝐧𝟐(𝟐𝒙)∞

𝟎

𝒅𝒙 = ∫ 𝒙−𝟐 ⋅ 𝒆−𝟒𝒙 ⋅𝟏 − 𝐜𝐨𝐬(𝟒𝒙)

𝟐

∞

𝟎

𝒅𝒙 =

=𝟏

𝟐∫ 𝒆−𝟒𝒙 ⋅

𝟏 − 𝐜𝐨𝐬(𝟒𝒙)

𝒙𝟐

∞

𝟎

𝒅𝒙 =𝟏

𝟐𝑳(𝟏 − 𝐜𝐨𝐬(𝟒𝒕)

𝒕𝟐)|𝒔=𝟒

=

=𝟏

𝟐(∫ ∫ 𝑳(𝟏 − 𝐜𝐨𝐬(𝟒𝒕))𝒅𝒔𝒅𝒔

∞

𝒔

∞

𝒔

)|𝒔=𝟒

=𝟏

𝟐(∫ ∫ (

𝟏

𝒔−

𝒔

𝒔𝟐 + 𝟏𝟔)𝒅𝒔𝒅𝒔

∞

𝒔

∞

𝒔

)|𝒔=𝟒

=

=𝟏

𝟐(∫ (𝐥𝐨𝐠 𝒔 −

𝟏

𝟐𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔))

𝒔

∞∞

𝒔

)|𝒔=𝟒

=𝟏

𝟐(∫ (

𝟏

𝟐𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) − 𝐥𝐨𝐠 𝒔)𝒅𝒙

∞

𝒔

)|𝒔=𝟒

=

=𝟏

𝟐((𝟏

𝟐𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) + 𝟒 𝐭𝐚𝐧−𝟏 (

𝒔

𝟒) − 𝒔 𝐥𝐨𝐠 𝒔)|

𝒔

∞

)|𝒔=𝟒

=

=𝟏

𝟐(𝟐𝝅 − (

𝟏

𝟐𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) + 𝟒 𝐭𝐚𝐧−𝟏 (

𝒔

𝟒) − 𝒔 𝐥𝐨𝐠 𝒔)|

𝒔=𝟒=

=𝟏

𝟐(𝟐𝝅 − (𝟏𝟎 𝐥𝐨𝐠 𝟐 + 𝝅 − 𝟖 𝐥𝐨𝐠 𝟐)) =

𝟏

𝟐(𝝅 − 𝟐 𝐥𝐨𝐠 𝟐) =

𝝅

𝟐− 𝐥𝐨𝐠𝟐.

Where,

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𝒊) ∫ 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔)𝒅𝒔 = 𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) − 𝟐∫𝒔𝟐

𝒔𝟐 + 𝟏𝟔𝒅𝒔 =

𝒖=𝐥𝐨𝐠(𝒔𝟐+𝟏𝟔)

= 𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) − 𝟐∫(𝟏 −𝟏𝟔

𝒔𝟐 + 𝟏𝟔)𝒅𝒔 = 𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) − 𝟐𝒔 + 𝟖 𝐭𝐚𝐧−𝟏 (

𝒔

𝟒) + 𝑪

𝒊𝒊) ∫ 𝐥𝐨𝐠 𝒔 𝒅𝒔 = 𝒔 𝐥𝐨𝐠 𝒔 − 𝒔 + 𝑪.


𝛀(𝒂) = ∫𝒙√𝒙

(𝒙𝟐 + 𝟏)(𝟏 + 𝒂𝟐𝒙𝟐)

∞

𝟎

𝒅𝒙, 𝒂 > 𝟎



𝛀(𝒂) = ∫𝒙√𝒙

(𝒙𝟐 + 𝟏)(𝟏 + 𝒂𝟐𝒙𝟐)

∞

𝟎

𝒅𝒙 = ∫√𝒙

(𝟏 + 𝒙𝟐)(𝒂𝟐 + 𝒙𝟐)𝒅𝒙

∞

𝟎

=

=𝟏

𝒂𝟐 − 𝟏(∫

√𝒙

𝟏 + 𝒙𝟐

∞

𝟎

𝒅𝒙⏟

𝑨

− ∫√𝒙

𝒂𝟐 + 𝒙𝟐

∞

𝟎

𝒅𝒙⏟

𝑩

)

𝑨 = ∫√𝒙

𝟏 + 𝒙𝟐

∞

𝟎

𝒅𝒙 =𝝅

𝟐𝐬𝐢𝐧 (𝟑𝝅𝟒 )

=𝝅

√𝟒

𝑩 =𝒙=𝒂𝒚 𝟏

√𝒂∫

√𝒚

𝟏 + 𝒚𝟐

∞

𝟎

𝒅𝒚 =𝝅

√𝟐𝒂

𝛀 =𝝅

(𝒂𝟐 − 𝟏)√𝟐(𝟏 −

𝟏

√𝒂) =

𝝅

(√𝒂 + 𝟏)(𝒂 + 𝟏)√𝟐𝒂

Therefore,

𝛀(𝒂) = ∫𝒙√𝒙

(𝒙𝟐 + 𝟏)(𝟏 + 𝒂𝟐𝒙𝟐)

∞

𝟎

𝒅𝒙 =𝝅

(√𝒂 + 𝟏)(𝒂 + 𝟏)√𝟐𝒂

1563. If 𝟎 < 𝒂 ≤ 𝒃 <𝝅

𝟖 then find:

𝛀(𝒂, 𝒃) = ∫ ∫(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧 (

𝝅𝟒− 𝒙 − 𝒚))

𝟏 + 𝐭𝐚𝐧 𝒙 ⋅ 𝐭𝐚𝐧 𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒− 𝒙 − 𝒚)

𝒅𝒙𝒃

𝒂

𝒅𝒚𝒃

𝒂


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𝐭𝐚𝐧 (𝝅

𝟒− 𝒙 − 𝒚) =

𝟏 − 𝐭𝐚𝐧(𝒙 + 𝒚)

𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)⇒ 𝟏 + 𝐭𝐚𝐧 (

𝝅

𝟒− 𝒙 − 𝒚) =

𝟐

𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)

(𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)) (𝐭𝐚𝐧(𝝅

𝟒− (𝒙 + 𝒚)) = 𝟏 − 𝐭𝐚𝐧(𝒙 + 𝒚)

= [𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)] [𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧 (𝝅

𝟒− (𝒙 + 𝒚))] =

= 𝟏 + 𝐭𝐚𝐧𝒙 𝐭𝐚𝐧𝒚 + 𝐭𝐚𝐧(𝒙 + 𝒚) − 𝐭𝐚𝐧𝒙 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧(𝒙 + 𝒚) =

= 𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧𝒚 + 𝐭𝐚𝐧 𝒙 + 𝐭𝐚𝐧 𝒚

Hence,

(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚))

𝟏 + 𝐭𝐚𝐧𝒙 ⋅ 𝐭𝐚𝐧 𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚)

=𝟐(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚)

𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧 (𝝅𝟒 −

(𝒙 + 𝒚))=

=𝟐(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧𝒚)

𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚=𝟐(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧𝒚)

(𝟏 + 𝐭𝐚𝐧 𝒚)(𝟏 + 𝐭𝐚𝐧 𝒙)= 𝟐

Therefore,

𝛀(𝒂, 𝒃) = ∫ ∫(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧 (

𝝅𝟒 − 𝒙 − 𝒚))

𝟏 + 𝐭𝐚𝐧 𝒙 ⋅ 𝐭𝐚𝐧𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚)

𝒅𝒙𝒃

𝒂

𝒅𝒚𝒃

𝒂

= 𝟐(𝒃 − 𝒂)𝟐


(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧𝒚) (𝟏 + 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚))

𝟏 + 𝐭𝐚𝐧𝒙 ⋅ 𝐭𝐚𝐧𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚)

=

=(𝟏 + 𝐭𝐚𝐧𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) ⋅

𝟐 − 𝟐 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 + 𝐭𝐚𝐧𝒙 + 𝐭𝐚𝐧𝒚

𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 ⋅𝟏 − 𝐭𝐚𝐧(𝒙 + 𝒚)𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)

=

= 𝟐 ⋅(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) ⋅

𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚𝟏 − 𝐭𝐚𝐧𝒙 𝐭𝐚𝐧 𝒚 + 𝐭𝐚𝐧𝒙 + 𝐭𝐚𝐧 𝒚

𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 ⋅𝟏 −

𝐭𝐚𝐧𝒙 + 𝐭𝐚𝐧𝒚𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚

𝟏 +𝐭𝐚𝐧𝒙 + 𝐭𝐚𝐧𝒚𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚

=

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= 𝟐 ⋅(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚)(𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚)

𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝟐 𝒚 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧 𝒚 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧𝟐 𝒚=

= 𝟐 ⋅𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧𝒚 + 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧𝟐 𝒚 + 𝐭𝐚𝐧 𝒙 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝒚 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝟐 𝒚

𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧𝒚 + 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧𝟐 𝒚 + 𝐭𝐚𝐧 𝒙 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝒚 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝟐 𝒚= 𝟐

Therefore,

𝛀(𝒂, 𝒃) = ∫ ∫(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧 (

𝝅𝟒 − 𝒙 − 𝒚))

𝟏 + 𝐭𝐚𝐧 𝒙 ⋅ 𝐭𝐚𝐧𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚)

𝒅𝒙𝒃

𝒂

𝒅𝒚𝒃

𝒂

= 𝟐(𝒃 − 𝒂)𝟐

1564.

𝑨 = ∫𝐥𝐨𝐠 𝒙

𝟏 + 𝒆𝒙 + 𝒆𝟐𝒙 + 𝒆𝟑𝒙𝒅𝒙

∞

𝟎

, 𝑩 = ∫𝒙 𝐥𝐨𝐠(𝟏 + 𝒙)

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

Prove that:

𝛀 = 𝑨+ 𝑩 =𝝅𝟐

𝟗𝟔−𝟏

𝟐𝐥𝐨𝐠𝟐 𝟐 −

𝝅

𝟖𝐥𝐨𝐠𝑹 and hence find the value of 𝑹.



𝑨 = ∫𝐥𝐨𝐠𝒙

𝟏 + 𝒆𝒙 + 𝒆𝟐𝒙 + 𝒆𝟑𝒙𝒅𝒙

∞

𝟎

=𝟏

𝟐(∫

𝐥𝐨𝐠𝒙

𝒆𝒙 + 𝟏𝒅𝒙

∞

𝟎⏟ 𝑨𝟏

+∫𝐥𝐨𝐠𝒙

𝒆𝟐𝒙 + 𝟏𝒅𝒙

∞

𝟎⏟ 𝑨𝟐

−∫𝒆𝒙 𝐥𝐨𝐠𝒙

𝒆𝟐𝒙 + 𝟏𝒅𝒙

∞

𝟎⏟ 𝑨𝟑

)

𝑨𝟏 = ∫𝐥𝐨𝐠 𝒙


∞

𝟎

=𝝏

𝝏𝒔{∫

𝒙𝒔−𝟏


∞

𝟎

}𝒔=𝟏

=𝝏

𝝏𝒔{𝜼(𝒔)𝚪(𝒔)}𝒔=𝟏 =

= 𝜼′(𝟏) + 𝜼(𝟏)𝚪(𝟏)𝝍(𝟏) = −𝟏

𝟐𝐥𝐨𝐠𝟐 𝟐

𝑨𝟐 =𝟏

𝟐∫

𝐥𝐨𝐠 𝒙 − 𝐥𝐨𝐠𝟐


∞

𝟎

=𝟏

𝟐∫

𝐥𝐨𝐠 𝒙


∞

𝟎⏟ 𝑨𝟏

−𝟏

𝟐𝐥𝐨𝐠𝟐∫

𝒅𝒙

𝒆𝒙 + 𝟏

∞

𝟎⏟ 𝜼(𝟏)=𝐥𝐨𝐠 𝟐

= −𝟑

𝟒𝐥𝐨𝐠𝟐 𝟐

𝑨𝟑 = ∫𝒆−𝒙 𝐥𝐨𝐠 𝒙

𝟏 + 𝒆−𝟐𝒙𝒅𝒙

∞

𝟎

=𝒕=𝒆−𝒙

∫𝐥𝐨𝐠 (𝐥𝐨𝐠 (

𝟏𝒕))

𝟏 + 𝒕𝟐𝒅𝒕

𝟏

𝟎

Using Malmsten’s integral:

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𝑰(𝝋) = ∫𝐥𝐨𝐠 (𝐥𝐨𝐠

𝟏𝒙)

𝟏 + 𝟐𝒙 𝐜𝐨𝐬𝝋 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

=𝝅

𝟐𝐬𝐢𝐧𝝋𝐥𝐨𝐠 {

(𝟐𝝅)𝝋𝝅 (𝟏𝟐 +

𝝋𝟐𝝅)

𝚪 (𝟏𝟐 −

𝝋𝟐𝝅)

}

𝑨𝟑 = 𝑰 (𝝅

𝟐) =

𝝅

𝟐𝐥𝐨𝐠{

√𝟐𝝅𝚪 (𝟑𝟒)

𝚪 (𝟏𝟒)

} =𝝅

𝟐𝐥𝐨𝐠 {

𝟐𝝅𝟑𝟐

𝚪𝟐 (𝟏𝟒)} =

𝝅

𝟒𝐥𝐨𝐠{

𝟒𝝅𝟑

𝚪𝟒 (𝟏𝟒)}

𝑨 = −𝟓

𝟖𝐥𝐨𝐠𝟐 𝟐 −

𝝅

𝟖𝐥𝐨𝐠(

𝟒𝝅𝟑

𝚪 (𝟏𝟒))

𝑩 = ∫𝒙 𝐥𝐨𝐠(𝟏 + 𝒙)

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

=∬𝒙𝟐

(𝟏 + 𝒙𝟐)(𝟏 + 𝒙𝒚)𝒅𝒙𝒅𝒚

𝟏

𝟎

=

= ∫𝟏

𝟏 + 𝒚𝟐

𝟏

𝟎

∫ (𝒙𝒚 − 𝟏

𝟏 + 𝒙𝟐+

𝟏

𝟏 + 𝒙𝒚)

𝟏

𝟎

𝒅𝒙𝒅𝒚 =

= ∫𝟏

𝟏 + 𝒚𝟐{𝟏

𝟐𝒚 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − 𝐭𝐚𝐧−𝟏 𝒙 +

𝟏

𝒚𝐥𝐨𝐠(𝟏 + 𝒙𝒚)}

𝟎

𝟏

𝒅𝒚𝟏

𝟎

=

= ∫𝟏

𝟏 + 𝒚𝟐(𝒚

𝟐𝐥𝐨𝐠 𝟐 −

𝝅

𝟒+𝐥𝐨𝐠(𝟏 + 𝒚)

𝒚)𝒅𝒚

𝟏

𝟎

=

=𝟏

𝟐𝐥𝐨𝐠 𝟐∫

𝒚

𝟏 + 𝒚𝟐𝒅𝒚

𝟏

𝟎

−𝝅

𝟒∫

𝒅𝒚

𝟏 + 𝒚𝟐

𝟏

𝟎

+∫𝐥𝐨𝐠(𝟏 + 𝒚)

𝒚𝒅𝒚

𝟏

𝟎

− 𝑩 =

=𝟏

𝟒𝐥𝐨𝐠𝟐 𝟐 −

𝝅𝟐

𝟏𝟔− 𝑳𝒊𝟐(−𝟏) − 𝑩 =

𝟏

𝟖𝐥𝐨𝐠𝟐 𝟐 −

𝝅𝟐

𝟑𝟐+𝝅𝟐

𝟐𝟒=𝟏

𝟖𝐥𝐨𝐠𝟐 𝟐 +

𝝅𝟐

𝟗𝟔

𝑨 + 𝑩 =𝟓

𝟖𝐥𝐨𝐠𝟐 𝟐 −

𝝅

𝟖𝐥𝐨𝐠(

𝟒𝝅𝟑

𝚪 (𝟏𝟒)) +

𝟏

𝟖𝐥𝐨𝐠𝟐 𝟐 +

𝝅𝟐

𝟗𝟔=

=𝝅𝟐

𝟗𝟔−𝟏

𝟐𝐥𝐨𝐠𝟐 𝟐 −

𝝅

𝟖𝐥𝐨𝐠(

𝟒𝝅𝟑

𝚪𝟒 (𝟏𝟒)) ⇒ 𝑹 =

𝟒𝝅𝟑

𝚪𝟒 (𝟏𝟒)

1565. If 𝟓 < 𝒂 ≤ 𝒃 then find:

𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑

𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏)𝒅𝒙

𝒃

𝒂


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𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑

𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏)𝒅𝒙

𝒃

𝒂

=𝒙=𝐭𝐚𝐧 𝒕

= ∫ 𝐬𝐞𝐜𝟐 𝒕 ⋅ 𝐭𝐚𝐧−𝟏 (𝟒 𝐭𝐚𝐧 𝒕 − 𝟒 𝐭𝐚𝐧𝟑 𝒕

𝐭𝐚𝐧𝟒 𝒕 − 𝟔 𝐭𝐚𝐧𝟐 𝒕 + 𝟏)

𝐭𝐚𝐧−𝟏 𝒃

𝐭𝐚𝐧−𝟏 𝒂

𝒅𝒕 =

= ∫ 𝐬𝐞𝐜𝟐 𝒕 ⋅ 𝐭𝐚𝐧−𝟏(𝐭𝐚𝐧𝟒𝒕) 𝒅𝒕𝐭𝐚𝐧−𝟏 𝒃


= 𝟒∫ 𝒕 ⋅ 𝐬𝐞𝐜𝟐 𝒕𝐭𝐚𝐧−𝟏 𝒃


𝒅𝒕 =

= [𝟒𝒕 ⋅ 𝐭𝐚𝐧 𝒕 + 𝟒 𝐥𝐨𝐠|𝐜𝐨𝐬 𝒕|]𝐭𝐚𝐧−𝟏 𝒂𝐭𝐚𝐧−𝟏 𝒃 − 𝟐𝝅(𝒃 − 𝒂) =

= 𝟒𝒃 ⋅ 𝐭𝐚𝐧−𝟏 𝒃 − 𝟒𝒂 ⋅ 𝐭𝐚𝐧−𝟏 𝒂 + 𝟒 𝐥𝐨𝐠𝐜𝐨𝐬(𝐭𝐚𝐧−𝟏 𝒃)

𝐜𝐨𝐬(𝐭𝐚𝐧−𝟏 𝒂)− 𝟐𝝅(𝒃 − 𝒂) =

= 𝟒 [𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏

𝟐𝐥𝐨𝐠 (

𝒂𝟐 + 𝟏

𝒃𝟐 + 𝟏)] − 𝟐𝝅(𝒃 − 𝒂)


∵ 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒙 = 𝝅 + 𝐭𝐚𝐧−𝟏 (𝟐𝒙

𝟏 − 𝒙𝟐) , 𝒙 ∈ (𝟏,∞)

𝟒 𝐭𝐚𝐧−𝟏 𝒙 = 𝟐𝝅 + 𝐭𝐚𝐧−𝟏 (𝟐𝒙

𝟏 − 𝒙𝟐) + 𝐭𝐚𝐧−𝟏 (

𝟐𝒙

𝟏 − 𝒙𝟐)

𝟒 𝐭𝐚𝐧−𝟏 𝒙 = 𝟐𝝅 + 𝐭𝐚𝐧−𝟏 (

𝟒𝒙𝟏− 𝒙𝟐

𝟏 −𝟒𝒙𝟐

(𝟏 − 𝒙𝟐)𝟐

)

𝟒 𝐭𝐚𝐧−𝟏 𝒙 = 𝟐𝝅 + 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑

𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏) , ∀𝒙 ∈ (𝟏,∞)

∫ 𝟒 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝒃

𝒂

= ∫ 𝟐𝝅𝒅𝒙𝒃

𝒂

+𝛀

𝛀 = 𝟒 [𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝟏

𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝟏)]

𝒂

𝒃

− 𝟐𝝅(𝒃 − 𝒂)

Therefore,

𝛀(𝒂, 𝒃) = 𝟒 [𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏

𝟐𝐥𝐨𝐠 (

𝒂𝟐 + 𝟏

𝒃𝟐 + 𝟏)] − 𝟐𝝅(𝒃 − 𝒂)

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Solution 3 by Ajetunmobi Abdulquoyyum-Nigeria

𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑

𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏)𝒅𝒙

𝒃

𝒂

− 𝟐𝝅∫ 𝒅𝒙𝒃

𝒂

𝑨 = ∫ 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑

𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏)𝒅𝒙 = ∫

𝟒𝒙(𝟏 − 𝒙𝟐)

(𝒙𝟐 − 𝟐𝒙 − 𝟏)(𝒙𝟐 + 𝟐𝒙 − 𝟏)𝒅𝒙 =

= ∫𝐭𝐚𝐧−𝟏(𝟏 − 𝒙𝟐) (𝟒𝒙

(𝒙𝟐 − 𝟐𝒙 − 𝟏)(𝒙𝟐 + 𝟐𝒙 − 𝟏))𝒅𝒙 =

= ∫𝐭𝐚𝐧−𝟏(𝟏 − 𝒙𝟐) (𝟏

𝒙𝟐 + 𝟐𝒙 − 𝟏−

𝟏

𝒙𝟐 + 𝟐𝒙 − 𝟏)𝒅𝒙 =

= ∫𝐭𝐚𝐧−𝟏 (𝟏 − 𝒙𝟐

𝟏 − 𝟐𝒙 − 𝒙𝟐−

𝟏 − 𝒙𝟐

𝟏 + 𝟐𝒙 − 𝒙𝟐)𝒅𝒙 =

= ∫𝐭𝐚𝐧−𝟏(𝟏

𝟏 −𝟐𝒙

𝟏 − 𝒙𝟐

−𝟏

𝟏 +𝟐𝒙

𝟏 − 𝒙𝟐

)𝒅𝒙 =𝐭𝐚𝐧 𝒕=𝒙

= ∫𝐬𝐞𝐜𝟐 𝒕 𝐭𝐚𝐧−𝟏 (𝟏

𝟏 −𝟐 𝐭𝐚𝐧 𝒕𝟏 − 𝐭𝐚𝐧𝟐 𝒕

−𝟏

𝟏 +𝟐 𝐭𝐚𝐧 𝒕𝟏 − 𝐭𝐚𝐧𝟐 𝒕

)𝒅𝒕 =

= ∫𝐬𝐞𝐜𝟐 𝒕 𝐭𝐚𝐧−𝟏 (𝟏

𝟏 − 𝐭𝐚𝐧 𝟐𝒕−

𝟏

𝟏 + 𝐭𝐚𝐧𝟐𝒕)𝒅𝒕 =

= ∫𝐬𝐞𝐜𝟐 𝒕 𝐭𝐚𝐧−𝟏𝟏 + 𝐭𝐚𝐧 𝟐𝒕 − 𝟏 + 𝐭𝐚𝐧𝟐𝒕

(𝟏 − 𝐭𝐚𝐧 𝟐𝒕)(𝟏 + 𝐭𝐚𝐧𝟐𝒕)𝒅𝒕 =

= ∫𝐬𝐞𝐜𝟐 𝒕 𝐭𝐚𝐧−𝟏(𝐭𝐚𝐧𝟒𝒕) 𝒅𝒕 = 𝟒∫𝒕 ⋅ 𝐬𝐞𝐜𝟐 𝒕 𝒅𝒕 =𝑰𝑩𝑷

= 𝟒(𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝟏

𝟐𝐥𝐨𝐠(𝟏 + 𝒙𝟐)) + 𝑪

𝛀(𝒂, 𝒃) = 𝟒 [𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏

𝟐𝐥𝐨𝐠 (

𝒂𝟐 + 𝟏

𝒃𝟐 + 𝟏)] − 𝟐𝝅(𝒃 − 𝒂)

1566. If 𝟎 < 𝒂 ≤ 𝒃 <𝝅

𝟐 then find:

𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬 𝟒𝒙

𝟏 − 𝐜𝐨𝐬 𝟒𝒙

𝒃

𝒂

𝒅𝒙


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𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬𝟒𝒙

𝟏 − 𝐜𝐨𝐬𝟒𝒙

𝒃

𝒂

𝒅𝒙 = 𝟒∫𝒅𝒙

𝟐𝐬𝐢𝐧𝟐 𝟐𝒙

𝒃

𝒂

−∫ 𝒅𝒙𝒃

𝒂

=

= 𝟐∫ 𝐜𝐬𝐜𝟐 𝟐𝒙𝒃

𝒂

𝒅𝒙 − (𝒃 − 𝒂) = −𝐜𝐨𝐭𝟐𝒙|𝒂𝒃 − (𝒃 − 𝒂) = 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭 𝟐𝒃 − (𝒃 − 𝒂)


𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬 𝟒𝒙


𝒃

𝒂

𝒅𝒙 = −∫−𝟑 − 𝐜𝐨𝐬 𝟒𝒙


𝒃

𝒂

𝒅𝒙 = −∫𝟏 − 𝐜𝐨𝐬 𝟒𝒙 − 𝟒

𝟏 − 𝐜𝐨𝐬 𝟒𝒙𝒅𝒙

𝒃

𝒂

=

= −∫ 𝒅𝒙𝒃

𝒂

+ ∫𝟒


𝒃

𝒂

= (𝒂 − 𝒃) +∫𝟐

𝐬𝐢𝐧𝟐 𝟐𝒙𝒅𝒙

𝒃

𝒂

=

= 𝒂 − 𝒃 − 𝐜𝐨𝐭𝟐𝒙|𝒂𝒃 = 𝒂 − 𝒃 + 𝐜𝐨𝐭𝟐𝒂 − 𝐜𝐨𝐭 𝟐𝒃

Solution 3 by Mohammad Hamed Nasery-Afghanistan

𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬𝟒𝒙


𝒃

𝒂

𝒅𝒙 = −∫−𝟑 − 𝐜𝐨𝐬 𝟒𝒙

𝟏 − 𝐜𝐨𝐬𝟒𝒙𝒅𝒙

𝒃

𝒂

=

= −∫𝟏 − 𝐜𝐨𝐬𝟒𝒙 − 𝟑 − 𝟏


𝒃

𝒂

= −∫𝟏 − 𝐜𝐨𝐬𝟒𝒙 − 𝟒


𝒃

𝒂

=

= 𝟒∫𝟏


𝒃

𝒂

−∫ 𝒅𝒙𝒃

𝒂

= 𝟐∫𝟏

𝐬𝐢𝐧𝟐 𝒙

𝒃

𝒂

𝒅𝒙 − (𝒃 − 𝒂) =

= 𝒂 − 𝒃 + 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭 𝟐𝒃

Solution 4 by Hussain Reza Zadah-Afghanistan

𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬 𝟒𝒙


𝒃

𝒂

𝒅𝒙 =𝒕=𝟒𝒙 𝟏

𝟒∫

𝟑 + 𝐜𝐨𝐬 𝒕

𝟏 − 𝐜𝐨𝐬 𝒕

𝟒𝒃

𝟒𝒂

𝒅𝒕 =𝐭𝐚𝐧

𝒕𝟐=𝒖

=𝟏

𝟒∫

𝟑 +𝟏 − 𝒖𝟐

𝟏 + 𝒖𝟐

𝟏 −𝟏 − 𝒖𝟐

𝟏 + 𝒖𝟐

⋅𝟐𝒅𝒖

𝟏 + 𝒖𝟐

𝐭𝐚𝐧 𝟐𝒃

𝐭𝐚𝐧 𝟐𝒂

=

=𝟏

𝟐∫

𝒖𝟐 + 𝟐

𝒖𝟐(𝟏 + 𝒖𝟐)



𝒅𝒖 =𝟏

𝟐∫

𝟐

𝒖𝟐



𝒅𝒖 − ∫𝟏

𝒖𝟐 + 𝟏𝒅𝒖



=

= [−𝟐

𝒖−𝟏

𝟐𝐭𝐚𝐧−𝟏 𝒖]




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𝛀 = ∫𝟑 + 𝐜𝐨𝐬𝟒𝒙

𝟏 − 𝐜𝐨𝐬𝟒𝒙𝒅𝒙 =

𝟏

𝟒∫𝟑 + 𝐜𝐨𝐬𝒙

𝟏 − 𝐜𝐨𝐬 𝒙𝒅𝒙 =

𝐭𝐚𝐧𝒙𝟐=𝒕

=𝟏

𝟒∫𝟑 +

𝟏 − 𝒕𝟐

𝟏 + 𝒕𝟐

𝟏 −𝟏 − 𝒕𝟐

𝟏 + 𝒕𝟐

⋅𝟐𝒅𝒕

𝟏 + 𝒕𝟐=𝟏

𝟐∫

𝟐𝒕𝟐 + 𝟒

𝟐𝒕𝟐(𝟏 + 𝒕𝟐)𝒅𝒕 =

𝟏

𝟐∫

𝒕𝟐 + 𝟐

𝒕𝟐(𝟏 + 𝒕𝟐)𝒅𝒕 =

=𝟏

𝟐∫𝟏

𝒕𝟐𝒅𝒕 +∫

𝟏

𝒕𝟐(𝒕𝟐 + 𝟏)𝒅𝒕 = ∫

𝟏

𝒕𝟐𝒅𝒕 − ∫

𝟏

𝒕𝟐 + 𝟏𝒅𝒕 =

= −𝟏

𝒕−𝟏

𝟐𝐭𝐚𝐧−𝟏 𝒕 = −

𝟏

𝐭𝐚𝐧 𝟐𝒙−𝟏

𝟐𝐭𝐚𝐧−𝟏(𝐭𝐚𝐧𝟐𝒙) = −𝐜𝐨𝐭𝟐𝒙 − 𝒙

Therefore,

𝛀(𝒂, 𝒃) = 𝒂 − 𝒃 + 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭𝟐𝒃

Solution 6 by Satyam Roy-India

By generalization:

𝛀 = ∫𝒎+ 𝐜𝐨𝐬 𝟒𝒙

𝒏 − 𝐜𝐨𝐬 𝟒𝒙𝒅𝒙 = ∫

𝒑

𝒏 − 𝐜𝐨𝐬𝟒𝒙

𝒃

𝒂

𝒅𝒙 + 𝒂 − 𝒃

If 𝒎,𝒏 ∈ ℕ,𝒑 = 𝒎 + 𝒏. Here 𝒎 = 𝟑,𝒏 = 𝟏,𝒑 = 𝟒

∫𝟒

𝟏 + 𝐜𝐨𝐬𝟒𝒙

𝒃

𝒂

𝒅𝒙 = ∫𝟐

𝐬𝐢𝐧𝟐 𝟐𝒙

𝒃

𝒂

𝒅𝒙 = 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭𝟐𝒃

Therefore,

𝛀(𝒂, 𝒃) = 𝒂 − 𝒃 + 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭𝟐𝒃

Solution 7 by Sujit Bhowmick-India

𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬𝟒𝒙


𝒃

𝒂

𝒅𝒙 = ∫𝟑(𝟏 + 𝐭𝐚𝐧𝟐 𝟐𝒙) + 𝟏 − 𝐭𝐚𝐧𝟐 𝟐𝒙

𝟏 + 𝐭𝐚𝐧𝟐 𝟐𝒙 − (𝟏 − 𝐭𝐚𝐧𝟐 𝟐𝒙)𝒅𝒙

𝒃

𝒂

=

= ∫𝟏 + 𝟐 𝐭𝐚𝐧𝟐 𝟐𝒙

𝟐 𝐭𝐚𝐧𝟐 𝟐𝒙

𝒃

𝒂

𝒅𝒙 = ∫𝟏 + 𝐬𝐞𝐜𝟐 𝟐𝒙

𝐭𝐚𝐧𝟐 𝟐𝒙𝒅𝒙

𝒃

𝒂

= 𝟐∫ 𝐜𝐬𝐜𝟐 𝟐𝒙𝒅𝒙𝒃

𝒂

−∫ 𝒅𝒙𝒃

𝒂

=


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1567. If 𝟎 < 𝒂 ≤ 𝒃 find a closed form:

𝛀(𝒂, 𝒃) = ∫

(

𝒙

𝟏 +𝒙𝟐

𝟑 +𝒙𝟐

𝟓 +𝒙𝟐

𝟕 +⋯)

𝒃

𝒂

𝒅𝒙


Solution by Naren Bhandari-Bajura-Nepal

Due to Lambert continued fraction (particular case of Gauss continued fraction)

𝐭𝐚𝐧𝒙 =𝒙

𝟏 + 𝕂𝒏=𝟏∞ −𝒙𝟐

𝟐𝒏 + 𝟏

Now we replace 𝒙 by 𝒊𝒙 giving us

𝐭𝐚𝐧𝐡 𝒙 =𝒙

𝟏 +𝕂𝒏=𝟏∞ 𝒙𝟐

𝟐𝒏 + 𝟏

So, we need to integrate 𝑰 + ∫ 𝐭𝐚𝐧𝐡 𝒙𝒅𝒙𝒃

𝒂 which is easy to see

𝑰 = ∫𝒅

𝒅𝒙𝐥𝐨𝐠(𝐜𝐨𝐬𝐡𝒙)

𝒃

𝒂

𝒅𝒙 = 𝐥𝐨𝐠 (𝐜𝐨𝐬𝐡𝒃

𝐜𝐨𝐬𝐡𝒂)

1568. Prove that:

∫ 𝐬𝐢𝐧 (𝒙

𝟐) 𝐭𝐚𝐧𝐡−𝟏(𝐬𝐢𝐧 𝟐𝒙)

𝝅𝟐

𝟎

𝒅𝒙 =

= 𝐥𝐨𝐠((𝟐√𝟐 − √𝟐 + 𝟐√𝟐 − 𝟏)

√𝟐+√𝟐

(𝟏 + 𝟐√𝟐 − 𝟐√𝟐 − √𝟐)

√𝟐−√𝟐

)



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𝑰 = ∫ 𝐬𝐢𝐧 (𝒙

𝟐) 𝐭𝐚𝐧𝐡−𝟏(𝐬𝐢𝐧𝟐𝒙)

𝝅𝟐

𝟎

𝒅𝒙 =𝑰𝑩𝑷𝟒∫

𝐜𝐨𝐬 (𝒙𝟐)

𝐜𝐨𝐬𝟐𝒙

𝝅𝟐

𝟎

𝒅𝒙 = 𝟒∫𝐜𝐨𝐬 (

𝒙𝟐)

𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝒙

𝝅𝟐

𝟎

𝒅𝒙

By the use of compound angle formula and subbing 𝐬𝐢𝐧 (𝒙

𝟐) = 𝒖 leads us to

∫𝟖𝒅𝒖

𝟏 − 𝟖𝒖𝟐(𝟏 − 𝒖𝟐)

𝟏

√𝟐

𝟎

= ∫𝟖𝒅𝒖

𝟖𝒖𝟒 − 𝟖𝒖𝟐 + 𝟏

𝟏

√𝟐

𝟎

= 𝐥𝐢𝐦𝒂→

𝟏

√𝟐

∫𝟖𝒅𝒖

𝑷(𝒖)

𝒂

𝟎

Note that the polynomial 𝑷(𝒖) is reducible over ℝ or

𝟐𝑷(𝒖) = (𝟒𝒖𝟐 − √𝟐 − 𝟐)(𝟒𝒖𝟐 + √𝟐 − 𝟐), with positive factors (√𝟐 + √𝟐⏟ 𝒓𝟏

, √𝟐 − √𝟐⏟ 𝒓𝟐

).

Therefore,

𝑰 = 𝟒√𝟐∫ (𝒅𝒖

𝟒𝒖𝟐 − √𝟐− 𝟐−

𝒅𝒖

𝟒𝒖𝟐 + √𝟐)

𝟏

√𝟐

𝟎

= −𝒓𝟐 𝐥𝐨𝐠 (𝒓𝟐 + 𝟏

𝟏 − 𝒓𝟐) + 𝒓𝟏 𝐥𝐨𝐠 (

𝒓𝟏 + 𝟏

𝒓𝟏 − 𝟏)

By putting the roots we have the result however, to get the desired result as presented in final form we observe that:

(𝒓𝟏 + 𝟏

𝒓𝟏 − 𝟏,𝒓𝟐 + 𝟏

𝟏 − 𝒓𝟐) =𝒓𝒂𝒕𝒊𝒐𝒏𝒂𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏

(𝟐√𝟐− 𝟏 + 𝟐√𝟐 + √𝟐,𝟏

𝟏 + 𝟐√𝟐 − 𝟐√𝟐 − √𝟐)

Hence,

∫ 𝐬𝐢𝐧 (𝒙

𝟐) 𝐭𝐚𝐧𝐡−𝟏(𝐬𝐢𝐧𝟐𝒙)

𝝅𝟐

𝟎

𝒅𝒙 =

= 𝐥𝐨𝐠((𝟐√𝟐− √𝟐 + 𝟐√𝟐− 𝟏)

√𝟐+√𝟐

(𝟏 + 𝟐√𝟐− 𝟐√𝟐− √𝟐)

√𝟐−√𝟐

)

1569. Prove that:

𝑰𝟐(𝒌) = ∫𝒙 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙

√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙

𝝅𝟐

𝟎

𝒅𝒙 = −𝝅

𝟐⋅𝒌′

𝒌𝟐+𝑬(𝒌)

𝒌𝟐

Proposed by Onikoyi Adeboye-Nigeria


𝑰𝟐(𝒌) = ∫𝒙 𝐬𝐢𝐧𝒙 𝐜𝐨𝐬 𝒙

√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙

𝝅𝟐

𝟎

𝒅𝒙 =

𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙=𝒖

𝒙=𝐬𝐢𝐧−𝟏√𝒖

𝒌𝟐 𝟏

𝟐𝒌𝟐∫ 𝐬𝐢𝐧−𝟏√

𝒖

𝒌𝟐𝒅𝒖

√𝟏 − 𝒌𝟐

𝒌𝟐

𝟎

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∫ 𝐬𝐢𝐧−𝟏√𝒖

𝒌𝟐𝒅𝒖

√𝟏 − 𝒌𝟐

𝒌𝟐

𝟎

= 𝑱(𝒌) ⇒

𝑱′(𝒌) = (∫ 𝐬𝐢𝐧−𝟏√𝒖

𝒌𝟐𝒅𝒖

√𝟏 − 𝒌𝟐

𝒌𝟐

𝟎

)

′

=𝐬𝐢𝐧−𝟏 𝟏

√𝟏 − 𝒌𝟐⋅ 𝟐𝒌 =

𝒌𝝅

√𝟏 − 𝒌𝟐

⇒ 𝑱(𝒌) = −𝝅√𝟏 − 𝒌𝟐 + 𝟐𝑬(𝒌) ⇒

𝑰𝟐(𝒌) =𝟏

𝟐𝒌𝟐(−𝝅√𝟏 − 𝒌𝟐 + 𝟐𝑬(𝒌)) = −

𝝅

𝟐⋅𝒌′

𝒌𝟐+𝑬(𝒌)

𝒌𝟐⇒ 𝒌′ = √𝟏 − 𝒌𝟐

For 𝒌 = 𝟏 ⇒ 𝑬(𝒙) = ∫ 𝒙 𝐬𝐢𝐧𝒙𝒅𝒙𝝅

𝟐𝟎

= 𝟏

Solution 2 by Sediqakbar Restheen-Afghanistan

𝑰𝟐(𝒌) = ∫𝒙𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙

√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙

𝝅𝟐

𝟎

𝒅𝒙 = −𝟏

𝒌𝟐∫−𝟐𝒌𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙

𝟐√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙

𝝅𝟐

𝟎

𝒅𝒙 =

= −𝟏

𝒌𝟐[𝒙√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙]

𝟎

𝝅𝟐+𝟏

𝒌𝟐∫ √𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙

𝝅𝟐

𝟎

= −𝝅√𝟏 − 𝒌𝟐

𝟐𝒌𝟐+𝑬(𝒌)

𝒌𝟐

Let 𝒌′ = √𝟏 − 𝒌𝟐 ⇒ 𝑰𝟐(𝒌) = −𝝅

𝟐⋅𝒌′

𝒌𝟐+𝑬(𝒌)

𝒌𝟐

1570. Find:

𝛀 = ∫ ∫ ∫𝒙𝒚𝒛

(𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙)

𝟏

𝟎

𝒅𝒙𝒅𝒚𝒅𝒛𝟏

𝟎

𝟏

𝟎



∵ (𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙) = ∑(𝒙𝟐𝒚 + 𝒙𝟐𝒛)

𝒄𝒚𝒄

+ 𝟐𝒙𝒚𝒛

𝑰 = ∫ ∫ ∫𝒙𝟐𝒚 + 𝒙𝟐𝒛 +

𝟐𝟑𝒙𝒚𝒛

(𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙)

𝟏

𝟎


𝟎

𝟏

𝟎

=𝟏

𝟑

∫ ∫ ∫𝒙𝟐

(𝒙 + 𝒚)(𝒙 + 𝒛)

𝟏

𝟎


𝟎

𝟏

𝟎

+𝟐

𝟑∫ ∫ ∫

𝒙𝒚𝒛

(𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙)

𝟏

𝟎


𝟎

𝟏

𝟎

=𝟏

𝟑

∫ ∫𝒙𝟐

𝒙 + 𝒚𝐥𝐨𝐠 (

𝒙 + 𝟏

𝒙)𝒅𝒙𝒅𝒚

𝟏

𝟎

𝟏

𝟎

+𝟐

𝟑𝛀 =

𝟏

𝟑

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∫ 𝒙𝟐 𝐥𝐨𝐠𝟐 (𝒙 + 𝟏

𝒙)𝒅𝒙

𝟏

𝟎

+𝟐

𝟑𝛀 =

𝟏

𝟑

∫ 𝒙𝟐 𝐥𝐨𝐠𝟐(𝒙 + 𝟏)𝒅𝒙𝟏

𝟎

+∫ 𝒙𝟐 𝐥𝐨𝐠𝟐 𝒙𝟏

𝟎

𝒅𝒙 − 𝟐∫ 𝒙𝟐 𝐥𝐨𝐠(𝒙 + 𝟏) 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

+𝟐

𝟑𝛀 =

𝟏

𝟑

𝑰𝟏 + 𝑰𝟐 − 𝟐𝑰𝟑 +𝟐

𝟑𝛀 =

𝟏

𝟑; (𝑰), where

𝑰𝟏 = ∫ 𝒙𝟐 𝐥𝐨𝐠𝟐(𝒙 + 𝟏)𝒅𝒙𝟏

𝟎

=𝒙+𝟏=𝒆𝒚

∫ (𝟐𝟐𝒚 − 𝟐𝒆𝒚 + 𝟏)𝒚𝟐𝒆𝒚𝐥𝐨𝐠 𝟐

𝟎

𝒅𝒚 =

= ∫ (𝒆𝟑𝒚 − 𝟐𝒆𝟐𝒚 + 𝒆𝒚)𝒚𝟐𝐥𝐨𝐠 𝟐

𝟎

𝒅𝒚 =𝟐

𝟑𝐥𝐨𝐠𝟐 𝟐 −

𝟏𝟔

𝟗𝐥𝐨𝐠 𝟐 +

𝟓𝟓

𝟓𝟒

𝑰𝟐 = ∫ 𝒙𝟐 𝐥𝐨𝐠𝟐 𝒙𝟏

𝟎

𝒅𝒙 =𝒙=𝒆𝒚

∫ 𝒆𝟑𝒚𝒚𝟐𝒅𝒚𝟎

−∞

=𝟐

𝟐𝟕

𝑰𝟑 = ∫ 𝒙𝟐 𝐥𝐨𝐠(𝒙 + 𝟏) 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

= −∫ 𝒙𝟐 𝐥𝐨𝐠 𝒙∑(−𝟏)𝒌𝒙𝒌

𝒌

∞

𝒌=𝟏

𝒅𝒙𝟏

𝟎

=

= −∑(−𝟏)𝒌

𝒌∫ 𝒙𝒌+𝟐 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

∞

𝒌=𝟏

=∑(−𝟏)𝒌

𝒌(𝒌 + 𝟑)𝟐

∞

𝒌=𝟏

= ∑(−𝟏)𝒌 (𝑨𝟏𝒌+

𝑨𝟐𝒌 + 𝟑

+𝑨𝟑

(𝒌 + 𝟑)𝟐)

∞

𝒌=𝟏

𝑨𝟏 =𝟏

𝟗, 𝑨𝟐 = −

𝟏

𝟑, 𝑨𝟑 = −

𝟏

𝟗

Hence,

𝑰𝟑 =𝟏

𝟗(− 𝐥𝐨𝐠𝟐 − (𝟏 −

𝟏

𝟐+𝟏

𝟑+∑

(−𝟏)𝒌−𝟏

𝒌

∞

𝒌=𝟒

) −𝟏

𝟑(𝟏

𝟏𝟐−𝟏

𝟐𝟐+𝟏

𝟑𝟐+∑

(−𝟏)𝒌

(𝒌 + 𝟑)𝟐

∞

𝒌=𝟏

) +𝟒𝟏

𝟏𝟎𝟖

= −𝟐

𝟗𝐥𝐨𝐠 𝟐 −

𝝅𝟐

𝟑𝟔+𝟒𝟏

𝟏𝟎𝟖

From (𝑰): 𝑰𝟏 + 𝑰𝟐 − 𝟐𝑰𝟑 +𝟐

𝟑𝛀 =

𝟏

𝟑 it follows:

𝟐

𝟑𝐥𝐨𝐠𝟐 𝟐 −

𝟏𝟔

𝟗𝐥𝐨𝐠𝟐 +

𝟓𝟓

𝟓𝟒+𝟐

𝟐𝟕− 𝟐 (−

𝟐

𝟗𝐥𝐨𝐠 𝟐 −

𝝅𝟐

𝟑𝟔+𝟒𝟏

𝟏𝟎𝟖) +

𝟐

𝟑𝛀 =

𝟏

𝟑

𝟐

𝟑𝐥𝐨𝐠𝟐 𝟐 −

𝟏𝟐

𝟗𝐥𝐨𝐠𝟐 +

𝝅𝟐

𝟏𝟖+𝟐

𝟑𝛀 = 𝟎 ⇔

𝟐 𝐥𝐨𝐠𝟐 𝟐 −𝟏𝟐

𝟑𝐥𝐨𝐠 𝟐 +

𝝅𝟐

𝟔+ 𝟐𝛀 = 𝟎 ⇔

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𝛀 = 𝟐 𝐥𝐨𝐠𝟐 − 𝐥𝐨𝐠𝟐 𝟐 −𝝅𝟐

𝟏𝟐

Therefore,

𝛀 = ∫ ∫ ∫𝒙𝒚𝒛

(𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙)

𝟏

𝟎


𝟎

𝟏

𝟎

= 𝟐 𝐥𝐨𝐠 𝟐 − 𝐥𝐨𝐠𝟐 𝟐 −𝝅𝟐

𝟏𝟐


𝛀 = ∫𝒙 𝐭𝐚𝐧−𝟏 𝒙

𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙

∞

𝟎




𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙

∞

𝟎

=𝒙=𝟏𝒙= ∫

𝒙 𝐭𝐚𝐧−𝟏 (𝟏𝒙)

𝟏 − 𝒙𝟐 + 𝒙𝟒

∞

𝟎

𝒅𝒙

∵ 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 (𝟏

𝒙) =

𝝅

𝟐

𝟐𝛀 =𝝅

𝟐∫

𝒙

𝟏 − 𝒙𝟐 + 𝒙𝟒

∞

𝟎

𝒅𝒙 =𝒙𝟐=𝒙 𝝅

𝟒∫

𝟏

𝟏 − 𝒙 + 𝒙𝟐

∞

𝟎

𝒅𝒙

𝛀 =𝝅

𝟖∫

𝟏

(𝒙 −𝟏𝟐)𝟐

+ (√𝟑𝟐)

𝟐

∞

𝟎

𝒅𝒙 =𝝅

𝟒√𝟑[𝐭𝐚𝐧−𝟏(

𝒙 −𝟏𝟐

√𝟑𝟐

)]

𝟎

∞

=𝝅

𝟒√𝟑(𝝅

𝟐+𝝅

𝟔)


𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙

∞

𝟎

=𝝅𝟐

𝟔√𝟑

Solution 2 by Sesiqakbar Restheen-Afghanistan

∵ 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 (𝟏

𝒙) =

𝝅

𝟐,∀𝒙 ∈ ℝ


𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙

∞

𝟎

=𝒖=𝟏𝒙∫

𝟏𝒖 𝐭𝐚𝐧

−𝟏 (𝟏𝒖)

𝟏𝒖𝟒−𝟏𝒖𝟐+ 𝟏

∞

𝟎

(−𝒅𝒖

𝒖𝟐) =

= ∫𝒖(𝝅𝟐 − 𝐭𝐚𝐧

−𝟏 𝒖)

𝒖𝟒 − 𝒖𝟐 + 𝟏𝒅𝒖

∞

𝟎

=𝝅

𝟒∫

𝒖

𝒖𝟒 − 𝒖𝟐 + 𝟏

∞

𝟎

𝒅𝒖

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∫𝒖

𝒖𝟒 − 𝒖𝟐 + 𝟏𝒅𝒖 =

𝟏

√𝟑𝐭𝐚𝐧−𝟏 [

𝟏

√𝟑(𝟐𝒙𝟐 − 𝟏)] + 𝑪

∫𝒖

𝒖𝟒 − 𝒖𝟐 + 𝟏

∞

𝟎

𝒅𝒖 =𝟏

√𝟑[𝝅

𝟐− 𝐭𝐚𝐧−𝟏 (−

𝟏

√𝟑)] =

𝟏

√𝟑[𝝅

𝟐—𝝅

𝟔) =

𝟐𝝅

𝟑√𝟑

Therefore,


𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙

∞

𝟎

=𝝅𝟐

𝟔√𝟑

Solution 3 by Ghuiam Naseri-Afghanistan

∵ 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 (𝟏

𝒙) =

𝝅

𝟐,∀𝒙 ∈ ℝ


𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙

∞

𝟎

=𝒙=𝟏𝒙∫

𝒙𝟐 𝐭𝐚𝐧−𝟏 (𝟏𝒙)

𝒙𝟒 − 𝒙𝟐 + 𝟏

∞

𝟎

𝒅𝒙

𝝅

𝟐∫

𝒙

𝒙𝟒 − 𝒙𝟐 + 𝟏

∞

𝟎

𝒅𝒙 =𝒙𝟐=𝒙 𝝅

𝟒∫

𝟏

𝒙𝟐 − 𝒙 + 𝟏𝒅𝒙

∞

𝟎

=

=𝝅

𝟒⋅𝟐

√𝟑𝐭𝐚𝐧−𝟏 (

𝟐𝒙 − 𝟏

√𝟑) =𝒙=𝒙𝟐 𝝅

𝟒⋅𝟐

√𝟑[𝐭𝐚𝐧−𝟏 (

𝟐𝒙𝟐 − 𝟏

√𝟑)]𝟎

∞

=𝝅𝟐

𝟔√𝟑

1572.

𝛀(𝒏) = ∫𝒙𝒏−𝟏(𝒙 − 𝒏) 𝐥𝐨𝐠 𝒙

𝒆𝒙𝒅𝒙

∞

𝟎

, 𝒏 ≥ 𝟏

Find:

𝛀 =∑𝟏

𝛀(𝒏)

∞

𝒏=𝟏



𝛀(𝒏) = ∫ 𝒆−𝒙𝒙𝒏 𝐥𝐨𝐠 𝒙∞

𝟎

𝒅𝒙 − 𝒏∫ 𝒙𝒏−𝟏𝒆−𝒙 𝐥𝐨𝐠 𝒙∞

𝟎

𝒅𝒙 =

=𝝏

𝝏𝒔|𝒔=𝟎(∫ 𝒆−𝒙𝒙(𝒏+𝒔+𝟏)−𝟏

∞

𝟎

𝒅𝒙 − 𝒏∫ 𝒙𝒏+𝒔−𝟏𝒆−𝒙∞

𝟎

𝒅𝒙) =

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=𝝏

𝝏𝒔|𝒔=𝟎

(𝚪(𝒏 + 𝒔 + 𝟏) − 𝒏𝚪(𝒏 + 𝒔)) = 𝚪′(𝒏 + 𝟏) − 𝒏𝚪′(𝒏) =

= ∑𝟏

𝚪(𝒏 + 𝟏)𝝍(𝒏+ 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏)

∞

𝒏=𝟏

= ∑𝟏

𝚪(𝒏 + 𝟏)𝝍(𝒏+ 𝟏) − 𝚪(𝒏 + 𝟏)𝝍(𝒏)

∞

𝒏=𝟏

=

= ∑𝟏

𝚪(𝒏 + 𝟏)(𝝍(𝒏 + 𝟏) − 𝝍(𝒏))

∞

𝒏=𝟏

= ∑𝟏

𝚪(𝒏 + 𝟏) (𝟏𝒏)

∞

𝒏=𝟏

= ∑𝒏

𝚪(𝒏 + 𝟏)

∞

𝒏=𝟏

= ∑𝟏

𝚪(𝒏)

∞

𝒏=𝟏

= 𝒆

Therefore,

𝛀 = ∑𝟏

𝛀(𝒏)

∞

𝒏=𝟏

= 𝒆


𝝍(𝒛) =𝟏

𝚪(𝒛)∫ 𝒙𝒛−𝟏𝒆−𝒙 𝐥𝐨𝐠 𝒙𝒅𝒙∞

𝟎

𝛀(𝒏) = ∫ 𝒙𝒏+𝟏𝒆−𝒙 𝐥𝐨𝐠 𝒙 𝒅𝒙∞

𝟎

− 𝒏∫ 𝒙𝒏−𝟏𝒆−𝒙 𝐥𝐨𝐠𝒙𝒅𝒙∞

𝟎

=

= 𝚪(𝒏 + 𝟏)𝝍(𝒏 + 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏) = 𝒏! [𝝍(𝒏+ !) − 𝝍(𝒏)] =

= 𝒏! (𝟏

𝒏) = (𝒏 − 𝟏)!

Therefore,

𝛀 = ∑𝟏

𝛀(𝒏)

∞

𝒏=𝟏

= ∑𝟏

(𝒏 − 𝟏)!

∞

𝒏−𝟏

= 𝒆

Solution 3 by Santiago Alvarez-Mexico

𝛀(𝒏) = ∫𝒙𝒏−𝟏(𝒙 − 𝒏) 𝐥𝐨𝐠𝒙

𝒆𝒙𝒅𝒙

∞

𝟎

= (𝝏

𝝏𝒔∫ 𝒙𝒏−𝟏+𝒔(𝒙 − 𝒏)𝒆−𝒙𝒅𝒙∞

𝟎

)𝒔=𝟎

=

= (𝝏

𝝏𝒔∫ 𝒙𝒏+𝒔𝒆−𝒙𝒅𝒙∞

𝟎

− 𝒏𝝏

𝝏𝒔∫ 𝒙𝒏+𝒔−𝟏𝒆−𝒙𝒅𝒙∞

𝟎

)𝒔=𝟎

=

= (𝝏

𝝏𝒔𝚪(𝒏 + 𝒔 + 𝟏) − 𝒏

𝝏

𝝏𝒔𝚪(𝒏 + 𝒔))

𝒔=𝟎

=

= (𝚪(𝒏 + 𝒔 + 𝟏)𝝍(𝒏 + 𝒔 + 𝟏) − 𝒏𝚪(𝒏 + 𝒔)𝝍(𝒏 + 𝒔))𝒔=𝟎

=

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= 𝚪(𝒏 + 𝟏)𝝍(𝒏+ 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏)

𝛀 = ∑𝟏

𝚪(𝒏 + 𝟏)𝝍(𝒏 + 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏)

∞

𝒏=𝟏

= ∑𝟏

𝒏𝚪(𝒏)𝝍(𝒏 + 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏)

∞

𝒏=𝟏

=

= ∑𝟏

𝒏𝚪(𝒏) (𝟏𝒏)

∞

𝒏=𝟏

= ∑𝟏

(𝒏 − 𝟏)!

∞

𝒏=𝟏

= ∑𝟏

𝒏!

∞

𝒏=𝟎

= 𝒆

1573. Find:

𝛀 = ∫𝒙𝟒

𝒙𝟒 𝐥𝐨𝐠𝟒 𝟒 + 𝟒(𝒙𝟑 𝐥𝐨𝐠𝟑 𝟒 + 𝟑𝒙𝟐 𝐥𝐨𝐠𝟐 𝟒 + 𝟔𝒙 𝐥𝐨𝐠 𝟒 + 𝟔 + 𝟔 ⋅ 𝟒𝒙)𝒅𝒙


Solution by Almas Babirov-Azerbaijan

𝛀 = ∫𝒙𝟒

𝒙𝟒 𝐥𝐨𝐠𝟒 𝟒 + 𝟒(𝒙𝟑 𝐥𝐨𝐠𝟑 𝟒 + 𝟑𝒙𝟐 𝐥𝐨𝐠𝟐 𝟒 + 𝟔𝒙 𝐥𝐨𝐠 𝟒 + 𝟔 + 𝟔 ⋅ 𝟒𝒙)𝒅𝒙 =

= ∫𝒙𝟒

𝒙𝟒 𝐥𝐨𝐠𝟒 𝟒 + 𝟒𝒙𝟑 𝐥𝐨𝐠𝟑 𝟒 + 𝟏𝟐𝒙𝟐 𝐥𝐨𝐠𝟐 𝟒 + 𝟐𝟒𝒙 𝐥𝐨𝐠 𝟒 + 𝟐𝟒 + 𝟐𝟒 ⋅ 𝟒𝒙𝒅𝒙 = (∗)

∵ 𝐥𝐨𝐠 𝟒𝒙 = 𝒕 ⇒ 𝟒𝒙 = 𝒆𝒕 ⇒ 𝒙 = 𝐥𝐨𝐠𝟒 𝒆𝒕 = 𝒕 𝐥𝐨𝐠𝟒 𝒆

(∗) = ∫𝒕𝟒 𝐥𝐨𝐠𝟒

𝟒 𝒆

𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕𝒅(𝒕 𝐥𝐨𝐠𝟒 𝒆) =

= 𝐥𝐨𝐠𝟒𝟓 𝒆 ⋅ ∫(

𝒕𝟒 − (𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕)

𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕+ 𝟏)𝒅𝒕 =

= 𝒕 𝐥𝐨𝐠𝟒𝟓 𝒆 − 𝐥𝐨𝐠𝟒

𝟓 𝒆 ⋅ ∫𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕

𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕𝒅𝒕 =

=𝒕

𝐥𝐨𝐠𝟓 𝟒−

𝟏

𝐥𝐨𝐠𝟓 𝟒⋅ ∫𝒅(𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕)

𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕=

=𝒕

𝐥𝐨𝐠𝟓 𝟒−

𝟏

𝐥𝐨𝐠𝟓 𝟒⋅ 𝐥𝐨𝐠(𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕) + 𝑪 =

=𝒙

𝐥𝐨𝐠𝟒 𝟒−

𝟏

𝐥𝐨𝐠𝟓 𝟒⋅ 𝐥𝐨𝐠(𝒙𝟒 𝐥𝐨𝐠𝟒 𝟒 + 𝟒𝒙𝟑 𝐥𝐨𝐠𝟑 𝟒 + 𝟏𝟐𝒙𝟐 𝐥𝐨𝐠𝟐 𝟒 + 𝟐𝟒𝒙 𝐥𝐨𝐠𝟒 + 𝟐𝟒 + 𝟐𝟒 ⋅ 𝟒𝒙) + 𝑪

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1574. If 𝟎 < 𝑎 < 4𝒃 then find:

𝛀 = ∫𝐬𝐢𝐧−𝟏(√𝒙)

𝒂𝒙𝟐 − 𝒂𝒙 + 𝒃

𝟏

𝟎

𝒅𝒙

Proposed by Marin Chirciu-Romania

Solution by Marian Ursărescu-Romania

𝛀 = ∫𝐬𝐢𝐧−𝟏(√𝒙)


𝟏

𝟎

𝒅𝒙 =𝒙=𝟏−𝒕

∫𝐬𝐢𝐧−𝟏(√𝟏 − 𝒕)

𝒂(𝟏 − 𝒕)𝟐 − 𝒂(𝟏 − 𝒕) + 𝒃

𝟎

𝟏

(−𝒅𝒕) =

= ∫𝐬𝐢𝐧−𝟏(√𝟏 − 𝒕)

𝒂 − 𝟐𝒂𝒕 + 𝒂𝒕𝟐 − 𝒂 + 𝒂𝒕 + 𝒃𝒅𝒕

𝟏

𝟎

= ∫𝐬𝐢𝐧−𝟏(√𝟏 − 𝒕)

𝒂𝒕𝟐 − 𝒂𝒕 + 𝒃𝒅𝒕

𝟏

𝟎

𝚪 = ∫𝐬𝐢𝐧−𝟏(√𝟏 − 𝒕)

𝒂𝒕𝟐 − 𝒂𝒕 + 𝒃𝒅𝒕

𝟏

𝟎

⇒ 𝛀 = 𝚪.

But 𝐬𝐢𝐧−𝟏(√𝒙) + 𝐬𝐢𝐧−𝟏(√𝟏 − 𝒙) =𝝅

𝟐; ∀𝒙 ∈ [𝟎, 𝟏], because

𝒇(𝒙) = 𝐬𝐢𝐧−𝟏(√𝒙) + 𝐬𝐢𝐧−𝟏(√𝟏 − 𝒙) ; 𝒇′(𝒙) = 𝟎 ⇒ 𝒇(𝒙) = 𝒄 =𝝅

𝟐⇒

𝛀 + 𝚪 = ∫𝐬𝐢𝐧−𝟏(√𝒙) + 𝐬𝐢𝐧−𝟏(√𝟏 − 𝒙)

𝒂𝒙𝟐 − 𝒂𝒙 + 𝒃𝒅𝒙

𝟏

𝟎

=𝝅

𝟐∫

𝒅𝒙


𝟏

𝟎

=

=𝝅

𝟐𝒂∫

𝒅𝒙

𝒙𝟐 − 𝒙 +𝒃𝒂

𝟏

𝟎

=𝝅

𝟐𝒂∫

𝒅𝒙

(𝒙 −𝟏𝟐)𝟐

+𝟒𝒃 − 𝒂𝟒𝒂

𝟏

𝟎

=𝝅

𝟐𝒂∫

𝒅𝒙

(𝒙 −𝟏𝟐)𝟐

+ (√𝟒𝒃− 𝒂𝟒𝒂 )

𝟐

𝟏

𝟎

𝝅

𝟐𝒂⋅ 𝟐√

𝒂

𝟒𝒃 − 𝒂⋅ 𝐭𝐚𝐧−𝟏

𝒙 −𝟏𝟐

√𝟒𝒃 − 𝒂𝒂

||

𝟎

𝟏

=𝝅

√𝒂(𝟒𝒃 − 𝒂)⋅ 𝟐 𝐭𝐚𝐧−𝟏√

𝒂

𝟒𝒃 − 𝒂; (𝟐)


𝛀 =𝝅

√𝒂(𝟒𝒃 − 𝒂)⋅ 𝟐 𝐭𝐚𝐧−𝟏√

𝒂

𝟒𝒃 − 𝒂

1575.

𝒇 ∈ 𝑪𝟏([𝟎, 𝟑]), 𝒇(𝟎) = 𝟒, 𝒇(𝟑) = 𝒌, 𝒇′(𝒙) = 𝐭𝐚𝐧−𝟏 𝒙

Find:

𝛀(𝒌) = ∫ 𝒙𝟐𝒇(𝒙)𝟑

𝟎

𝒅𝒙


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𝛀(𝒌) = ∫ 𝒙𝟐𝒇(𝒙)𝟑

𝟎

𝒅𝒙 =𝒙𝟑

𝟑𝒇(𝒙)|

𝟎

𝟑

−𝟏

𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑

𝟎

𝒅𝒙 = 𝟗𝒇(𝟑) −𝟏

𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑

𝟎

𝒅𝒙 =

= 𝟗𝒌 −𝟏

𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑

𝟎

𝒅𝒙

𝛀(𝒌) = 𝟗𝒌 −𝒙𝟒 𝐭𝐚𝐧−𝟏 𝒙

𝟏𝟐|𝟎

𝟑

+𝟏

𝟏𝟐∫

𝒙𝟒

𝟏 + 𝒙𝟐𝒅𝒙

𝟑

𝟎

=

= 𝟗𝒌 −𝟐𝟕

𝟒𝐭𝐚𝐧−𝟏 𝟑 +

𝟏

𝟏𝟐∫ (𝒙𝟐 − 𝟏 +

𝟏

𝒙𝟐 + 𝟏)𝒅𝒙

𝟑

𝟎

=

= 𝟗𝒌−𝟐𝟕

𝟒𝐭𝐚𝐧−𝟏 𝟑 +

𝟏

𝟏𝟐⋅𝒙𝟑

𝟑|𝟎

𝟑

−𝟏

𝟏𝟐⋅ 𝒙|

𝟎

𝟑

+𝟏

𝟏𝟐𝐭𝐚𝐧−𝟏 𝒙|

𝟎

𝟑

=

= 𝟗𝒌 −𝟐𝟕

𝟒𝐭𝐚𝐧−𝟏 𝟑 +

𝟑

𝟒−𝟏

𝟒+𝟏

𝟏𝟐𝐭𝐚𝐧−𝟏 𝟑 = 𝟗𝒌 −

𝟐𝟎

𝟑𝐭𝐚𝐧−𝟏 𝟑 +

𝟏

𝟐

Solution 2 by Fayssal Abdelli-Bejaia-Algerie

𝛀(𝒌) = ∫ 𝒙𝟐𝒇(𝒙)𝟑

𝟎

𝒅𝒙 =𝒙𝟑

𝟑𝒇(𝒙)|

𝟎

𝟑

−𝟏

𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑

𝟎

𝒅𝒙 =

𝟗𝒌𝟐 −𝟏

𝟑𝚪;

𝚪 = ∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑

𝟎

𝒅𝒙 =𝒙𝟒

𝟒𝐭𝐚𝐧−𝟏 𝒙|

𝟎

𝟑

−𝟏

𝟒∫

𝒙𝟒

𝟏 + 𝒙𝟐𝒅𝒙

𝟑

𝟎

=

=𝟖𝟏

𝟒𝐭𝐚𝐧−𝟏 𝒙 −

𝟏

𝟒∫(𝒙𝟐 − 𝟏)(𝒙𝟐 + 𝟏)

𝒙𝟐 + 𝟏𝒅𝒙

𝟑

𝟎

−∫𝒅𝒙

𝟏 + 𝒙𝟐

𝟑

𝟎

=

=𝟖𝟏

𝟒𝐭𝐚𝐧−𝟏 𝟑 −

𝟏

𝟒𝐭𝐚𝐧−𝟏 𝒙|

𝟎

𝟑

−𝟏

𝟒⋅ (𝒙𝟑

𝟑− 𝒙)|

𝟎

𝟑

= 𝟐𝟎 𝐭𝐚𝐧−𝟏 𝟑 −𝟑

𝟐

Therefore,

𝛀(𝒌) = 𝟗𝒌 −𝟐𝟎

𝟑𝐭𝐚𝐧−𝟏 𝟑 +

𝟏

𝟐

Solution 3 by Soumitra Mandal-Chandar Nagore-India

𝒇′(𝒙) = 𝐭𝐚𝐧−𝟏 𝒙 ⇒ ∫ 𝒇′(𝒙)𝟑

𝟎

𝒅𝒙 = ∫ 𝐭𝐚𝐧−𝟏 𝒙𝟑

𝟎

𝒅𝒙 ⇒

𝒇(𝟑) − 𝒇(𝟎) = 𝒙 𝐭𝐚𝐧−𝟏 𝒙|𝟎

𝟑−∫

𝒙

𝟏 + 𝒙𝟐𝒅𝒙

𝟑

𝟎

= 𝟑 𝐭𝐚𝐧−𝟏 𝟑 −𝟏

𝟐𝐥𝐨𝐠(𝟏 + 𝒙𝟐)|

𝟎

𝟑

=

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= 𝟑 𝐭𝐚𝐧−𝟏 𝟑 −𝟏

𝟐⇒ 𝒌 − 𝟒 = 𝟑 𝐭𝐚𝐧−𝟏 𝟑 −

𝟏

𝟐⇒ 𝒌 =

𝟕

𝟐+ 𝟑 𝐭𝐚𝐧−𝟏 𝟑

𝛀(𝒌) = ∫ 𝒙𝟐𝒇(𝒙)𝟑

𝟎

𝒅𝒙 = 𝟗𝒇(𝟑) −𝟏

𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝟑

𝟎

=

= 𝟗𝒇(𝟑) −𝟏

𝟏𝟐𝒙𝟒 𝐭𝐚𝐧−𝟏 𝒙|

𝟎

𝟑

+𝟏

𝟏𝟐∫

𝒙𝟒

𝟏 + 𝒙𝟐𝒅𝒙

𝟑

𝟎

=

= 𝟗𝒌 −𝟐𝟕

𝟒𝐭𝐚𝐧−𝟏 𝟑 +

𝟏

𝟏𝟐∫ 𝒙𝟐 (𝟏 −

𝟏

𝟏 + 𝒙𝟐)

𝟑

𝟎

𝒅𝒙 =

= 𝟗𝒌−𝟐𝟕

𝟒𝐭𝐚𝐧−𝟏 𝟑 +

𝟏

𝟑𝟔(𝟑𝟑 − 𝟎) −

𝟏

𝟏𝟐∫

𝒙𝟐

𝟏 + 𝒙𝟐𝒅𝒙

𝟑

𝟎

=

= 𝟗(𝟕

𝟐+ 𝟑 𝐭𝐚𝐧−𝟏 𝟑) −

𝟐𝟕

𝟒𝐭𝐚𝐧−𝟏 𝟑 +

𝟑

𝟒−𝟏

𝟏𝟐∫ (𝟏 −

𝟏

𝟏 + 𝒙𝟐)

𝟑

𝟎

𝒅𝒙 =

=𝟔𝟑

𝟐+𝟖𝟏

𝟒𝐭𝐚𝐧−𝟏 𝟑 +

𝟑

𝟒−𝟏

𝟒+𝟏

𝟏𝟐𝐭𝐚𝐧−𝟏 𝒙|

𝟎

𝟑

=

= 𝟑𝟐 +𝟏𝟐𝟏

𝟔𝐭𝐚𝐧−𝟏 𝟑

1576. If 𝟎 < 𝒂 ≤ 𝒃 <𝝅

𝟐 then:

∫𝐭𝐚𝐧 (

𝝅 − 𝟐𝒙𝟒

) (𝟏 + 𝐬𝐢𝐧 𝒙)

𝐬𝐢𝐧 𝒙𝒅𝒙

𝒃

𝒂

≤𝐬𝐢𝐧 𝒃 − 𝐬𝐢𝐧 𝒂

𝐬𝐢𝐧 𝒂


Solution by Ravi Prakash-New Delhi-India

𝐭𝐚𝐧 (𝝅− 𝟐𝒙𝟒

)(𝟏 + 𝐬𝐢𝐧𝒙)

𝐬𝐢𝐧 𝒙=

𝟏 − 𝐭𝐚𝐧𝒙𝟐

𝟏 + 𝐭𝐚𝐧𝒙𝟐

(𝐜𝐨𝐬𝒙𝟐 + 𝐬𝐢𝐧

𝒙𝟐)𝟐

𝐬𝐢𝐧𝒙=

=

𝐜𝐨𝐬𝒙𝟐 − 𝐬𝐢𝐧

𝒙𝟐

𝐜𝐨𝐬𝒙𝟐 + 𝐬𝐢𝐧

𝒙𝟐

(𝐜𝐨𝐬𝒙𝟐 + 𝐬𝐢𝐧

𝒙𝟐)𝟐

𝐬𝐢𝐧 𝒙=(𝐜𝐨𝐬

𝒙𝟐 − 𝐬𝐢𝐧

𝒙𝟐) (𝐜𝐨𝐬

𝒙𝟐 + 𝐬𝐢𝐧

𝒙𝟐)

𝐬𝐢𝐧 𝒙=

=𝐜𝐨𝐬𝟐

𝒙𝟐 − 𝐬𝐢𝐧

𝟐 𝒙𝟐

𝐬𝐢𝐧 𝒙=𝐜𝐨𝐬 𝒙

𝐬𝐢𝐧𝒙

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Hence,

∫𝐭𝐚𝐧 (

𝝅− 𝟐𝒙𝟒 )(𝟏 + 𝐬𝐢𝐧𝒙)

𝐬𝐢𝐧 𝒙𝒅𝒙

𝒃

𝒂

= 𝐥𝐨𝐠(𝐬𝐢𝐧 𝒙)|𝒂𝒃 = 𝐥𝐨𝐠 (

𝐬𝐢𝐧𝒃

𝐬𝐢𝐧𝒂) = 𝐥𝐨𝐠 (𝟏 −

𝐬𝐢𝐧𝒃 − 𝐬𝐢𝐧𝒂

𝐬𝐢𝐧𝒂)

≤

≤𝐬𝐢𝐧 𝒃 − 𝐬𝐢𝐧 𝒂

𝐬𝐢𝐧𝒂; (∵ 𝐥𝐨𝐠(𝟏 + 𝒙) ≤ 𝒙, ∀𝒙 ≥ 𝟎)

1577.

𝛀(𝒂) = ∫ 𝐥𝐨𝐠(𝟏 + 𝒙) ⋅ 𝐭𝐚𝐧−𝟏(√𝒙)𝒅𝒙𝒂

𝟎

, 𝒂 > 𝟎

Prove that:

𝛀(𝒂) + 𝛀(𝒃) + 𝛀(𝒄) < (𝒂 + 𝒃 + 𝒄) (𝒂 + 𝒃 + 𝒄 +𝟏

𝟐) , ∀𝒂, 𝒃, 𝒄 > 𝟎



We have: 𝒆𝒙 ≥ 𝒙 + 𝟏,∀𝒙 > 𝟎 ⇔ 𝐥𝐨𝐠(𝟏 + 𝒙) ≤ 𝒙, ∀𝒙 > 𝟎 and 𝐭𝐚𝐧−𝟏(√𝒙) ≤𝝅

𝟐, ∀𝒙 > 𝟎.

Therefore, 𝛀(𝒂) ≤𝝅

𝟐∫ 𝒙𝒃

𝒂𝒅𝒙 ⇔ 𝛀(𝒂) ≤

𝝅

𝟒𝒂𝟐. Hence,

𝛀(𝒂) + 𝛀(𝒃) + 𝛀(𝒄) ≤𝝅

𝟒(𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐) < 𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐 <

< (𝒂+ 𝒃 + 𝒄)𝟐 < (𝒂 + 𝒃 + 𝒄)𝟐 +𝟏

𝟐(𝒂 + 𝒃 + 𝒄) = (𝒂 + 𝒃 + 𝒄) (𝒂 + 𝒃 + 𝒄 +

𝟏

𝟐)


First, we prove that: 𝐥𝐨𝐠(𝟏 + 𝒙) 𝐭𝐚𝐧−𝟏(√𝒙) < 𝟐𝒙 +𝟏

𝟐.

If 𝒇(𝒙) = 𝐥𝐨𝐠(𝟏 + 𝒙) 𝐭𝐚𝐧−𝟏(√𝒙) − 𝟐𝒙 −𝟏

𝟐, then 𝒇′(𝒙) =

𝐭𝐚𝐧−𝟏(√𝒙)

𝟏+𝒙+𝐥𝐨𝐠(𝟏+𝒙)

𝟐(𝟏+𝒙)√𝒙− 𝟐

𝐭𝐚𝐧−𝟏(√𝒙) < √𝒙 < 𝟐√𝒙 < 𝟏 + 𝒙 ⇒𝐭𝐚𝐧−𝟏(√𝒙)

𝟏 + 𝒙< 𝟏; (𝟏)

𝐥𝐨𝐠(𝟏 + 𝒙) < 𝒙 <(∗)

𝟐(𝟏 + 𝒙)√𝒙 ⇔ √𝒙 < 𝟐(𝟏 + 𝒙)

√𝒙 ≤𝒙+𝟏

𝟐< 𝟐(𝒙 + 𝟏) ⇔ (∗) true. Thus,

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𝐥𝐨𝐠(𝟏 + 𝒙)

𝟐(𝟏 + 𝒙)√𝒙< 𝟏; (𝟐)

From (1),(2) we have 𝒇′(𝒙) < 𝟎 ⇒ 𝒇(𝒙) < 𝒇(𝟎) = −𝟏

𝟐< 𝟎.

Now, integrating the inequality: 𝐥𝐨𝐠(𝟏 + 𝒙) 𝐭𝐚𝐧−𝟏(√𝒙) < 𝟐𝒙 +𝟏

𝟐, we get:

𝛀(𝒂) < ∫ (𝟐𝒙 +𝟏

𝟐)𝒅𝒙

𝒂

𝟎

= 𝒂𝟐 +𝟏

𝟐𝒂

Therefore,

𝛀(𝒂) + 𝛀(𝒃) + 𝛀(𝒄) < 𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐 +𝟏

𝟐(𝒂 + 𝒃 + 𝒄) <

< (𝒂 + 𝒃 + 𝒄)𝟐 +𝟏

𝟐(𝒂 + 𝒃 + 𝒄) = (𝒂 + 𝒃 + 𝒄) (𝒂 + 𝒃 + 𝒄 +

𝟏

𝟐)


𝛀(𝒂) = ∫ 𝐥𝐨𝐠(𝟏 + 𝒙) ⋅𝒂

𝟎

𝐭𝐚𝐧−𝟏(√𝒙)𝒅𝒙 = ∫𝐥𝐨𝐠(𝟏 + 𝒙)

√𝒙

𝒂

𝟎

⋅ √𝒙 𝐭𝐚𝐧−𝟏(√𝒙)𝒅𝒙

Let be the function 𝒇: ℝ → ℝ, 𝒇(𝒙) = 𝐭𝐚𝐧−𝟏 𝒙 −𝟏

𝒙𝟐+𝟏⋅ 𝐭𝐚𝐧−𝟏 𝒙

𝒇′(𝒙) =𝟐𝒙⋅𝐭𝐚𝐧−𝟏 𝒙+𝒙𝟐

(𝒙𝟐+𝟏)𝟐 ≥ 𝟎 ⇒ 𝒇 −increasing on [𝟎,∞) ⇒ 𝒇(𝒙) ≥ 𝒇(𝟎) = 𝟎

𝟎 <𝒙𝟐

𝒙𝟐 + 𝟏⋅ 𝐭𝐚𝐧−𝟏 𝒙 < 𝟏 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 < 𝒙;∀𝒙 ≥ 𝟎 ⇒

𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙

𝒙𝟐 + 𝟏< 𝟏 ⟺

𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 < 𝒙𝟐 + 𝟏 ⟺ √𝒙 ⋅ 𝐭𝐚𝐧−𝟏 √𝒙 < 𝒙 + 𝟏;∀𝒙 > 𝟎; (𝟏)

It is well-known: 𝐥𝐨𝐠(𝟏 + 𝒙) ≤𝒙

√𝟏+𝒙; ∀𝒙 ≥ 𝟎 ⇒

𝐥𝐨𝐠(𝟏 + 𝒙)

√𝒙≤

√𝒙

√𝒙 + 𝟏;∀𝒙 ≥ 𝟎; (𝟐)


𝛀(𝒂) = ∫𝐥𝐨𝐠(𝟏 + 𝒙)

√𝒙

𝒂

𝟎

⋅ √𝒙 𝐭𝐚𝐧−𝟏(√𝒙)𝒅𝒙 < ∫√𝒙

√𝒙 + 𝟏⋅ (𝒙 + 𝟏)

𝒂

𝟎

𝒅𝒙 =

= ∫ √𝒙(𝒙 + 𝟏)𝒂

𝟎

𝒅𝒙 ≤𝑨𝑴−𝑮𝑴

∫ (𝒙 +𝟏

𝟐)𝒅𝒙

𝒂

𝟎

=𝟏

𝟐(𝒂𝟐 + 𝒂)

Therefore,

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𝛀(𝒂) + 𝛀(𝒃) + 𝛀(𝒄) <𝟏

𝟐(𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐) +

𝟏

𝟐(𝒂 + 𝒃 + 𝒄) < (𝒂 + 𝒃 + 𝒄)𝟐 +

𝟏

𝟐(𝒂 + 𝒃 + 𝒄)

= (𝒂 + 𝒃 + 𝒄) (𝒂 + 𝒃 + 𝒄 +𝟏

𝟐)

1578. If 𝟎 < 𝒂 ≤ 𝒃 then:

∫𝟏

√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (

𝒌𝝅

𝟐𝒏 + 𝟏)

[𝒙]

𝒌=𝟏

𝒅𝒙𝒃

𝒂

≥𝟏

𝟐𝒂−𝟏

𝟐𝒃, [∗] − 𝑮𝑰𝑭.


Solution by Asmat Qatea-Afghanistan

∫𝟏

√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (

𝒌𝝅

𝟐𝒏 + 𝟏)

[𝒙]

𝒌=𝟏

𝒅𝒙𝒃

𝒂

≥𝟏

𝟐𝒂−𝟏

𝟐𝒃⇔

∫𝟏

√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (

𝒌𝝅

𝟐𝒏 + 𝟏)

[𝒙]

𝒌=𝟏

𝒅𝒙𝒃

𝒂

≥(?)

𝐥𝐨𝐠𝟐∫ 𝟐−𝒙𝒃

𝒂

𝒅𝒙

𝟏

√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (

𝒌𝝅

𝟐𝒏 + 𝟏)

[𝒙]

𝒌=𝟏

≥(?)

𝟐−𝒙 𝐥𝐨𝐠 𝟐 ⇒ 𝒙 ∈ [𝟎,∞)

Case 1. If 𝒙 ∈ [𝟎, 𝟏) then:

𝟏

√𝟐 ⋅ 𝟎 + 𝟏⋅∏𝐬𝐢𝐧 (

𝒌𝝅

𝟐𝒏 + 𝟏)

𝟎

𝒌=𝟏

≥(?)

𝟐−𝒙 𝐥𝐨𝐠 𝟐 ⇒ 𝟏 ≥ 𝟐−𝒙 𝐥𝐨𝐠𝟐 ⇒ 𝟐𝒙 ≥ 𝐥𝐨𝐠 𝟐 − 𝐭𝐫𝐮𝐞.

Case 2. If 𝒙 ∈ [𝒏, 𝒏 + 𝟏), 𝒏 ∈ ℕ then:

𝟏

√𝟐 ⋅ 𝒏 + 𝟏⋅∏𝐬𝐢𝐧 (

𝒌𝝅

𝟐𝒏 + 𝟏)

𝒏

𝒌=𝟏

≥(?)

𝟐−𝒙 𝐥𝐨𝐠 𝟐

∵∏𝐬𝐢𝐧 (𝒌𝝅

𝟐𝒏 + 𝟏)

𝒏

𝒌=𝟏

=√𝟐𝒏 + 𝟏

𝟐𝒏

𝟏

𝟐𝒏≥(?)

𝟐−𝒙 𝐥𝐨𝐠𝟐 ⇒ 𝟐𝒙−𝒏 ≥ 𝐥𝐨𝐠𝟐 − 𝐭𝐫𝐮𝐞, 𝐛𝐞𝐜𝐚𝐮𝐬𝐞 𝒙 − 𝒏 ≥ 𝟎

Therefore,

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∫𝟏

√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (

𝒌𝝅

𝟐𝒏 + 𝟏)

[𝒙]

𝒌=𝟏

𝒅𝒙𝒃

𝒂

≥𝟏

𝟐𝒂−𝟏

𝟐𝒃

1579.

𝒙𝒏 = (𝒏

𝟎)𝒑𝒏 + (

𝒏

𝟑)𝒑𝒏−𝟑 + (

𝒏

𝟔)𝒑𝒏−𝟔 +⋯ ;𝒚𝒏 = (

𝒏

𝟏)𝒑𝒏−𝟏 + (

𝒏

𝟒)𝒑𝒏−𝟒 + (

𝒏

𝟕)𝒑𝒏−𝟕

𝒛𝒏 = (𝒏

𝟐)𝒑𝒏−𝟐 + (

𝒏

𝟓)𝒑𝒏−𝟓 + (

𝒏

𝟖)𝒑𝒏−𝟖 +⋯ ;𝒏 ∈ ℕ, 𝒑 ≥ 𝟐. 𝐅𝐢𝐧𝐝:

𝛀(𝒑) = 𝐥𝐢𝐦𝒏→∞

√𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏𝒏

Proposed by Marian Ursărescu-Romania


(𝒏

𝒌) = (

𝒏

𝒏 − 𝒌) ⇒ 𝒙𝒏 = (

𝒏

𝒏)𝒑𝒏 + (

𝒏

𝒏 − 𝟑)𝒑𝒏−𝟑 + (

𝒏

𝒏 − 𝟔)𝒑𝒏−𝟔 +⋯

𝒚𝒏 = (𝒏

𝒏 − 𝟏)𝒑𝒏−𝟏 + (

𝒏

𝒏 − 𝟒)𝒑𝒏−𝟒 + (

𝒏

𝒏 − 𝟕)𝒑𝒏−𝟕 +⋯

𝒛𝒏 = (𝒏

𝒏 − 𝟐)𝒑𝒏−𝟐 + (

𝒏

𝒏 − 𝟓)𝒑𝒏−𝟓 + (

𝒏

𝒏 − 𝟖)𝒑𝒏−𝟖 +⋯

Let 𝜺 be root by three order of unity, hence 𝜺𝟐 + 𝜺+ 𝟏 = 𝟎 and 𝜺𝟑 = 𝟏.

(𝟏 + 𝒑)𝒏 = (𝒏

𝟎) + (

𝒏

𝟏)𝒑 + (

𝒏

𝟐)𝒑𝟐 + (

𝒏

𝟑)𝒑𝟑 +⋯ ; (𝟏)

(𝟏 + 𝜺𝒑)𝒏 = (𝒏

𝟎) + (

𝒏

𝟏) 𝜺𝒑 + (

𝒏

𝟐) (𝜺𝒑)𝟐 + (

𝒏

𝟑) (𝜺𝒑)𝟑 +⋯ ; (𝟐)

(𝟏 + 𝜺𝟐𝒑)𝒏 = (𝒏

𝟎) + (

𝒏

𝟏) (𝜺𝟐𝒑) + (

𝒏

𝟐) (𝜺𝟐𝒑)𝟐 + (

𝒏

𝟑) (𝜺𝟐𝒑)𝟑 +⋯ ; (𝟑)

Adding (1),(2),(3) it follows that:

(𝟏 + 𝒑)𝒏 + (𝟏 + 𝜺𝒑)𝒏 + (𝟏 + 𝜺𝟐𝒑)𝒏 =(𝟏 + 𝒑)𝒏 + (𝟏 + 𝜺𝒑)𝒏 + (𝟏 + 𝜺𝟐𝒑)𝒏

𝟑= 𝒙𝒏

(𝟏 + 𝒑)𝒏 + 𝜺𝟐(𝟏 + 𝜺𝒑)𝒏 + 𝜺(𝟏 + 𝜺𝟐𝒑)𝒏 =(𝟏 + 𝒑)𝒏 + 𝜺𝟐(𝟏 + 𝜺𝒑)𝒏 + 𝜺(𝟏 + 𝜺𝟐𝒑)𝒏

𝟑= 𝒚𝒏

(𝟏 + 𝒑)𝒏 + 𝜺(𝟏 + 𝜺𝒑)𝒏 + 𝜺𝟐(𝟏 + 𝜺𝟐𝒑)𝒏 =(𝟏 + 𝒑)𝒏 + 𝜺(𝟏 + 𝜺𝒑)𝒏 + 𝜺𝟐(𝟏 + 𝜺𝟐𝒑)𝒏

𝟑= 𝒛𝒏

Let us denote: 𝒂 = (𝟏 + 𝒑)𝒏, 𝒃 = (𝟏 + 𝜺𝒑)𝒏, 𝒄 = (𝟏 + 𝜺𝟐𝒑)𝒏

𝒙𝒏𝒚𝒏 =(𝒂 + 𝒃 + 𝒄)(𝒂 + 𝜺𝟐𝒃 + 𝜺𝒄)

𝟗

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𝒙𝒏𝒛𝒏 =(𝒂+ 𝒃 + 𝒄)(𝒂 + 𝜺𝒃 + 𝜺𝟐𝒄)

𝟗

𝒚𝒏𝒛𝒏 =(𝒂 + 𝜺𝟐𝒃 + 𝜺𝒄)(𝒂 + 𝜺𝒃 + 𝜺𝟐𝒄)

𝟗

𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏 =𝟑𝒂𝟐 + 𝟑𝒃𝒄(𝜺 + 𝜺𝟐)

𝟗=𝒂𝟐 − 𝒃𝒄

𝟑

=(𝟏 + 𝒑)𝟐𝒏 − (𝟏 + 𝜺𝒑)𝒏(𝟏 + 𝜺𝟐𝒑)𝒏

𝟑

=(𝟏 + 𝒑)𝟐𝒏 − (𝟏 + 𝜺𝟐𝒑 + 𝜺𝒑 + 𝒑𝟐)𝒏

𝟑=(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏

𝟑


√𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏𝒏 = 𝐥𝐢𝐦

𝒏→∞√(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏

𝟑

𝒏

=𝑪−𝑫


(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏= 𝐥𝐢𝐦𝒏→∞

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 [𝟏 −(𝒑𝟐 − 𝒑 + 𝟏)𝒏

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏]

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 [𝟏 −(𝒑𝟐 − 𝒑 + 𝟏)𝒏

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏]=

= (𝒑 + 𝟏)𝟐 𝐥𝐢𝐦𝒏→∞

𝟏 − 𝒂𝒏+𝟏

𝟏 − 𝒂𝒏= (𝒑 + 𝟏)𝟐; (𝒂 =

𝒑𝟐 − 𝒑 + 𝟏

𝒑𝟐 + 𝟐𝒑 + 𝟏< 1 ⇒ 𝒂𝒏 → 𝟎)


(𝒏

𝒌) = (

𝒏

𝒏 − 𝒌) ⇒ 𝒙𝒏 = (

𝒏

𝒏)𝒑𝒏 + (

𝒏

𝒏 − 𝟑)𝒑𝒏−𝟑 + (

𝒏

𝒏 − 𝟔)𝒑𝒏−𝟔 +⋯

𝒚𝒏 = (𝒏

𝒏 − 𝟏)𝒑𝒏−𝟏 + (

𝒏

𝒏 − 𝟒)𝒑𝒏−𝟒 + (

𝒏

𝒏 − 𝟕)𝒑𝒏−𝟕 +⋯

𝒛𝒏 = (𝒏

𝒏 − 𝟐)𝒑𝒏−𝟐 + (

𝒏

𝒏 − 𝟓)𝒑𝒏−𝟓 + (

𝒏

𝒏 − 𝟖)𝒑𝒏−𝟖 +⋯

Let 𝜺 be root by three order of unity, hence 𝜺𝟐 + 𝜺+ 𝟏 = 𝟎 and 𝜺𝟑 = 𝟏.

𝒙𝒏 + 𝒚𝒏 + 𝒛𝒏 = (𝒑 + 𝟏)𝒏

𝒙𝒏 + 𝜺𝒚𝒏 + 𝜺𝟐𝒛𝒏 = (𝒑 + 𝜺)

𝒏

𝒙𝒏 + 𝜺𝟐𝒚𝒏 + 𝜺𝒛𝒏 = (𝒑+ 𝜺

𝟐)𝒏

𝒙𝒏 =𝟏

𝟑[(𝒑 + 𝟏)𝒏 + (𝒑 + 𝜺)𝒏 + (𝒑 + 𝜺𝟐)𝒏]

𝒚𝒏 =𝟏

𝟑[(𝒑 + 𝟏)𝒏 + 𝜺𝟐(𝒑 + 𝜺)𝒏 + 𝜺(𝒑 + 𝜺𝟐)𝒏]

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𝒛𝒏 =𝟏

𝟑[(𝒑 + 𝟏)𝒏 + 𝜺𝟐(𝒑 + 𝜺)𝒏 + 𝜺(𝒑 + 𝜺𝟐)𝒏]

Let 𝒂 = (𝒑 + 𝟏)𝒏, 𝒃 = (𝒑 + 𝜺)𝒏, 𝒄 = (𝒑 + 𝜺𝟐)𝒏. Hence,

(𝒂 + 𝒃 + 𝒄)(𝒂 + 𝒃𝜺 + 𝒄𝜺𝟐) + (𝒂 + 𝒃 + 𝒄)(𝒂 + 𝒃𝜺𝟐 + 𝒄𝜺)

+ (𝒂 + 𝒃𝜺 + 𝒄𝜺𝟐)(𝒂 + 𝒃𝜺𝟐 + 𝒄𝜺)

= (𝒂 + 𝒃 + 𝒄)(𝟐𝒂 − 𝒃 − 𝒄) + 𝒂𝟐 + 𝒂𝒃𝜺 + 𝒂𝒄𝜺𝟐 + 𝒂𝒃𝜺𝟐 + 𝒃𝟐 + 𝒃𝒄𝜺 + +𝒂𝒄𝜺 + 𝒄𝟐 + 𝒃𝒄𝜺𝟐

= 𝟐𝒂𝟐 − 𝒂𝒃 − 𝒂𝒄 − 𝒃𝟐 − 𝒃𝒄 + 𝟐𝒂𝒃 + 𝟐𝒂𝒄 − 𝒃𝒄 − 𝒄𝟐 + 𝒂𝟐 − 𝒂𝒃 − 𝒂𝒄 + 𝒃𝟐 + 𝒄𝟐 − 𝒃𝒄

= 𝟑𝒂𝟐 − 𝟑𝒃𝒄 = 𝟑(𝒑 + 𝟏)𝒏 − 𝟑(𝒑𝟐 + 𝒑𝜺 + 𝒑𝜺𝟐 + 𝟏)𝒏 =

= 𝟑[(𝒑 + 𝟏)𝟐𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏]. Thus,

𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏 =𝟏

𝟑[(𝒑 + 𝟏)𝒏 − (𝒑𝟏 − 𝒑 + 𝟏)𝒏]

Therefore,


√𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏𝒏 =

𝟏

𝟑𝐥𝐢𝐦𝒏→∞

[(𝒑 + 𝟏)𝒏 − (𝒑𝟏 − 𝒑 + 𝟏)𝒏] =


(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 [𝟏 −(𝒑𝟐 − 𝒑 + 𝟏)𝒏

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏]

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 [𝟏 −(𝒑𝟐 − 𝒑 + 𝟏)𝒏

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏]= (𝒑 + 𝟏)𝟐;

(𝐰𝐡𝐞𝐫𝐞 𝒑𝟐 − 𝒑 + 𝟏

𝒑𝟐 + 𝟐𝒑 + 𝟏< 1 ⇒

(𝒑𝟐 − 𝒑 + 𝟏)𝒏

(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏→ 𝟎)

1580. Find:


√∑𝒌𝟐 (𝒏

𝒌 − 𝟏) (𝒏

𝒌)

𝒏

𝒌=𝟏

𝒏

Proposed by Marian Ursărescu-Romania


𝒌𝟐 (𝒏

𝒌 − 𝟏) (𝒏

𝒌) = 𝒌𝟐 (

𝒏

𝒌)(𝒏

𝒌) = 𝒏𝟐 (

𝒏 − 𝟏

𝒌 − 𝟏)𝟐

⇒

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∑𝒌𝟐 (𝒏

𝒌 − 𝟏) (𝒏

𝒌)

𝒏

𝒌=𝟏

= 𝒏𝟐∑(𝒏− 𝟏

𝒌− 𝟏)𝟐𝒏

𝒌=𝟏

= 𝒏𝟐∑(𝒏− 𝟏

𝒌)𝟐𝒏−𝟏

𝒌=𝟎

= 𝒏𝟐 (𝟐𝒏 − 𝟐

𝒏 − 𝟏)

= 𝒏𝟐𝒂𝒏; (𝒂𝒏 = (𝟐𝒏 − 𝟐

𝒏 − 𝟏))


√∑𝒌𝟐 (𝒏

𝒌 − 𝟏) (𝒏

𝒌)

𝒏

𝒌=𝟏

𝒏


√𝒏𝟐𝒂𝒏𝒏

=𝑪−𝑫


(𝟐𝒏)!

𝒏!𝒏!⋅(𝒏 − 𝟏)! (𝒏 − 𝟏)!

(𝟐𝒏 − 𝟐)!=


𝟐𝒏(𝟐𝒏 − 𝟏)

𝒏𝟐= 𝟒


𝒌

𝒏(𝒏

𝒌) = (

𝒏 − 𝟏

𝒌 − 𝟏) ⇒ 𝒌𝟐 (

𝒏

𝒌 − 𝟏) (𝒏

𝒌) = 𝒌 (

𝒏

𝒌 − 𝟏) ⋅ 𝒏(

𝒏 − 𝟏

𝒌 − 𝟏) =

= (𝒌 − 𝟏 + 𝟏) (𝒏

𝒌 − 𝟏) ⋅ 𝒏 (

𝒏 − 𝟏

𝒌 − 𝟏) = (𝒌 − 𝟏) (

𝒏

𝒌 − 𝟏) ⋅ 𝒏 (

𝒏 − 𝟏

𝒌 − 𝟏) + 𝒏(

𝒏

𝒌 − 𝟏) ⋅ (

𝒏 − 𝟏

𝒌 − 𝟏)

=

= 𝒏(𝒏 − 𝟏

𝒌 − 𝟐) ⋅ 𝒏(

𝒏 − 𝟏

𝒌 − 𝟏) + 𝒏 (

𝒏

𝒌 − 𝟏) ⋅ (

𝒏 − 𝟏

𝒌 − 𝟏)

∑𝒌𝟐 (𝒏

𝒌 − 𝟏) (𝒏

𝒌)

𝒏

𝒌=𝟏

= 𝒏𝟐∑(𝒏− 𝟏

𝒌− 𝟐)

𝒏

𝒌=𝟏

(𝒏 − 𝟏

𝒌 − 𝟏) + 𝒏∑(

𝒏

𝒌− 𝟏)(𝒏 − 𝟏

𝒌 − 𝟏)

𝒏

𝒌=𝟏

𝐋𝐞𝐭: (𝟏 + 𝒙)𝒏−𝟏 = (𝒏 − 𝟏

𝟎) + 𝒙(

𝒏 − 𝟏

𝟏) + 𝒙𝟐 (

𝒏 − 𝟏

𝟐) +⋯+ 𝒙𝒏−𝟏 (

𝒏 − 𝟏

𝒏 − 𝟏)

(𝒙 + 𝟏)𝒏 = 𝒙𝒏−𝟏 (𝒏 − 𝟏

𝟎) + 𝒙𝒏−𝟐 (

𝒏 − 𝟏

𝟏) + ⋯+ (

𝒏 − 𝟏

𝒏 − 𝟏)

𝐓𝐡𝐞𝐧 𝑺𝟏 =∑(𝒏− 𝟏

𝒌 − 𝟐)

𝒏

𝒌=𝟏

(𝒏 − 𝟏

𝒌 − 𝟏) = (

𝟐𝒏 − 𝟐

𝒏 − 𝟐)

𝑺𝟐 = 𝒏∑(𝒏

𝒌− 𝟏)(𝒏 − 𝟏

𝒌 − 𝟏)

𝒏

𝒌=𝟏

=∑((𝒏 − 𝟏

𝒌 − 𝟏) + (

𝒏 − 𝟏

𝒌 − 𝟐)) (

𝒏 − 𝟏

𝒌 − 𝟏)

𝒏

𝒌=𝟏

=

=∑(𝒏− 𝟏

𝒌 − 𝟏)𝟐𝒏

𝒌=𝟏

+∑(𝒏− 𝟏

𝒌 − 𝟐) (𝒏 − 𝟏

𝒌 − 𝟏)

𝒏

𝒌=𝟏

= (𝟐𝒏 − 𝟐

𝒏 − 𝟏) + (

𝟐𝒏 − 𝟐

𝒏 − 𝟐) = (

𝟐𝒏 − 𝟏

𝒏 − 𝟏)

Hence,

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∑𝒌𝟐 (𝒏

𝒌 − 𝟏)(𝒏

𝒌)

𝒏

𝒌=𝟏

= 𝒏𝟐 (𝟐𝒏 − 𝟐

𝒏 − 𝟏) + 𝒏 (

𝟐𝒏 − 𝟏

𝒏 − 𝟏) = 𝒏𝟐 ⋅

(𝟐𝒏 − 𝟐)!

(𝒏 − 𝟐)!𝒏!+ 𝒏 ⋅

(𝟐𝒏 − 𝟏)!

(𝒏 − 𝟏)!𝒏!=

=(𝟐𝒏 − 𝟐)! (𝒏𝟑 + 𝒏𝟐 − 𝒏)

(𝒏 − 𝟏)! 𝒏!

Therefore,


√∑𝒌𝟐 (𝒏

𝒌 − 𝟏) (𝒏

𝒌)

𝒏

𝒌=𝟏

𝒏


√(𝟐𝒏 − 𝟐)! (𝒏𝟑 + 𝒏𝟐 − 𝒏)

(𝒏 − 𝟏)! 𝒏!

𝒏

=𝑪−𝑫


(𝟐𝒏)! [(𝒏 + 𝟏)𝟑 + (𝒏 + 𝟏)𝟐 − (𝒏 + 𝟏)]

𝒏! (𝒏 + 𝟏)!⋅

𝒏! (𝒏 − 𝟏)!

(𝟐𝒏 − 𝟐)! (𝒏𝟑 + 𝒏𝟐 − 𝒏)=


(𝟐𝒏)!𝒏! (𝒏 − 𝟏)!

𝒏! (𝒏 + 𝟏)! (𝟐𝒏 − 𝟐)!= 𝐥𝐢𝐦𝒏→∞

(𝟐𝒏 − 𝟏) ⋅ 𝟐𝒏

𝒏(𝒏 + 𝟏)= 𝟒

1581. Find:


∑

(

𝐬𝐢𝐧−𝟏 (

𝟐𝟖𝒆𝟏+𝐥𝐨𝐠𝒌

𝒌 + 𝒏𝒌)

𝐥𝐨𝐠 (𝟏 +𝒏

𝒌 + 𝒏𝒌√𝒏!𝒏 )

𝒏𝟐

)

𝒏

𝒌=𝟏

Proposed by Ruxandra Daniela Tonilă-Romania

Solution by Adrian Popa-Romania


∑

(

𝐬𝐢𝐧−𝟏 (

𝟐𝟖𝒆𝟏+𝐥𝐨𝐠 𝒌

𝒌 + 𝒏𝒌)

𝐥𝐨𝐠 (𝟏 +𝒏


𝒏𝟐

)

𝒏

𝒌=𝟏


∑

(

𝐬𝐢𝐧−𝟏 (𝟐𝟖𝒆𝒌𝒌 + 𝒏𝒌

)

𝟐𝟖𝒆𝒌𝒌 + 𝒏𝒌

⋅𝟐𝟖𝒆𝒌𝒌 + 𝒏𝒌

𝒏𝟐 ⋅𝐥𝐨𝐠 (𝟏 +

𝒏


𝒏

𝒌 + 𝒏𝒌√𝒏!𝒏

⋅𝒏

𝒌 + 𝒏𝒌√𝒏!𝒏

)

𝒏

𝒌=𝟏


∑𝟐𝟖𝒆𝒌

𝒌 + 𝒏𝒌⋅(𝒌 + 𝒏𝒌)√𝒏!

𝒏

𝒏𝟑

𝒏

𝒌=𝟏

=

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𝟐𝟖𝒆√𝒏!𝒏

𝒏𝟑⋅∑𝒌

𝒏

𝒌=𝟏


𝟐𝟖𝒆√𝒏!𝒏

⋅ 𝒏(𝒏 + 𝟏)

𝟐𝒏𝟑= 𝟏𝟒𝒆 𝐥𝐢𝐦

𝒏→∞√𝒏! (𝒏 + 𝟏)𝒏

𝒏𝟐𝒏

𝒏

=𝑪−𝑫

= 𝟏𝟒𝒆 𝐥𝐢𝐦𝒏→∞

(𝒏 + 𝟏)! (𝒏 + 𝟐)𝒏+𝟏

(𝒏 + 𝟏)𝟐𝒏+𝟐⋅

𝒏𝟐𝒏

𝒏! (𝒏 + 𝟏)𝒏= 𝟏𝟒𝒆 𝐥𝐢𝐦

𝒏→∞

𝒏 + 𝟐

𝒏 + 𝟏⋅ (𝒏 + 𝟐

𝒏 + 𝟏)𝒏

⋅ (𝒏

𝒏 + 𝟏)𝟐𝒏

=

= 𝟏𝟒𝒆 𝐥𝐢𝐦𝒏→∞

(𝟏 +𝟏

𝒏 + 𝟏)𝒏

(𝟏 −𝟏

𝒏 + 𝟏)𝟐𝒏

= 𝟏𝟒𝒆 ⋅ 𝒆 ⋅ 𝒆−𝟐 = 𝟏𝟒

1582. If 𝒂, 𝒃, 𝒄 > 𝟏 then find:

𝛀 = 𝐥𝐢𝐦𝒙→𝟎(𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄

𝒃 ⋅ 𝒆−𝒙)))) (𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙))))

−𝟏



(𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙))))

′=

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙))

′

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠 𝒂=

=(𝐥𝐨𝐠𝒄(𝒄

𝒃 ⋅ 𝒆−𝒙))′

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒄(𝒄

𝒃 ⋅ 𝒆−𝒙) ⋅ 𝐥𝐨𝐠 𝒂 𝐥𝐨𝐠𝒃=

=−𝒆𝒃𝒆−𝒙

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙) ⋅ 𝒄𝒃𝒆−𝒙 𝐥𝐨𝐠𝒂 𝐥𝐨𝐠𝒃 𝐥𝐨𝐠 𝒄=

=−𝟏

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙) ⋅ 𝐥𝐨𝐠 𝒂 𝐥𝐨𝐠 𝒃 𝐥𝐨𝐠 𝒄

(𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙))))

′=

(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)))

′

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠 𝒄=

=(𝐥𝐨𝐠𝒂(𝒂

𝒃 ⋅ 𝒆−𝒙))′

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙) 𝐥𝐨𝐠 𝒃 𝐥𝐨𝐠 𝒄=

=−𝒂𝒃𝒆−𝒙

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)) 𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙) 𝒂𝒃𝒆−𝒙 𝐥𝐨𝐠𝒂 𝐥𝐨𝐠𝒃 𝐥𝐨𝐠 𝒄=

=−𝟏

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)) 𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙) 𝐥𝐨𝐠 𝒂 𝐥𝐨𝐠 𝒃 𝐥𝐨𝐠 𝒄



−𝟏=𝑳′𝑯


𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒂(𝒂

𝒃𝒆−𝒙)

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒄(𝒄𝒃𝒆−𝒙)=𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂

𝒃)) ⋅ 𝐥𝐨𝐠𝒂(𝒂𝒃)

𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃)) ⋅ 𝐥𝐨𝐠𝒄(𝒄𝒃)= 𝟏

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Solution 2 by Florentin Vişescu-Romania



−𝟏=

𝒚=𝒆−𝒙

= 𝐥𝐢𝐦𝒚→𝟏(𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄

𝒃 ⋅ 𝒚)))) (𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒚))))

−𝟏

𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒚))) = 𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝒃 + 𝐥𝐨𝐠𝒄 𝒚)) =

= 𝐥𝐨𝐠𝒂 (𝐥𝐨𝐠𝒃 (𝒃 (𝟏 +𝐥𝐨𝐠𝒄 𝒚

𝒃))) = 𝐥𝐨𝐠𝒂 (𝟏 + 𝐥𝐨𝐠𝒃 (𝟏 +

𝐥𝐨𝐠𝒄 𝒚

𝒃))

Similarly,

𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙))) = 𝐥𝐨𝐠𝒄 (𝟏 + 𝐥𝐨𝐠𝒃 (𝟏 +

𝐥𝐨𝐠𝒂 𝒚

𝒃))

𝛀 = 𝐥𝐢𝐦𝒚→𝟏

𝐥𝐨𝐠𝒂 (𝟏 + 𝐥𝐨𝐠𝒃 (𝟏 +𝐥𝐨𝐠𝒄 𝒚𝒃 ))

𝐥𝐨𝐠𝒄 (𝟏 + 𝐥𝐨𝐠𝒃 (𝟏 +𝐥𝐨𝐠𝒂 𝒚𝒃 ))

= 𝐥𝐢𝐦𝒚→𝟏

𝐥𝐨𝐠 𝒄

𝐥𝐨𝐠 𝒂⋅𝐥𝐨𝐠𝒃 (𝟏 +

𝐥𝐨𝐠𝒄 𝒚𝒃 )

𝐥𝐨𝐠𝒃 (𝟏 +𝐥𝐨𝐠𝒂 𝒚𝒃 )

=

= 𝐥𝐢𝐦𝒚→𝟏

𝐥𝐨𝐠 𝒄

𝐥𝐨𝐠 𝒂⋅𝐥𝐨𝐠𝒄 𝒚

𝒃⋅𝒃

𝐥𝐨𝐠𝒂 𝒚= 𝐥𝐢𝐦𝒚→𝟏

𝐥𝐨𝐠 𝒄

𝐥𝐨𝐠𝒂⋅𝐥𝐨𝐠𝒚

𝐥𝐨𝐠 𝒄⋅𝐥𝐨𝐠 𝒂

𝐥𝐨𝐠 𝒚= 𝟏

1583. 𝒂𝟏 = 𝟒, 𝒂𝟐 = 𝟐, 𝒂𝒏 = 𝒂𝒏+𝟏

𝟑𝒏+𝟑

𝟕𝒏 ⋅ 𝒂𝒏+𝟐

𝟒𝒏+𝟖

𝟕𝒏 . Find:


(∑(𝒂𝒌𝒂𝒌+𝟏

)𝒌

𝒏

𝒌=𝟏

)(𝟏

𝒏!∑𝒌 ⋅ 𝒌!

𝒏+𝟏

𝒌=𝟏

)

−𝟏

Proposed by Ruxandra Daniela Tonilă-Romania

Solution by George Florin Şerban-Romania

𝟏

𝒏!∑𝒌 ⋅ 𝒌!

𝒏+𝟏

𝒌=𝟏

=𝟏

𝒏!∑((𝒌 + 𝟏)! − 𝒌!)

𝒏+𝟏

𝒌=𝟏

=𝟏

𝒏!((𝒏 + 𝟐)! − 𝟏)

𝒂𝒏𝟕𝒏 = 𝒂𝒏+𝟏

𝟑(𝒏+𝟏) ⋅ 𝒂𝒏+𝟐𝟒(𝒏+𝟐). Let 𝒙𝒏 = 𝒂𝒏

𝒏 ⇒ 𝒙𝒏𝟕 = 𝒙𝒏+𝟏

𝟑 ⋅ 𝒙𝒏+𝟐𝟒 , 𝒙𝟏 = 𝒂𝟏 = 𝟒, 𝒙𝟐 = 𝒂𝟐

𝟐 = 𝟒.

𝒙𝟏𝟕 = 𝒙𝟐

𝟑 ⋅ 𝒙𝟑𝟒

𝒙𝟐𝟕 = 𝒙𝟑

𝟑 ⋅ 𝒙𝟒𝟒

…………

𝒙𝒏𝟕 = 𝒙𝒏+𝟏

𝟑 ⋅ 𝒙𝒏+𝟐𝟒

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𝑷𝒏 = 𝒙𝟏𝒙𝟐…𝒙𝒏 ⇒ 𝑷𝒏𝟕 =

𝑷𝒏𝟑

𝒙𝟏𝟑 ⋅ 𝒙𝒏+𝟏

𝟑 ⋅𝑷𝒏𝟒

𝒙𝟏𝟒𝒙𝟐𝟒 ⋅ 𝒙𝒏+𝟏

𝟒 𝒙𝒏+𝟐𝟒

We prove that 𝑷(𝒏): 𝒙𝒏 = 𝟒,∀𝒏 ≥ 𝟎 (by mathematical induction).

(I): 𝑷(𝟎): 𝟒𝟏𝟏 = 𝒙𝟏𝟕𝒙𝟐𝟒 = 𝟒𝟕 ⋅ 𝟒𝟒 = 𝟒𝟏𝟏 true.

(II): Suppose that: 𝑷(𝟎),𝑷(𝟏),… , 𝑷(𝒏 + 𝟏) are true, then

𝒙𝒏+𝟐𝟒 =

𝟒𝟏𝟏

𝒙𝒏+𝟏𝟕 =

𝟒𝟏𝟏

𝟒𝟕= 𝟒𝟒, because 𝒂𝒏, 𝒙𝒏 > 𝟎 ⇒ 𝒙𝒏+𝟐 = 𝟒 ⇒ 𝒂𝒏

𝒏 = 𝟒 ⇒ 𝒂𝒏 = √𝟒𝒏.



)𝒌

𝒏

𝒌=𝟏

)(𝟏

𝒏!∑𝒌 ⋅ 𝒌!

𝒏+𝟏

𝒌=𝟏

)

−𝟏


(∑(√𝟒𝒌

√𝟒𝒌+𝟏 )

𝒌𝒏

𝒌=𝟏

) ⋅𝒏!

(𝒏 + 𝟐)! − 𝟏=


(𝟏

𝒏 + 𝟏(∑ √𝟒

𝒌+𝟏

𝒏

𝒌=𝟏

) ⋅𝒏! (𝒏 + 𝟏)

(𝒏 + 𝟐)! − 𝟏) =𝑪−𝑺


√𝟒𝒏+𝟐

𝒏 + 𝟐 −𝟏

(𝒏 + 𝟏)!

= 𝟎

Therefore,



)𝒌

𝒏

𝒌=𝟏

)(𝟏

𝒏!∑𝒌 ⋅ 𝒌!

𝒏+𝟏

𝒌=𝟏

)

−𝟏

= 𝟎

1584. 𝑺(𝒑) = {(𝒙, 𝒚)|𝒙𝟑 + 𝒚𝟑 ≤ 𝒑𝟑, 𝒙 ≥ 𝟎, 𝒚 ≥ 𝟎, 𝒑 ≥ 𝟎}

Find:

𝛀 = 𝐥𝐢𝐦𝒑→∞

𝑨𝒓𝒆𝒂(𝑺(𝒑))

𝒑𝟐


Solution by Ty Halpen-Florida-USA

Rewrite the area bounded by 𝑺(𝒑) as 𝒚 ≤ (𝒑𝟑 − 𝒙𝟑)𝟏

𝟑 and integrate it from 𝟎 ≤ 𝒙 ≤ 𝒑:

𝑨𝒓𝒆𝒂(𝑺(𝒑)) = ∫ √𝒑𝟑 − 𝒙𝟑𝟑

𝒑

𝟎

𝒅𝒙 =

𝒕=𝒙𝟑

𝒑𝟑 𝒑𝟐

𝟑∫ 𝒕−

𝟐𝟑√𝟏 − 𝒕𝟑

𝟏

𝟎

𝒅𝒕 =

=𝒑𝟑

𝟔⋅𝚪 (𝟏𝟑)𝚪(

𝟒𝟑)

𝚪 (𝟓𝟑)

=𝒑𝟐

𝟔⋅𝚪𝟐 (

𝟏𝟑)

𝚪 (𝟐𝟑)

=𝒑𝟐

𝟔⋅

(√𝟑𝚪(𝟏𝟑))𝚪

𝟐 (𝟏𝟑)

𝟐𝝅=𝒑𝟐𝚪𝟑 (

𝟏𝟑)

𝟒𝝅√𝟑

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Therefore,

𝛀 = 𝐥𝐢𝐦𝒑→∞

𝑨𝒓𝒆𝒂(𝑺(𝒑))

𝒑𝟐=𝚪𝟑 (

𝟏𝟑)

𝟒𝝅√𝟑

1585. Prove that:

∑𝟏

𝒏+ 𝟏

∞

𝒏=𝟎

𝐜𝐨𝐬 (𝝅𝒏

𝟒) =

√𝟐

𝟏𝟔(𝟑𝝅 + 𝟒 𝐥𝐨𝐠(𝟏 + √𝟐) − 𝐥𝐨𝐠 𝟒)


Solution by Mohammad Rostami-Afghanistan

First we prove that:

∵ ∑𝐬𝐢𝐧(𝒌𝜽)

𝒌

∞

𝒌=𝟏

=𝝅 − 𝜽

𝟐, (𝟎 < 𝜽 < 𝟐𝝅)

∑𝐬𝐢𝐧(𝒌𝜽)

𝒌

∞

𝒌=𝟏

= ∑

𝟏𝟐𝒊(𝒆𝒊𝒌𝜽 − 𝒆−𝒊𝒌𝜽)

𝒌

𝒏

𝒌=𝟏

=

=𝟏

𝟐𝒊(∑𝒆𝒊𝒌𝜽 ∫ 𝒆−𝒊𝒌

∞

𝟎

𝒅𝒙 −∑𝒆−𝒊𝒌𝜽∞

𝒌=𝟏

∫ 𝒆−𝒊𝒌𝒅𝒙∞

𝟎

∞

𝒌=𝟏

) =

=𝟏

𝟐𝒊[𝐥𝐨𝐠(𝟏 − 𝒆𝒊𝜽−𝒙) − 𝐥𝐨𝐠(𝟏 − 𝒆𝒊𝜽−𝒙)]

𝟎

∞=𝟏

𝟐𝒊[𝐥𝐨𝐠(𝟏 − 𝒆−𝒊𝜽) − 𝐥𝐨𝐠(𝟏 + 𝒆𝒊𝜽)] =

=𝟏

𝟐𝒊𝐥𝐨𝐠 (

𝟏 − 𝒆−𝒊𝜽

𝟏 − 𝒆𝒊𝜽) =

𝟏

𝟐𝒊𝐥𝐨𝐠 (

𝒆𝒊𝜽 − 𝟏

𝒆𝒊𝜽(𝟏 − 𝒆𝒊𝜽)) =

𝟏

𝟐𝒊𝐥𝐨𝐠(−𝒆−𝒊𝜽) =

=𝟏

𝟐𝒊(𝐥𝐨𝐠(−𝟏) + 𝐥𝐨𝐠(𝒆−𝒊𝜽)) =

{ 𝒆𝝅𝒊=−𝟏𝝅𝒊=𝐥𝐨𝐠(−𝟏) 𝟏

𝟐𝒊(𝝅𝒊 − 𝜽𝒊) =

𝒊(𝝅 − 𝜽)

𝟐𝒊=𝝅 − 𝜽

𝟐

Next, we prove that:

∵ ∑𝐜𝐨𝐬(𝒌𝜽)

𝒌

∞

𝒌=𝟏

= −𝟏

𝟐𝐥𝐨𝐠(𝟐 − 𝟐𝐜𝐨𝐬 𝜽) ; (𝜽 ∈ ℝ)

∑𝐜𝐨𝐬(𝒌𝜽)

𝒌

∞

𝒌=𝟏

=∑

𝟏𝟐(𝒆𝒊𝒌𝜽 + 𝒆−𝒊𝒌𝜽)

𝒌

𝒏

𝒌=𝟏

=

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=𝟏

𝟐(∑𝒆𝒊𝒌𝜽∫ 𝒆−𝒊𝒌

∞

𝟎

𝒅𝒙 +∑𝒆−𝒊𝒌𝜽∞

𝒌=𝟏

∫ 𝒆−𝒊𝒌𝒅𝒙∞

𝟎

∞

𝒌=𝟏

) =

=𝟏

𝟐(∫ ∑(𝒆𝒊𝜽−𝒙)

𝒌∞

𝒌=𝟏

𝒅𝒙∞

𝟎

+∫ ∑(𝒆−𝒊𝜽−𝒙)𝒌

𝒏

𝒌=𝟏

𝒅𝒙∞

𝟎

) =

=𝟏

𝟐(∫

𝒆𝒊𝜽−𝒙

𝟏 − 𝒆𝒊𝜽−𝒙

∞

𝟎

𝒅𝒙 +∫𝒆−𝒊𝜽−𝒙

𝟏 − 𝒆𝒊𝜽−𝒙𝒅𝒙

∞

𝟎

) =

=𝟏

𝟐[𝐥𝐨𝐠(𝟏 − 𝒆𝒊𝜽−𝒙) + 𝐥𝐨𝐠(𝟏 − 𝒆𝒊𝜽−𝒙)]

𝟎

∞=𝟏

𝟐[𝐥𝐨𝐠(𝟏 − 𝒆−𝒊𝜽) − 𝐥𝐨𝐠(𝟏 + 𝒆𝒊𝜽)] =

= −𝟏

𝟐𝐥𝐨𝐠(

(𝟏 − 𝒆𝒊𝜽)(𝒆𝒊𝜽 − 𝟏)

𝒆𝒊𝜽) = −

𝟏

𝟐𝐥𝐨𝐠 (−𝒆−𝒊𝜽(𝟏 − 𝟐𝒆𝒊𝜽 + 𝒆𝟐𝒊𝜽)) =

= −𝟏

𝟐𝐥𝐨𝐠(−𝒆−𝒊𝜽 + 𝟐 − 𝒆𝒊𝜽) = −

𝟏

𝟐𝐥𝐨𝐠 (𝟐 − (𝒆𝒊𝜽 + 𝒆−𝒊𝜽)) = −

𝟏

𝟐𝐥𝐨𝐠(𝟐 − 𝟐 𝐜𝐨𝐬𝜽)

Hence, we have:

∑𝟏

𝒏 + 𝟏

∞

𝒏=𝟎


𝟒) = ∑

𝟏

𝒏𝐜𝐨𝐬 [

𝝅

𝟒(𝒏 − 𝟏)] =

∞

𝒏=𝟏

∑𝟏

𝒏𝐜𝐨𝐬 [(

𝝅

𝟒𝒏) −

𝝅

𝟒] =

∞

𝒏=𝟏

= ∑𝟏

𝒏[𝐜𝐨𝐬

𝝅

𝟒𝐜𝐨𝐬 (

𝝅

𝟒𝒏) + 𝐬𝐢𝐧

𝝅

𝟒𝐬𝐢𝐧 (

𝝅

𝟒𝒏)]

∞

𝒏=𝟏

=

=√𝟐

𝟐[∑

𝐜𝐨𝐬 (𝝅𝟒 𝒏)

𝒏

∞

𝒌=𝟏

+∑𝐬𝐢𝐧 (

𝝅𝟒 𝒏)

𝒏

∞

𝒏=𝟏

] =√𝟐

𝟐(−𝟏

𝟐𝐥𝐨𝐠(𝟐 − √𝟐) +

𝟑𝝅

𝟖) =

=√𝟐

𝟏𝟔[𝟑𝝅 − 𝟒 𝐥𝐨𝐠(𝟐 − √𝟐)] =

√𝟐

𝟏𝟔[𝟑𝝅 + 𝟒 𝐥𝐨𝐠(

𝟏(𝟐 + √𝟐)√𝟐

(𝟐 − √𝟐)(𝟐 + √𝟐)√𝟐) =

=√𝟐

𝟏𝟔[𝟑𝝅 + 𝟒 𝐥𝐨𝐠 (

𝟐√𝟐 + 𝟐

𝟐√𝟐)] =

√𝟐

𝟏𝟔[𝟑𝝅 + 𝟒 𝐥𝐨𝐠 (

𝟏 + √𝟐

𝟐)] =

=√𝟐

𝟏𝟔[𝟑𝝅 + 𝟒 𝐥𝐨𝐠(𝟏 + √𝟐) − 𝟒 𝐥𝐨𝐠 √𝟐] =

√𝟐

𝟏𝟔(𝟑𝝅 + 𝟒 𝐥𝐨𝐠(𝟏 + √𝟐) − 𝐥𝐨𝐠 𝟒)

⇒∑𝟏

𝒏 + 𝟏

∞

𝒏=𝟎


𝟒) =

√𝟐

𝟏𝟔(𝟑𝝅 + 𝟒 𝐥𝐨𝐠(𝟏 + √𝟐) − 𝐥𝐨𝐠 𝟒)

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1586.

𝛀𝟏(𝒏) = ∫ 𝒙𝒏+𝟏√𝒙𝒏√𝒙𝒏−𝟏…√𝒙√𝟖𝒏

√𝟐𝒏

𝒅𝒙,𝛀𝟐(𝒏) = ∫ 𝒙√𝒙𝟐√𝒙𝟑…√𝒙𝒏+𝟏𝟐

𝟏

𝒅𝒙, 𝒏 ∈ ℕ∗ , 𝒏 ≥ 𝟐

Find:


𝟐𝟎𝟐𝟏𝛀𝟏(𝒏)𝛀𝟐(𝒏)

Proposed by Costel Florea-Romania

Solution by Kamel Gandouli Rezgui-Tunisia

𝒇(𝒙) = 𝒙𝒏+𝟏√𝒙𝒏√𝒙𝒏−𝟏…√𝒙 = 𝒙𝒏+𝟏 ⋅ 𝒙𝒏𝟐 ⋅ … ⋅ 𝒙

𝒏−𝒑

𝟐𝒑+𝟏 ⋅ … ⋅ 𝒙𝟏𝟐𝒏 = 𝒙𝒗𝒏

𝒗𝒏 = 𝒏+ 𝟏 +𝒏

𝟐+𝒏 − 𝟏

𝟐+𝒏 − 𝟐

𝟖+⋯+

𝟏

𝟐𝒏= 𝒏∑

𝟏

𝟐𝒌

𝒏

𝒌=𝟎

−∑𝒌− 𝟏

𝟐𝒌

𝒏

𝒌=𝟐

=

=∑𝟏

𝟐𝒌(𝒏 + 𝟏)

𝒏

𝒌=𝟐

+ 𝟏 +𝟏

𝟐−∑

𝒌

𝟐𝒌

𝒏

𝒌=𝟐

∑𝒌

𝟐𝒌

𝒏

𝒌=𝟐

=𝟑

𝟐− (𝒏 + 𝟐) (

𝟏

𝟐)𝒏

⇒ 𝒗𝒏 =𝟏

𝟒⋅𝟏 − (

𝟏𝟐)𝒏−𝟏

𝟏 −𝟏𝟐

⋅ (𝒏 + 𝟏) +𝟑

𝟐− (𝟑

𝟐− (𝒏 + 𝟐) (

𝟏

𝟐)𝒏

) =

= (𝟏

𝟐− (𝟏

𝟐)𝒏

)(𝒏 + 𝟏) + (𝒏 + 𝟐) (𝟏

𝟐)𝒏

= (𝟏

𝟐)𝒏

+𝟏

𝟐(𝒏 + 𝟏)

𝒈(𝒙) = 𝒙√𝒙𝟐√𝒙𝟑…√𝒙𝒏+𝟏 = 𝒙𝟏 ⋅ 𝒙𝟐𝟐 ⋅ 𝒙

𝟑𝟒 ⋅ 𝒙

𝟒

𝟐𝟑 ⋅ … ⋅ 𝒙𝒏+𝟏𝟐𝒏 = 𝒙𝒘𝒏 ,

𝒘𝒏 = 𝟏 +𝟐

𝟐+𝟑

𝟐𝟐+⋯+

𝒏 + 𝟏

𝟐𝒏=∑

𝒌+ 𝟏

𝟐𝒌

𝒏

𝒌=𝟎

== 𝟐 − (𝒏 + 𝟐) (𝟏

𝟐)𝒏

+ 𝟐 − (𝟏

𝟐)𝒏

= 𝟒 − (𝟏

𝟐)𝒏

(𝒏 + 𝟑) → 𝟒

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𝛀𝟏(𝒏) = ∫ 𝒙𝒏+𝟏√𝒙𝒏√𝒙𝒏−𝟏…√𝒙√𝟖𝒏

√𝟐𝒏

𝒅𝒙 = ∫ 𝒙𝒗(𝒏)√𝟖𝒏

√𝟐𝒏

𝒅𝒙 = [𝒙𝒗𝒏+𝟏

𝒗𝒏 + 𝟏]√𝟐𝒏

√𝟖𝒏

=

=𝟐𝟑(

𝒗𝒏+𝟏𝒏) − 𝟐

𝒗𝒏+𝟏𝒏

𝒗𝒏 + 𝟏

𝛀𝟐(𝒏) = ∫ 𝒙√𝒙𝟐√𝒙𝟑…√𝒙𝒏+𝟏𝟐

𝟏

𝒅𝒙 = ∫ 𝒙𝒘𝒏𝟐

𝟏

𝒅𝒙 = [𝒙𝒘𝒏+𝟏

𝒘𝒏 + 𝟏]𝟏

𝟐

=𝟐𝒘𝒏+𝟏 − 𝟏

𝒘𝒏 + 𝟏

Hence,

𝛀𝟏(𝒏)

𝛀𝟐(𝒏)=

𝟐𝟑(𝒗𝒏+𝟏𝒏) − 𝟐

𝒗𝒏+𝟏𝒏

𝒗𝒏 + 𝟏

𝟐𝒘𝒏+𝟏 − 𝟏𝒘𝒏 + 𝟏

=𝒘𝒏 + 𝟏

𝒗𝒏 + 𝟏⏟ →𝟎

⋅𝟐𝟑(

𝒗𝒏+𝟏𝒏) − 𝟐

𝒗𝒏+𝟏𝒏

𝟐𝟏+𝒘𝒏 − 𝟏⏟ 𝟐√𝟐/𝟑𝟏

> 0

Therefore,


𝟐𝟎𝟐𝟏𝛀𝟏(𝒏)𝛀𝟐(𝒏) = 𝟏.

1587. Prove that:

∑(𝟏

(𝟔𝒌 − 𝟔)!+

𝟏

(𝟔𝒌 − 𝟓)!−

𝟏

(𝟔𝒌 − 𝟑)!−

𝟏

(𝟔𝒌 − 𝟐)!)

∞

𝒌=𝟏

= √𝟒𝒆

𝟑𝐬𝐢𝐧 (

𝟐𝝅 + 𝟑√𝟑

𝟔)


Solution by Felix Marin-Romania

𝑯 −Hankel Contour.

𝜶 ∈ {𝟐, 𝟑, 𝟓, 𝟔},∑𝟏

(𝟔𝒌 − 𝜶)!

∞

𝒌=𝟏

=∑𝟏

𝚪(𝟔𝒌 − 𝜶 + 𝟏)!

∞

𝒌=𝟏

=∑∮𝒆𝒕

𝒕𝟔𝒌−𝜶+𝟏𝒅𝒕

𝟐𝝅𝒊𝑯

∞

𝒌=𝟏

=

= ∑∫𝒆𝒕

𝒕𝟔𝒌−𝜶+𝟏

𝟏++∞𝒊

𝟏+−∞𝒊

𝒅𝒕

𝟐𝝅𝒊

∞

𝒌=𝟏

= ∫ 𝒆𝒕𝒕𝜶−𝟏𝟏++∞𝒊

𝟏+−∞𝒊

∑(𝟏

𝒕𝟔)𝒌∞

𝒌=𝟏

𝒅𝒕

𝟐𝝅𝒊= ∫

𝒆𝒕𝒕𝜶−𝟏

𝒕𝟔 − 𝟏

𝒅𝒕

𝟐𝝅𝒊

𝟏++∞𝒊

𝟏+−∞𝒊

=

= ∑𝒆𝒑𝒏𝒑𝒏

𝜶−𝟏

𝟔𝒑𝒏𝟓|𝒑𝒏=𝒆

𝒏𝝅𝒊𝟑

𝟓

𝒏=𝟎

=𝟏

𝟔∑𝒆𝒑𝒏𝒑𝒏

𝜶

𝟓

𝒏=𝟎

Therefore,

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∑(𝟏

(𝟔𝒌 − 𝟔)!+

𝟏

(𝟔𝒌 − 𝟓)!−

𝟏

(𝟔𝒌 − 𝟑)!−

𝟏

(𝟔𝒌 − 𝟐)!)

∞

𝒌=𝟏

=𝟏

𝟔∑𝒆𝒑𝒏(𝒑𝒏

𝟔 + 𝒑𝒏𝟓 − 𝒑𝒏

𝟑 − 𝒑𝒏𝟐)

𝟓

𝒏=𝟎

=

= √𝒆 [𝐜𝐨𝐬 (√𝟑

𝟐) +

√𝟑

𝟑𝐬𝐢𝐧(

√𝟑

𝟐)] ≅ 𝟏. 𝟕𝟗𝟑𝟑

1588. Find:


𝟏

𝒏!∫

𝒙𝒏 𝐬𝐢𝐧 (𝒙 +𝝅𝟒)

𝒆𝒙

∞

𝟎

𝒅𝒙


Solution by Mohammad Rostami-Afghanistan


𝟏

𝒏!∫


𝒆𝒙

∞

𝟎


𝟏

𝒏!∫

𝒙𝒏

𝒆𝒙⋅𝒆𝒊𝒙+

𝝅𝟒𝒊 − 𝒆−𝒊𝒙−

𝝅𝟒𝒊

𝟐𝒊𝒅𝒙

∞

𝟎

=


𝟏

𝒏!(𝒆𝝅𝟒𝒊

𝟐𝒊∫ 𝒙𝒏𝒆−(𝟏−𝒊)𝒙∞

𝟎

𝒅𝒙 −𝒆−𝒑𝒊𝟒𝒊

𝟐𝒊∫ 𝒙𝒏𝒆−(𝟏+𝒊)𝒙∞

𝟎

𝒅𝒙) =(𝟏±𝒙)𝒊=𝒖


𝟏

𝒏!(𝒆𝝅𝟒𝒊

𝟐𝒊∫

𝒖(𝒏+𝟏)−𝟏𝒆−𝒖

(𝟏 − 𝒊)𝒏+𝟏

∞

𝟎

𝒅𝒖 −𝒆−𝝅𝟒𝒊

𝟐𝒊∫

𝒖(𝒏+𝟏)−𝟏𝒆−𝒖

(𝟏 + 𝒊)𝒏+𝟏

∞

𝟎

𝒅𝒖) =


𝟏

𝚪(𝒏 + 𝟏)⋅ 𝚪(𝒏 + 𝟏) [

𝒆𝝅𝟒𝒊

𝟐𝒊(𝟏 − 𝒊)𝒏+𝟏−

𝒆−𝝅𝟒𝒊

𝟐𝒊(𝒊 + 𝟏)𝒏+𝟏] =


[𝒆𝝅𝟒𝒊

𝟐𝒊 (√𝟐𝒆−𝝅𝟒𝒊)𝒏+𝟏 −

𝒆−𝝅𝟒𝒊

𝟐𝒊 (√𝟐𝒆𝝅𝟒𝒊)𝒏+𝟏] =


𝟏

(√𝟐)𝒏+𝟏(𝒆𝝅𝟐𝒊+𝝅𝟒𝒏𝒊

𝟐𝒊−𝒆−𝝅𝟐𝒊−𝝅𝟒𝒏𝒊

𝟐𝒊) =


𝟏

(√𝟐)𝒏+𝟏

[(𝐜𝐨𝐬

𝝅𝟐 + 𝒊 𝐬𝐢𝐧

𝝅𝟐) 𝒆

(𝝅𝟒𝒏)𝒊

𝟐𝒊−(𝐜𝐨𝐬 (−

𝝅𝟐) + 𝒊 𝒔𝒊𝒏 (–

𝝅𝟐))𝒆

(−𝝅𝟒𝒏)𝒊

𝟐𝒊] =

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𝟏

(√𝟐)𝒏+𝟏(𝒆(𝝅𝟒𝒏)𝒊 + 𝒆(−

𝝅𝟒𝒏)𝒊

𝟐) = 𝐥𝐢𝐦

𝒏→∞

𝟏

(√𝟐)𝒏+𝟏 𝐜𝐨𝐬 (

𝝅

𝟒𝒏) = 𝟎

Solution 2 by Syed Shahabudeen-India


𝟏

𝒏!∫


𝒆𝒙

∞

𝟎

𝒅𝒙 =𝟏

√𝟐𝐥𝐢𝐦𝒏→∞

𝟏

𝒏!∫

𝒙𝒏(𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙)

𝒆𝒙

∞

𝟎

𝒅𝒙 =

=𝟏


𝟏

𝒏!(𝑰𝒎∫ 𝒙𝒏𝒆−𝒙(𝟏−𝒊)

∞

𝟎

𝒅𝒙 + 𝑹𝒆∫ 𝒙𝒏𝒆−𝒙(𝟏−𝒊)∞

𝟎

𝒅𝒙) =

=𝟏


𝟏

𝒏!(𝑰𝒎𝑴(𝒆−𝒙(𝟏−𝒊)) + 𝑹𝒆𝑴(𝒆−𝒙(𝟏−𝒊))) ; (𝒂𝒑𝒑𝒍𝒚 𝑴𝒆𝒍𝒍𝒊𝒏 𝑻𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎)

=𝟏


(𝑰𝒎(𝟏 − 𝒊)−(𝒏+𝟏) +𝑹𝒆(𝟏 − 𝒊)−(𝒏+𝟏)) =


𝟏

(√𝟐)𝒏+𝟐 (𝑰𝒎(𝒆

𝒊𝝅(𝒏+𝟏)𝟒 ) + 𝑹𝒆 (𝒆𝒊

𝝅(𝒏+𝟏)𝟒 )) =


𝟏

(√𝟐)𝒏+𝟐 (𝐬𝐢𝐧(

𝝅(𝒏 + 𝟏)

𝟒) + 𝐜𝐨𝐬 (

𝝅(𝒏 + 𝟏)

𝟒)) = 𝐥𝐢𝐦

𝒏→∞

𝐜𝐨𝐬 (𝝅𝒏𝟒 )

(√𝟐)𝒏+𝟏 = 𝟎, 𝐛𝐞𝐜𝐚𝐮𝐬𝐞

𝟎 ← −𝟏

(√𝟐)𝒏+𝟏 ≤

𝐜𝐨𝐬 (𝝅𝒏𝟒 )

(√𝟐)𝒏+𝟏 ≤

𝟏

(√𝟐)𝒏+𝟏 → 𝟎



𝟏

𝒏!∫


𝒆𝒙

∞

𝟎

𝒅𝒙 ; (𝟏)

𝝎 = ∫𝒙𝒏 𝐬𝐢𝐧 (𝒙 +

𝝅𝟒)

𝒆𝒙

∞

𝟎

𝒅𝒙 = 𝑰𝒎{∫ 𝒙𝒏𝒆𝒙(𝒊−𝟏)+

𝒊𝝅𝟒

∞

𝟎

𝒅𝒙} =

= 𝑰𝒎{𝒆𝒊𝝅𝟒∫ 𝒙𝒏𝒆𝒙(𝒊−𝟏)

∞

𝟎

𝒅𝒙} =𝒖=−𝒙(𝒊−𝟏)

𝑰𝒎{𝒆𝒊𝝅𝟒

(𝟏 − 𝒊)𝒏−𝟏∫ 𝒖𝒏𝒆−𝒖∞

𝟎

𝒅𝒖}

= 𝚪(𝒏 + 𝟏)𝑰𝒎{𝒆𝒊𝝅𝟒

𝟏

(𝟏 − 𝒊)𝒏−𝟏} = 𝒏! 𝑰𝒎{𝒆𝒊

𝝅𝟒(𝟏 + 𝒊)𝒏−𝟏

𝟐𝒏−𝟏} =

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=𝒏!

𝟐𝒏−𝟏𝑰𝒎{𝒆𝒊

𝝅𝟒𝟐𝒏−𝟏𝟐 𝒆𝒊

𝝅𝟒(𝒏−𝟏)} =

𝒏!

𝟐𝒏−𝟏𝟐

𝑰𝒎(𝒆𝒊𝒏𝝅𝟒)

𝛀 =𝒏!

𝟐𝒏−𝟏𝟐


𝟒)(𝟏)⇒


𝟏

𝒏!⋅𝒏!

𝟐𝒏−𝟏𝟐


𝟒) = √𝟐 𝐥𝐢𝐦

𝒏→∞

𝐬𝐢𝐧 (𝒏𝝅𝟒 )

√𝟐𝒏= √𝟐 𝐥𝐢𝐦

𝒏→∞

𝒏𝝅

𝟐⋅𝟏

√𝟐𝒏= 𝟎

Solution 4 by Ajenikoko Gbolahan-Nigeria


𝟏

𝒏!∫


𝒆𝒙

∞

𝟎


𝟏

𝒏!⋅𝟏

√𝟐∫

𝒙𝒏 𝐬𝐢𝐧 𝒙 + 𝒙𝒏 𝐜𝐨𝐬 𝒙

𝒆𝒙

∞

𝟎

𝒅𝒙 =


𝟏

𝒏!⋅𝟏

√𝟐∫ [𝑰𝒎(𝒙𝒏𝒆𝒊𝒙−𝒙) + 𝑹𝒆(𝒙𝒏𝒆𝒊𝒙−𝒙)]𝒅𝒙∞

𝟎

=


𝟏

𝒏!⋅𝟏

√𝟐∫ [𝑰𝒎(𝒙𝒏𝒆−(𝟏−𝒊)𝒙) + 𝑹𝒆(𝒙𝒏𝒆(𝟏−𝒊)𝒙)]∞

𝟎

𝒅𝒙 =


𝟏

𝒏!⋅𝟏

√𝟐[𝑰𝒎𝓛{𝒙𝒏}𝒔=𝟏−𝒊 + 𝑹𝒆𝓛{𝒙

𝒏}𝒔=𝟏−𝒊] =


𝟏

𝒏!⋅𝟏

√𝟐(𝑰𝒎(

𝒏!

(𝟏 − 𝒊)𝒏+𝟏) + 𝑹𝒆 (

𝒏!

(𝟏 − 𝒊)𝒏+𝟏)) =


𝟏

√𝟐[(𝟏

√𝟐)𝒏+𝟏

(−𝐬𝐢𝐧(𝝅(−𝒏 − 𝟏)

𝟒) + 𝐜𝐨𝐬 (

𝝅(−𝒏 − 𝟏)

𝟒)) =


𝟏

√𝟐𝒏+𝟐(− 𝐬𝐢𝐧(

𝝅(𝒏 − 𝟏)

𝟒) + 𝐜𝐨𝐬 (

𝝅(−𝒏 − 𝟏)

𝟒)) = 𝟎

1589. Prove that:

∑𝚪(𝒏 +

𝟏𝟑)𝝍(𝒏 +

𝟏𝟑)

𝟑𝒏𝒏!

∞

𝒏=𝟎

= −√𝟑

𝟐

𝟑

𝚪 (𝟏

𝟑) {𝟏

𝟐𝐥𝐨𝐠 (

𝟏𝟎𝟖

𝟗) + 𝜸 +

𝝅

𝟐√𝟑}


Solution by Dawid Bialek-Poland

𝛀 = ∑𝚪(𝒏 +

𝟏𝟑)𝝍(𝒏 +

𝟏𝟑)

𝟑𝒏𝒏!

∞

𝒏=𝟎

= ∑𝚪′ (𝒏 +

𝟏𝟑)

𝟑𝒏𝒏!

∞

𝒏=𝟎

= ∑𝟏

𝟑𝒏𝒏!⋅𝝏

𝝏𝒏(∫ 𝒆−𝒕 ⋅ 𝒕𝒏−

𝟐𝟑

∞

𝟎

𝒅𝒕)

∞

𝒏=𝟎

=

= ∑𝟏

𝟑𝒏𝒏!⋅ ∫ 𝒆−𝒕 ⋅ 𝒕𝒏−

𝟐𝟑 ⋅ 𝐥𝐨𝐠 𝒕

∞

𝟎

𝒅𝒕

∞

𝒏=𝟎

= ∫ 𝒆−𝒕 ⋅ 𝐥𝐨𝐠 𝒕 ⋅ 𝒕−𝟐𝟑∑

(𝒕𝟑)𝒏

𝒏!

∞

𝒏=𝟎

𝒅𝒕∞

𝟎

=

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= ∫ 𝒆−𝒕 ⋅ 𝐥𝐨𝐠 𝒕 ⋅ 𝒕−𝟐𝟑 ⋅ 𝒆

𝒕𝟑

∞

𝟎

𝒅𝒕 = ∫ 𝒕−𝟐𝟑 ⋅ 𝐥𝐨𝐠 𝒕 ⋅ 𝒆−

𝟐𝟑𝒕

∞

𝟎

𝒅𝒕 =

=𝝏

𝝏𝒔∫ 𝒕𝒔 ⋅ 𝒆−

𝟐𝟑𝒕𝒅𝒕|

𝒔=−𝟐𝟑

∞

𝟎

=𝝏

𝝏𝒔(𝚪(𝒔 + 𝟏)

(𝟐𝟑)𝒔+𝟏 )

𝒔=−𝟐𝟑

=

= ((𝟐

𝟑)−𝒔−𝟏

⋅ 𝚪(𝒔 + 𝟏) ⋅ (𝝍𝟎(𝒔 + 𝟏) − 𝐥𝐨𝐠 (𝟐

𝟑))𝒔=−

𝟐𝟑

=

= (𝟐

𝟑)−𝟏𝟑⋅ 𝚪 (

𝟏

𝟑) (𝝍𝟎 (

𝟏

𝟑) + 𝐥𝐨𝐠 (

𝟐

𝟑))

𝝍𝟎 (𝟏

𝟑) = −

𝝅

𝟐√𝟑− 𝜸 −

𝟑

𝟐𝐥𝐨𝐠 𝟑

𝛀 = −√𝟑

𝟐

𝟑

𝚪(𝟏

𝟑) {𝟑

𝟐𝐥𝐨𝐠 𝟑 − 𝐥𝐨𝐠 (

𝟑

𝟐) + 𝜸 +

𝝅

𝟐√𝟑} =

= −√𝟑

𝟐

𝟑

𝚪(𝟏

𝟑) {𝟑

𝟐𝐥𝐨𝐠𝟑 − 𝐥𝐨𝐠𝟑 + 𝐥𝐨𝐠 𝟐 + 𝜸 +

𝝅

𝟐√𝟑} =

= −√𝟑

𝟐

𝟑

𝚪(𝟏

𝟑) {𝟑

𝟐𝐥𝐨𝐠𝟑 − 𝐥𝐨𝐠𝟑 +

𝟏

𝟐𝐥𝐨𝐠 𝟒 + 𝜸 +

𝝅

𝟐√𝟑} =

= −√𝟑

𝟐

𝟑

𝚪(𝟏

𝟑) {𝜸 +

𝝅

𝟐√𝟑+𝟏

𝟐𝐥𝐨𝐠 𝟏𝟐}

1590. Find:


∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)

𝒌

𝒊=𝟏

]

−𝟏𝒏

𝒌=𝟏



𝒔𝒌 =∑(𝒊 + 𝟏)(𝒌 + 𝟐)

𝒌

𝒊=𝟏

−∑𝒊(𝒊 + 𝟏)

𝒌

𝒊=𝟏

= (𝒌 + 𝟐)∑𝒊

𝒌

𝒊=𝟏

+ 𝒌(𝒌 + 𝟐) −∑𝒊𝟐𝒌

𝒊=𝟏

−∑𝒊

𝒌

𝒊=𝟏

=

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=𝒌(𝒌 + 𝟏)𝟐

𝟐+ 𝒌(𝒌 + 𝟐) −∑𝒊𝟐

𝒌

𝒊=𝟏

=𝒌𝟑 + 𝟗𝒌𝟐 + 𝟏𝟒𝒌

𝟔⇒

⇒𝟏

𝒔𝒌=

𝟔

𝒌𝟑 + 𝟗𝒌𝟐 + 𝟏𝟒𝒌=

𝟔

𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)=

𝟔

𝟑𝟓(𝒌 + 𝟕)+𝟑

𝟕𝒌−

𝟑

𝟓(𝒌 + 𝟐)

⇒∑𝟏

𝒔𝒌

𝒏

𝒌=𝟏

=∑(𝟔

𝟑𝟓(𝒌 + 𝟕)+𝟑

𝟕𝒌−

𝟑

𝟓(𝒌 + 𝟐))

𝒏

𝒌=𝟏

=

=∑𝟔

𝟑𝟓(𝒌 + 𝟕)

𝒏

𝒌=𝟏

+∑𝟑

𝟕𝒌

𝒏

𝒌=𝟖

−∑𝟑

𝟓(𝒌 + 𝟐)

𝒏

𝒌=𝟔

+∑𝟑

𝟕𝒌

𝟕

𝒌=𝟏

−∑𝟑

𝟓(𝒌 + 𝟐)

𝟓

𝒌=𝟏

𝟔

𝟑𝟔+𝟑

𝟕−𝟑

𝟓= 𝟎 ⇒ ∑

𝟔

𝟑𝟓(𝒌 + 𝟕)

𝒏

𝒌=𝟏

+∑𝟑

𝟕𝒌

𝒏

𝒌=𝟖

−∑𝟑

𝟓(𝒌 + 𝟐)

𝒏

𝒌=𝟔

→ −


∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)

𝒌

𝒊=𝟏

]

−𝟏𝒏

𝒌=𝟏

= ∑𝟑

𝟕𝒌

𝟕

𝒌=𝟏

−∑𝟑

𝟓(𝒌 + 𝟐)

𝟓

𝒌=𝟏

=

= ∑𝟑

𝟕𝒌

𝟕

𝒌=𝟏

−∑𝟑

𝟓𝒌

𝟕

𝒌=𝟑

=𝟑

𝟕𝑯𝟕 −

𝟑

𝟓𝑯𝟕 +

𝟑

𝟓+𝟑

𝟏𝟎= −

𝟔

𝟑𝟓𝑯𝟕 +

𝟗

𝟏𝟎

Therefore,


∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)

𝒌

𝒊=𝟏

]

−𝟏𝒏

𝒌=𝟏

= −𝟔

𝟑𝟓𝑯𝟕 +

𝟗

𝟏𝟎


∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)

𝒌

𝒊=𝟏

=∑[𝒊(𝒌 + 𝟏) − 𝒊𝟐 + (𝒌 + 𝟐)]

𝒌

𝒊=𝟏

=

=(𝒌 + 𝟏)(𝒌 + 𝟏)𝒌

𝟐−𝟏

𝟔𝒌(𝒌 + 𝟏)(𝟐𝒌 + 𝟏) + (𝒌 + 𝟐)𝒌 =

=𝒌

𝟔(𝒌𝟐 + 𝟗𝒌+ 𝟏𝟒) =

𝟏

𝟔𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)

∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)

𝒌

𝒊=𝟏

]

−𝟏𝒏

𝒌=𝟏

=∑𝟔

𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)

𝒏

𝒌=𝟏

= ∑[𝟑

𝟕𝒌−

𝟑

𝟓(𝒌 + 𝟐)+

𝟔

𝟑𝟓(𝒌 + 𝟕)]

𝒏

𝒌=𝟏

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=𝟏𝟓

𝟑𝟓∑(

𝟏

𝒌−

𝟏

𝒌 + 𝟐)

𝒏

𝒌=𝟏

−𝟔

𝟑𝟓∑(

𝟏

𝒌 + 𝟐−

𝟏

𝒌 + 𝟕)

𝒏

𝒌=𝟏

Therefore,


∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)

𝒌

𝒊=𝟏

]

−𝟏𝒏

𝒌=𝟏

=𝟏𝟓

𝟑𝟓(𝟏 +

𝟏

𝟐) −

𝟔

𝟑𝟓(𝟏

𝟑+𝟏

𝟒+𝟏

𝟓+𝟏

𝟔+𝟏

𝟕) =

𝟓𝟓𝟖

𝟏𝟐𝟐𝟓.


𝑺𝒌 =∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)

𝒌

𝒊=𝟏

= (𝒌 + 𝟏)∑𝒊

𝒌

𝒊=𝟏

−∑𝒊𝟐𝒌

𝒊=𝟏

+ (𝒌 + 𝟐)∑𝟏

𝒌

𝒊=𝟏

=

=𝒌(𝒌 + 𝟏)𝟐

𝟐−𝒌(𝒌 + 𝟏)(𝟐𝒌 + !)

𝟔+ 𝒌(𝒌 + 𝟐) =

𝟏

𝟔𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)

𝑺′𝒏 =∑𝟏

𝑺𝒌

𝒏

𝒌=𝟏

=∑𝟔

𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)

𝒏

𝒌=𝟏

=∑𝟑

𝟕𝒌

𝒏

𝒌=𝟏

−∑𝟑

𝟓(𝒌 + 𝟐)

𝒏

𝒌=𝟏

+∑𝟔

𝟑𝟓(𝒌 + 𝟕)

𝒏

𝒌=𝟏

=

=𝟑

𝟕𝑯𝒏 −

𝟑

𝟓(𝑯𝒏 − 𝑯𝟐) +

𝟔

𝟑𝟓(𝑯𝒏 − 𝑯𝟕)

Therefore,


∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)

𝒌

𝒊=𝟏

]

−𝟏𝒏

𝒌=𝟏

=𝟗

𝟏𝟎−𝟏𝟎𝟖𝟏

𝟐𝟒𝟓𝟎=𝟓𝟓𝟖

𝟏𝟐𝟐𝟓

1591. Prove that:


(√𝒏𝟐 − 𝒏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏))

𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏)

= 𝒆−𝟏𝟖




(√𝒏𝟐 − 𝒏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏))



((𝟏 + 𝒖 − 𝟏)𝟏𝒖−𝟏)

(𝒖−𝟏)(𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏))

=

= 𝒆(𝒖−𝟏)(𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏)) = 𝒆𝑺

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𝑺 = 𝐥𝐢𝐦𝒏→∞

(√𝒏𝟐 − 𝒏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏) − 𝟏) (𝒏𝟐 + 𝐬𝐢𝐧(𝟑𝒏)) =


(𝒏𝟐√𝟏 −𝟏

𝒏+𝒏𝟐

𝟒𝐬𝐢𝐧 (

𝟐

𝒏) − 𝒏𝟐) + 𝐬𝐢𝐧(𝟑𝒏)(√𝟏 −

𝟏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏) − 𝟏)

⏟ 𝟎

=


(𝒏𝟐√𝟏 −𝟏

𝒏+𝒏

𝟐− 𝒏𝟐) =

√𝟏−𝟏𝒏=𝒖

𝐥𝐢𝐦𝒖→𝟏

𝟐 − 𝟐𝒖

𝟒(𝟏 − 𝒖𝟐)(−𝟐𝒖)= −

𝟏

𝟖

Therefore,


(√𝒏𝟐 − 𝒏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏))


= 𝒆−𝟏𝟖


(√𝒏𝟐 − 𝒏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏))


= (√𝟏 −𝟏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏))

𝒏𝟐+𝐬𝐢𝐧𝟑𝒏

=

= 𝒆𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏) 𝐥𝐨𝐠(√𝟏−

𝟏𝒏+𝟏𝟒𝐬𝐢𝐧(

𝟐𝒏))

𝐥𝐨𝐠 (√𝟏 −𝟏𝒏 +

𝟏𝟒𝐬𝐢𝐧 (

𝟐𝒏))

√𝟏 −𝟏𝒏 +

𝟏𝟒𝐬𝐢𝐧 (

𝟐𝒏) − 𝟏

→ 𝟏; (𝒏 → ∞)


(𝒏𝟐 + 𝐬𝐢𝐧(𝟑𝒏))(√𝟏 −𝟏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏) − 𝟏) =


𝒏𝟐 + 𝐬𝐢𝐧(𝟑𝒏)

𝒏𝟐(√𝟏 −

𝟏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏) − 𝟏) ⋅ 𝒏𝟐

Now, we want to find:

𝝎 = 𝐥𝐢𝐦𝒏→∞

(√𝟏 −𝟏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏) − 𝟏) ⋅ 𝒏𝟐

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√𝟏 − 𝒏 ≅ 𝟏 −𝒏

𝟐−𝒏𝟐

𝟖−𝒏𝟑

𝟏𝟔

𝐬𝐢𝐧 (𝟐

𝒏) ≅(𝟐𝒏 −

𝟒𝒏𝟑

𝟑)

𝟏

𝒏𝟐(𝟏 −

𝒏

𝟐−𝒏𝟐

𝟖−𝒏𝟑

𝟏𝟔+𝟏

𝟒(𝟐𝒏 −

𝟒𝒏𝟑

𝟑) − 𝟏) =; (𝒏 → 𝟎)

𝟏

𝒏𝟐(−𝒏𝟐

𝟖−𝒏𝟑

𝟏𝟔+ (−

𝒏𝟑

𝟑)) ⋅ 𝒏 = −

𝟏

𝟖−𝒏

𝟏𝟔−𝒏

𝟑

Hence,

𝝎 = 𝐥𝐢𝐦𝒏→∞

(√𝟏 −𝟏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏) − 𝟏) ⋅ 𝒏𝟐 = −

𝟏

𝟖

Therefore,


(√𝒏𝟐 − 𝒏

𝒏+𝟏

𝟒𝐬𝐢𝐧 (

𝟐

𝒏))


= 𝒆−𝟏𝟖

1592. Find:


√𝐜𝐨𝐬𝟐𝒏𝝅

𝟕− 𝟐𝟏−𝟐𝒏 ∑ (

𝒏

𝒊) (𝒏

𝒋) 𝐜𝐨𝐬

𝟐(𝒋 − 𝒊)𝝅

𝟕𝟎≤𝒊<𝑗≤𝑛

𝒏


Solution 1 by Surjeet Singhania-India

𝐃𝐞𝐧𝐨𝐭𝐞:𝑿𝒏 = ∑ (𝒏

𝒊) (𝒏

𝒋) 𝐜𝐨𝐬

𝟐𝝅(𝒋 − 𝒊)

𝟕𝟎≤𝒊<𝑗≤𝑛

Observe that 𝟎 ≤ 𝒊 < 𝑗 ≤ 𝑛 it meand 𝒊 can take value from 𝟎 to 𝒏 − 𝟏 and 𝒋 take value

from 𝒊 + 𝟏 to 𝒏,

𝑿𝒏 = ∑ (𝒏

𝒊) (𝒏

𝒋) 𝐜𝐨𝐬

𝟐(𝒋 − 𝒊)𝝅

𝟕𝟎≤𝒊<𝑗≤𝑛

= ∑ ∑ (𝒏

𝒊) (𝒏

𝒋) 𝐜𝐨𝐬

𝟐(𝒋 − 𝒊)𝝅

𝟕

𝒏

𝒋=𝒊+𝟏

=

𝒏−𝟏

𝒊=𝟎

= ∑(𝒏

𝒋)

𝒏−𝟏

𝒊=𝟎

∑ (𝒏

𝒋) 𝐜𝐨𝐬

𝟐𝝅(𝒋 − 𝒊)

𝟕

𝒏

𝒋=𝒊+𝟏

=𝒋−𝒊=𝒌

∑∑(𝒏

𝒊) (

𝒏

𝒌 + 𝒊) 𝐜𝐨𝐬

𝟐𝒌𝝅

𝟕

𝒏

𝒌=𝟏

𝒏−𝟏

𝒊=𝟎

=

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= ∑∑(𝒏

𝒊)(

𝒌

𝒌 + 𝒊) 𝐜𝐨𝐬

𝟐𝒌𝝅

𝟕

𝒏−𝟏

𝒊=𝟏

𝒏

𝒌=𝟏

=𝟏

𝟐𝝅𝒊∑∮

(𝟏 + 𝒛)𝒏

𝒛𝒌+𝟏∑(𝒏𝒊)𝟏

𝒛𝒊𝒅𝒛

𝒏−𝟏

𝒊=𝟎

𝐜𝐨𝐬 (𝟐𝒌𝝅

𝟕)

𝒏

𝒌=𝟏

=

=𝟏

𝟐𝝅𝒊∑𝐜𝐨𝐬 (

𝟐𝒌𝝅

𝟕)∮(

(𝟏 + 𝒛)𝟐𝒏

𝒛𝒏+𝒌+𝟏−(𝟏 + 𝒛)𝒏

𝒛𝒌+𝟏+𝒏)𝒅𝒛

𝒏

𝒌=𝟏

=∑(𝟐𝒏

𝒏 + 𝒌) 𝐜𝐨𝐬 (

𝟐𝒌𝝅

𝟕)

𝒏

𝒌=𝟏

=

=∑(𝟐𝒏

𝒏 − 𝒌)

𝒏

𝒌=𝟏

𝐜𝐨𝐬 (𝟐𝒌𝝅

𝟕) = ∑(

𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕

𝒏−𝟏

𝒌=𝟏⏟ 𝑨

𝑨𝒍𝒔𝒐,𝑿𝒏 =∑(𝟐𝒏

𝒏 + 𝒌)

𝒏

𝒌=𝟏

𝐜𝐨𝐬𝟐𝒌𝝅

𝟕= ∑ (

𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕

𝟐𝒏

𝒌=𝒏+𝟏

⇒ 𝑿𝒏 =𝟏

𝟐∑(

𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕

𝟐𝒏

𝒌=𝟎

−𝟏

𝟐(𝟐𝒏

𝒏)

𝑿𝒏 =𝟏

𝟐𝕽∑(

𝟐𝒏

𝒌) 𝐞𝐱𝐩 (

𝟐𝝅(𝒏 − 𝒌)

𝟕)

𝟐𝒏

𝒌=𝟎

−𝟏

𝟐(𝟐𝒏

𝒏) =

=𝟏

𝟐𝕽(𝐞𝐱𝐩 (

𝟐𝝅𝒊𝒏

𝟕)∑(

𝟐𝒏

𝒌)

𝟐𝒏

𝒌=𝟎

𝐞𝐱𝐩 (−𝟐𝝅𝒊𝒌

𝟕))−

𝟏

𝟐(𝟐𝒏

𝒏) =

=𝟏

𝟐𝕽(𝐞𝐱𝐩 (

𝟐𝝅𝒊

𝟕) (𝟏 + 𝐞𝐱𝐩 (−

𝟐𝝅

𝟕))𝟐𝒏

) −𝟏

𝟐(𝟐𝒏

𝒏)

𝑿𝒏 = ∑ (𝒏

𝒊) (𝒏

𝒋) 𝐜𝐨𝐬

𝟐𝝅(𝒋 − 𝒊)

𝟕𝟎≤𝒊<𝑗≤𝑛

= 𝟐𝟐𝒏−𝟏 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅

𝟕) −

𝟏

𝟐(𝟐𝒏

𝒏)


√𝐜𝐨𝐬𝟐𝒏 (𝝅

𝟕) − 𝟐𝟏−𝟐𝒏𝑿𝒏

𝒏= 𝐥𝐢𝐦𝒏→∞

√𝟒−𝒏 (𝟐𝒏

𝒏)

𝒏


√𝟏

√𝒏𝝅

𝒏

= 𝟏

Therefore,



𝟕− 𝟐𝟏−𝟐𝒏 ∑ (

𝒏

𝒊) (𝒏

𝒋) 𝐜𝐨𝐬

𝟐(𝒋 − 𝒊)𝝅

𝟕𝟎≤𝒊<𝑗≤𝑛

𝒏= 𝟏.


𝑺 = ∑ (𝒏

𝒊) (𝒏

𝒋) 𝐜𝐨𝐬

𝟐(𝒋 − 𝒊)𝝅

𝟕𝟎≤𝒊<𝑗≤𝑛

=

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= ∑ (𝒏

𝒊) (𝒏

𝒋) 𝐜𝐨𝐬

𝟐(𝒋 − 𝒊)𝝅

𝟕𝟎≤𝒊<𝑗≤𝑛

+ ∑ (𝒏

𝒊)(𝒏

𝒋) 𝐜𝐨𝐬

𝟐(𝒊 − 𝒋)𝝅

𝟕𝟎≤𝒊<𝑗≤𝑛

=

= 𝑹𝒆 [∑(𝒏

𝒊) (𝒏

𝒋) 𝒆

(𝒊−𝒋)𝝅𝟕

𝒊≠𝒋

] ⇒∑(𝒏

𝒋)𝟐

𝒏

𝒋=𝟎

+ 𝑺 =

= 𝑹𝒆 {[(𝒏

𝟎)+ (

𝒏

𝟏)𝒆

𝒊𝝅𝟕 + (

𝒏

𝟐)𝒆

𝟐𝝅𝒊𝟕 +⋯+ (

𝒏

𝒏)𝒆

𝒏𝝅𝒊𝟕 ]

⋅ [(𝒏

𝟎) + (

𝒏

𝟏)𝒆−

𝒊𝝅𝟕 + (

𝒏

𝟐)𝒆−

𝟐𝝅𝒊𝟕 +⋯+ (

𝒏

𝒏)𝒆−

𝒏𝝅𝒊𝟕 ]}

= 𝑹𝒆 [(𝟏 + 𝒆𝒊𝝅𝟕 )

𝒏

(𝟏 + 𝒆−𝒊𝝅𝟕 )

𝒏

] = 𝑹𝒆 [(𝟏 + 𝟐𝐜𝐨𝐬𝝅𝒊

𝟕+ 𝟏)

𝒏

] =

= 𝑹𝒆[𝟐𝒏 (𝟏 + 𝐜𝐨𝐬𝝅

𝟕)𝒏

] = 𝟐𝒏 (𝟏 + 𝐜𝐨𝐬𝝅

𝟕)𝒏

= 𝟐𝒏 (𝟐 𝐜𝐨𝐬𝟐𝟐𝝅

𝟕)𝒏

= 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅

𝟕)

⇒ 𝑺 = 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅

𝟕)−∑(

𝒏

𝒋)𝟐

𝒏

𝒋=𝟎

= 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅

𝟕) − (

𝟐𝒏

𝒏)

⇒ 𝟐−𝟐𝒏 = 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅

𝟕) −

𝟏

𝟐𝟐𝒏(𝟐𝒏

𝒏)

⇒ 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅

𝟕) − 𝟐𝟏−𝟐𝒏 ∑ (

𝒏

𝒊)(𝒏

𝒋) 𝐜𝐨𝐬

(𝒋 − 𝒊)𝝅

𝟕𝟎≤𝒊<𝑗≤𝑛

=𝟏

𝟐𝟐𝒏(𝟐𝒏

𝒏)

Therefore,



𝟕− 𝟐𝟏−𝟐𝒏 ∑ (

𝒏

𝒊)(𝒏

𝒋) 𝐜𝐨𝐬

𝟐(𝒋 − 𝒊)𝝅

𝟕𝟎≤𝒊<𝑗≤𝑛

𝒏= 𝐥𝐢𝐦𝒏→∞

√𝟏

𝟐𝟐𝒏(𝟐𝒏

𝒏)

𝒏

=

=𝑪−𝑫 𝟏

𝟒𝐥𝐢𝐦𝒏→∞

(𝟐𝒏 + 𝟐)(𝟐𝒏 + 𝟏)

(𝒏 + 𝟏)(𝒏 + 𝟏)= 𝟏

1593. Find:


𝟏

𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌) (𝐭𝐚𝐧−𝟏 (

𝟏

√𝒌))𝟐

𝒏

𝒌=𝟏




𝟏

𝒏𝟐∑𝐥𝐨𝐠(𝟏 +

𝟏

𝒌)(𝐭𝐚𝐧−𝟏 (

𝟏

√𝒌))𝟐

𝒏

𝒌=𝟏

=𝑪−𝑺


∑ 𝐥𝐨𝐠 (𝟏 +𝟏𝒌) (𝐭𝐚𝐧

−𝟏 (𝟏

√𝒌))𝟐

𝒏+𝟏𝒌=𝟏 –∑ 𝐥𝐨𝐠 (𝟏 +

𝟏𝒌)(𝐭𝐚𝐧

−𝟏 (𝟏

√𝒌))𝟐

𝒏𝒌=𝟏

(𝒏 + 𝟏)𝟐 − 𝒏𝟐=

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𝐥𝐨𝐠 (𝟏 +𝟏

𝒏 + 𝟏)(𝐭𝐚𝐧−𝟏 (

𝟏

√𝒏 + 𝟏))𝟐

𝟐𝒏 + 𝟏= 𝟎

Therefore,


𝟏

𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌) (𝐭𝐚𝐧−𝟏 (

𝟏

√𝒌))𝟐

𝒏

𝒌=𝟏

= 𝟎.

Solution 2 Ruxandra Daniela Tonilă-Romania

We have: |𝐭𝐚𝐧−𝟏 𝒙| ≤𝝅

𝟐, ∀𝒙 ∈ ℝ ⇔ (𝐭𝐚𝐧−𝟏 𝒙)𝟐 ≤

𝝅𝟐

𝟐, ∀𝒙 ∈ ℝ and 𝛀 ≥ 𝟎. Thus,

𝟎 ≤ 𝛀 ≤𝝅𝟐


𝟏

𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌)

𝒏

𝒌=𝟏

𝟎 ≤ 𝛀 ≤ 𝐥𝐢𝐦𝒏→∞

𝟏

𝒏𝟐𝐥𝐨𝐠 (∏

𝒌+ 𝟏

𝒌

𝒏

𝒌=𝟏

) ⇔ 𝟎 ≤ 𝛀 ≤ 𝐥𝐢𝐦𝒏→∞

𝟏

𝒏𝟐𝐥𝐨𝐠 (

(𝒏 + 𝟏)!

𝒏!) ⇔

𝟎 ≤ 𝛀 ≤𝝅𝟐


𝐥𝐨𝐠(𝒏 + 𝟏)

𝒏𝟐= 𝟎

Therefore,


𝟏

𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌) (𝐭𝐚𝐧−𝟏 (

𝟏

√𝒌))𝟐

𝒏

𝒌=𝟏

= 𝟎.


For 𝒙 > 𝟎, we have: 𝟎 < 𝐭𝐚𝐧−𝟏 𝒙 < 𝒙 and 𝟎 < 𝐥𝐨𝐠(𝟏 + 𝒙) < 𝒙. Thus,

𝟎 < 𝐭𝐚𝐧−𝟏 (𝟏

√𝒌) <

𝟏

√𝒌, ∀𝒌 > 𝟎 and 𝟎 < 𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌) <

𝟏

𝒌, ∀𝒌 > 𝟎.

𝟎 < 𝐥𝐨𝐠 (𝟏 +𝟏

𝒌) (𝐭𝐚𝐧−𝟏 (

𝟏

√𝒌))𝟐

≤𝟏

𝒌𝟐≤ 𝟏,∀𝒌 ≥ 𝟏

Hence,

𝟎 ≤∑𝐥𝐨𝐠 (𝟏 +𝟏

𝒌) (𝐭𝐚𝐧−𝟏 (

𝟏

√𝒌))𝟐

𝒏

𝒌=𝟏

≤ 𝒏, ∀𝒏 ∈ ℕ,𝒏 ≥ 𝟏

𝟎 ≤𝟏

𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌)(𝐭𝐚𝐧−𝟏 (

𝟏

√𝒌))𝟐

𝒏

𝒌=𝟏

≤𝟏

𝒏,∀𝒏 ∈ ℕ,𝒏 ≥ 𝟏

Therefore,

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𝟏

𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +

𝟏

𝒌) (𝐭𝐚𝐧−𝟏 (

𝟏

√𝒌))𝟐

𝒏

𝒌=𝟏

= 𝟎.

1594. Find:


√𝐬𝐢𝐧𝟐𝒏𝝅

𝟕− 𝟐𝟏−𝟐𝒏∑(−𝟏)𝒏−𝒌 (

𝟐𝒏

𝒏) 𝐜𝐨𝐬

(𝟐𝒏 − 𝟐𝒌)𝝅

𝟕

𝒏−𝟏

𝒌=𝟎

𝒏


Solution by Kamel Gandouli Rezgui-Tunisia

𝒌𝒏 = ∑(−𝟏)𝒏−𝒌 (𝟐𝒏

𝒏) 𝐜𝐨𝐬

(𝟐𝒏 − 𝟐𝒌)𝝅

𝟕

𝒏−𝟏

𝒌=𝟎

=∑(−𝟏)𝒌 (𝟐𝒏

𝒏 − 𝒌) 𝐜𝐨𝐬

𝟐𝒌𝝅

𝟕

𝒏

𝒌=𝟏

=

∑(−𝟏)𝒌 (𝟐𝒏

𝒏 + 𝒌) 𝐜𝐨𝐬

𝟐𝒌𝝅

𝟕

𝒏

𝒌=𝟏

= ∑ (−𝟏)𝒌−𝒏 (𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕

𝟐𝒏

𝒌=𝒏+𝟏

=

= ∑(−𝟏)𝒏−𝒌 (𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕=

𝟐𝒏

𝒌=𝟎

= ∑(−𝟏)𝒏−𝒌 (𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕

𝒏−𝟏

𝒌=𝟎

+ ∑ (−𝟏)𝒏−𝒌 (𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕

𝟐𝒏

𝒌=𝒏+𝟏

+ (𝟐𝒏

𝒏)

= 𝒌𝒏 + 𝒌𝒏 + (𝟐𝒏

𝒏)

⇒ 𝒌𝒏 =𝟏

𝟐(∑(−𝟏)𝒏−𝒌 (

𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕

𝟐𝒏

𝒌=𝟎

− (𝟐𝒏

𝒏))

∑(−𝟏)𝒏−𝒌 (𝟐𝒏

𝒌) 𝐜𝐨𝐬

𝟐𝝅(𝒏 − 𝒌)

𝟕

𝟐𝒏

𝒌=𝟎

= 𝑹𝒆(∑(−𝟏)𝒏−𝒌 (𝟐𝒏

𝒌) 𝒆

𝒊𝟐𝝅(𝒏−𝒌)𝟕

𝟐𝒏

𝒏=𝟎

) =

= 𝑹𝒆(∑(−𝟏)𝒏−𝒌𝒆𝟐𝒊𝝅𝒌𝟕 (

𝟐𝒏

𝒌) 𝒆𝒊⋅(−

𝟐𝝅𝒌𝟕)

𝟐𝒏

𝒌=𝟎

) = (−𝟏)𝒏𝑹(∑𝒆𝟐𝒊𝝅𝒏𝟕

𝟐𝒏

𝒌=𝟎

(𝟐𝒏

𝒌) 𝒆𝒊⋅

𝟓𝝅𝒌𝟕 ) =

= (−𝟏)𝒏𝑹𝒆(𝒆𝟐𝒊𝝅𝒏𝟕 (𝟏 + 𝒆𝒊(

𝟓𝒌𝝅𝟕))𝟐𝒏

) = (−𝟏)𝒏𝑹𝒆(𝒆𝟐𝒊𝝅𝒏𝟕 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏

𝟓𝝅

𝟕𝒆𝟓𝒊𝝅𝟏𝟒 )

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= (−𝟏)𝒏𝑹𝒆 (𝒆𝟐𝒊𝝅𝒏𝟕 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏

𝟓𝝅

𝟏𝟒𝒆𝟓𝒏𝒊𝝅𝟕 ) = (−𝟏)𝟏+𝒏𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏

𝟓𝝅

𝟕

𝒌𝒏 = (−𝟏)𝟏+𝒏𝟐𝟐𝒏−𝟏 𝐜𝐨𝐬𝟐𝒏

𝟓𝝅

𝟕−𝟏

𝟐(𝟐𝒏

𝒏)

𝟐𝟏−𝟐𝒏𝒌𝒏 = (−𝟏)𝟏+𝒏 𝐜𝐨𝐬𝟐𝒏

𝟓𝝅

𝟕−𝟏

𝟐𝟐𝒏(𝟐𝒏

𝒏)

⇒ 𝐬𝐢𝐧𝟐𝒏𝝅

𝟕− 𝟐𝟏−𝟐𝒏𝒌𝒏 = 𝐬𝐢𝐧

𝟐𝒏𝝅

𝟕+ (−𝟏)𝒏 𝐜𝐨𝐬𝟐𝒏

𝝅

𝟕⏟ →𝟎

+ 𝟐−𝟐𝒏 (𝟐𝒏

𝒏)


√(𝟐𝒏)!

𝟐𝟐𝒏(𝒏!)𝟐𝒏

= 𝐞𝐱𝐩 (𝐥𝐢𝐦𝒏→∞

𝟏

𝒏(𝐥𝐨𝐠(𝟐𝒏)! − 𝐥𝐨𝐠(𝟐𝟐𝒏) − 𝟐 𝐥𝐨𝐠(𝒏!))) =

= 𝐞𝐱𝐩 (𝐥𝐢𝐦𝒏→∞

𝟏

𝒏(∑ 𝐥𝐨𝐠 𝒌 −∑𝐥𝐨𝐠𝒌

𝒏

𝒌=𝟏

𝟐𝒏

𝒌=𝟏

)) = 𝐞𝐱𝐩(𝐥𝐢𝐦𝒏→∞

𝟏

𝒏∑𝐥𝐨𝐠 (𝟏 +

𝒌

𝒏)

𝒏

𝒌=𝟏

) =

= 𝐞𝐱𝐩 (∫ 𝐥𝐨𝐠 𝒙𝒅𝒙𝟐

𝟏

) =𝟏

𝒆

1595. −𝟏 < 𝒂 ≤ 𝒃 < 𝟏, 𝒏 ∈ ℕ∗, 𝑷𝒏 −Legendre’s polynomials. Find:

𝛀(𝒂, 𝒃) = 𝐥𝐢𝐦𝒏→∞

𝟏

𝒏∫

𝑷𝒏′ (𝒙)

𝑷𝒏−𝟏(𝒙) − 𝒙𝑷𝒏(𝒙)𝒅𝒙

𝒃

𝒂


Solution by Amrit Awasthi-Punjab-India

It is known that:

𝒅

𝒅𝒙𝑷𝒏(𝒙) =

𝒏

𝒙𝟐 − 𝟏(𝑷𝒏−𝟏(𝒙) − 𝒙𝑷𝒏(𝒙))

Rearrange and integrating:

∫𝑷𝒏′ (𝒙)


𝒃

𝒂

= ∫𝒏

𝟏 − 𝒙𝟐𝒅𝒙

𝒃

𝒂

=𝒏

𝟐𝒍𝒐𝒈(

(𝟏 + 𝒃)(𝟏 − 𝒂)

(𝟏 + 𝒂)(𝟏 − 𝒃))

𝛀(𝒂, 𝒃) = 𝐥𝐢𝐦𝒏→∞

𝟏

𝒏∫

𝑷𝒏′ (𝒙)


𝒃

𝒂

=𝟏

𝟐𝒍𝒐𝒈 (

(𝟏 + 𝒃)(𝟏 − 𝒂)

(𝟏 + 𝒂)(𝟏 − 𝒃))

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Note by editors (Daniel Sitaru, Florică Anastase)

∑𝑷𝒏(𝒙)𝒕𝒏

∞

𝒏=𝟎

= (𝟏 − 𝟐𝒕𝒙 + 𝒕𝟐)−𝟏𝟐 = 𝑲(𝒙, 𝒕)

−𝟏

𝟐⋅ (−𝟐𝒙 + 𝟐𝒕)(𝟏 − 𝟐𝒕𝒙 + 𝒕𝟐)−

𝟑𝟐 =

𝝏𝑲

𝝏𝒕

(𝒙 − 𝒕)(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)−𝟏 ⋅ 𝑲(𝒙, 𝒕) =𝝏𝑲

𝝏𝒕

(𝒙 − 𝒕)𝑲(𝒙, 𝒕) = (𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)𝝏𝑲

𝝏𝒕

(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)𝝏𝑲

𝝏𝒕+ (𝒕 − 𝒙)𝑲(𝒙, 𝒕) = 𝟎

(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)∑𝒏𝑷𝒏(𝒙)𝒕𝒏−𝟏

∞

𝒏=𝟎

+ (𝒕 − 𝒙)∑𝑷𝒏(𝒙)𝒕𝒏

∞

𝒏=𝟎

= 𝟎

Coefficient of 𝒕𝒏 is:

∑𝒏𝑷𝒏(𝒙)𝒕𝒏−𝟏

∞

𝒏=𝟎

− 𝟐𝒙𝒏∑𝑷𝒏(𝒙)𝒕𝒏

∞

𝒏=𝟎

+ 𝒏𝒙𝟐∑𝑷𝒏(𝒙)𝒕𝒏−𝟏

∞

𝒏=𝟎

+∑𝑷𝒏(𝒙)𝒕𝒏+𝟏

∞

𝒏=𝟎

− 𝒙∑𝑷𝒏(𝒙)𝒕𝒏

∞

𝒏=𝟎

= 𝟎

−𝟐𝒙𝒏𝑷𝒏(𝒙) − 𝒙𝑷𝒏(𝒙) + 𝒏𝑷𝒏−𝟏(𝒙) + (𝒏 + 𝟏)𝑷𝒏+𝟏(𝒙) = 𝟎

(𝒏 + 𝟏)𝑷𝒏+𝟏(𝒙) − (𝟐𝒏 + 𝟏)𝒙𝑷𝒏(𝒙) + 𝒏𝑷𝒏−𝟏(𝒙) = 𝟎; (𝟏)

(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)𝝏𝑲

𝝏𝒙− 𝒕𝑲(𝒙, 𝒕) =

= (𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐) ⋅ (−𝟏

𝟐) ⋅ (−𝟐𝒕) ⋅ (𝟏 − 𝟐𝒕𝒙 + 𝒕𝟐)−

𝟑𝟐 − 𝒕𝑲(𝒙, 𝒕) =

= 𝒕 ⋅ 𝑲(𝒙, 𝒕) − 𝒕 ⋅ 𝑲(𝒙, 𝒕) = 𝟎

(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐) ⋅ ∑𝑷𝒏′ (𝒙)𝒕𝒏

∞

𝒏=𝟎

− 𝒕 ⋅ ∑𝑷𝒏(𝒙)𝒕𝒏

∞

𝒏=𝟎

= 𝟎

Coefficient of 𝒕𝒏+𝟏 is:

𝑷𝒏+𝟏′ (𝒙) − 𝟐𝒙𝑷𝒏

′ (𝒙) + 𝑷𝒏−𝟏′ (𝒙) − 𝑷𝒏(𝒙) = 𝟎; (𝟐)

Derivative of (1):

(𝒏 + 𝟏)𝑷𝒏+𝟏′ (𝒙) − (𝟐𝒏 + 𝟏)𝑷𝒏(𝒙) − (𝟐𝒏 + 𝟏)𝒙𝑷𝒏

′ (𝒙) + 𝒏𝑷𝒏−𝟏′ (𝒙) = 𝟎; (𝟑)

By (2): 𝑷𝒏−𝟏′ (𝒙) = 𝑷𝒏(𝒙) + 𝟐𝒙𝑷𝒏

′ (𝒙) − 𝑷𝒏+𝟏′ (𝒙).

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Replace in (3):

(𝒏 + 𝟏)𝑷𝒏+𝟏′ (𝒙) − (𝟐𝒏 + 𝟏)𝑷𝒏(𝒙) − (𝟐𝒏 + 𝟏)𝒙𝑷𝒏

′ (𝒙) + 𝒏𝑷𝒏(𝒙) + 𝟐𝒏𝒙𝑷𝒏′ (𝒙) − 𝒏𝑷𝒏+𝟏

′ = 𝟎

𝑷𝒏+𝟏′ (𝒙) − 𝒙𝑷𝒏

′ (𝒙) = (𝒏 + 𝟏)𝑷𝒏(𝒙)

By (2): 𝑷𝒏+𝟏′ (𝒙) = 𝑷𝒏(𝒙) + 𝟐𝒙𝑷𝒏

′ (𝒙) − 𝑷𝒏−𝟏′ (𝒙)

Replace in (3):

(𝒏 + 𝟏)(𝑷𝒏(𝒙) + 𝟐𝒙𝑷𝒏′ (𝒙) − 𝑷𝒏−𝟏

′ (𝒙)) − (𝟐𝒏 + 𝟏)𝑷𝒏(𝒙) − (𝟐𝒏 + 𝟏)𝒙𝑷𝒏′ (𝒙) + 𝒏𝑷𝒏−𝟏

′ (𝒙) = 𝟎

𝒙𝑷𝒏′ (𝒙) − 𝑷𝒏−𝟏

′ (𝒙) = 𝒏𝑷𝒏(𝒙), 𝑷𝒏−𝟏′ (𝒙) = 𝒙𝑷𝒏

′ (𝒙) + 𝒏𝑷𝒏(𝒙)

𝑷𝒏′ (𝒙) − 𝒙(𝒙𝑷𝒏

′ (𝒙) − 𝒏𝑷𝒏(𝒙)) = 𝒏𝑷𝒏−𝟏(𝒙)

(𝟏 − 𝒙𝟐)𝑷𝒏′ (𝒙) = 𝒏(𝑷𝒏−𝟏(𝒙) − 𝒙𝑷𝒏(𝒙))

1596. Find:


𝟏

𝒏𝟐(𝟏 + 𝟐∑

𝟏

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟏

)

𝒏



𝟐∑𝟏

𝟐𝒌+ 𝟓

𝒏

𝒌=𝟏

=𝟐

𝟕+𝟐

𝟗+𝟐

𝟏𝟏+𝟐

𝟏𝟑+𝟐

𝟏𝟓+𝟐

𝟏𝟕+∑

𝟐

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟕

𝟐∑𝟏

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟏

=𝟖𝟑𝟖𝟏𝟗𝟐

𝟕𝟔𝟓𝟕𝟔𝟓⏟ >𝟏

+∑𝟐

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟕

; ∀𝒏 ≥ 𝟔

Let put 𝑹 = 𝐥𝐢𝐦𝒏→∞

𝟏

𝒏𝟐(𝟏 + 𝟏)𝒏; (𝛀 > 𝑹)

𝑹 = 𝐥𝐢𝐦𝒏→∞

𝟐𝒏

𝒏𝟐=𝑳′𝑯𝐥𝐢𝐦𝒏→∞

𝟐𝒏 𝐥𝐨𝐠𝟐

𝟐𝒏= 𝐥𝐢𝐦𝒏→∞

𝟐𝒏 𝐥𝐨𝐠𝟐 𝟐

𝟐= +∞

Therefore,


𝟏

𝒏𝟐(𝟏 + 𝟐∑

𝟏

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟏

)

𝒏

= ∞


Using the definition of digamma function, we have:

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𝝍(𝒏 + 𝟏 + 𝒛) = 𝝍(𝒛 + 𝟏) +∑𝟏

𝒌 + 𝒛

𝒏

𝒌=𝟏

;

𝝍(𝒛) = 𝐥𝐨𝐠 𝒛 −𝟏

𝟐𝒛−

𝟏

𝟏𝟐𝒛𝟐+

𝟏

𝟏𝟐𝟎𝒛𝟒−

𝟏

𝟐𝟓𝟐𝒛𝟔+⋯

For 𝒛 =𝟓

𝟐⇒

𝟐∑𝟏

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟏

= 𝝍(𝒏 +𝟕

𝟐) −𝝍(

𝟕

𝟐)


𝟏

𝒏𝟐(𝟏 + 𝟐∑

𝟏

𝟐𝒌+ 𝟓

𝒏

𝒌=𝟏

)

𝒏


𝟏

𝒏𝟐(𝟏 + 𝝍(𝒏 +

𝟕

𝟐) − 𝝍(

𝟕

𝟐))

𝒏

=𝑳′𝑯


(𝐥𝐨𝐠 (𝒏 +𝟕𝟐) −

𝟏

𝟐(𝒏 +𝟕𝟐)−

𝟏

𝟏𝟐(𝒏 +𝟕𝟐)𝟐 +

𝟏

𝟏𝟐𝟎(𝒏 +𝟕𝟐)𝟒 −

𝟏

𝟐𝟓𝟐(𝒏 +𝟕𝟐)𝟔+. . )

𝒏−𝟏

𝟐


(𝐥𝐨𝐠 (𝒏 +𝟕𝟐))

𝒏−𝟏

𝟐= ∞



𝟏

𝒏𝟐(𝟏 + 𝟐∑

𝟏

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟏

)

𝒏


𝟏

𝒏𝟐(𝟏 + 𝝍(𝒏 +

𝟕

𝟐) − 𝝍(

𝟕

𝟐))

𝒏

∵ 𝝍(𝒙) > 𝐥𝐨𝐠 (𝒙 +𝟏

𝟐) −

𝟏

𝒙⇒

𝛀 > 𝐥𝐢𝐦𝒏→∞

𝟏

𝒏𝟐(𝟏 + 𝐥𝐨𝐠(𝒏 + 𝟒) −

𝟐

𝟐𝒏 + 𝟕+ 𝝍(

𝟕

𝟐))

𝒏

; (𝒍𝒆𝒕 𝟏 + 𝝍(𝟕

𝟐) = 𝒄)

> 𝐥𝐢𝐦𝒏→∞

𝟏

𝒏𝟐(𝐥𝐨𝐠(𝒏 + 𝟒) −

𝟐

𝟐𝒏 + 𝟕+ 𝒄)

𝒏

; (∵ 𝐥𝐨𝐠(𝟏 + 𝒙) >𝒙

𝟏 + 𝒙)


𝟏

𝒏𝟐(𝒏 + 𝟑

𝒏 + 𝟒−

𝟐

𝟐𝒏 + 𝟕+ 𝒄)

𝒏


(

𝒏 + 𝟑𝒏 + 𝟒 −

𝟐𝟐𝒏 + 𝟕

𝒏𝟐𝒏

+𝒄

𝒏𝟐𝒏

)

𝒏

>


(𝟏 + 𝒄)𝒏

Therefore,

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𝟏

𝒏𝟐(𝟏 + 𝟐∑

𝟏

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟏

)

𝒏

= ∞


Let 𝒂𝒏 =𝟏

𝒏𝟐(𝟏 + 𝟐∑

𝟏

𝟐𝒌+𝟓

𝒏𝒌=𝟏 )

𝒏

.


𝒂𝒏+𝟏𝒂𝒏


𝟏(𝒏 + 𝟏)𝟐

(𝟏 + 𝟐∑𝟏

𝟐𝒌 + 𝟓𝒏+𝟏𝒌=𝟏 )

𝒏

𝟏𝒏𝟐(𝟏 + 𝟐∑

𝟏𝟐𝒌 + 𝟓

𝒏𝒌=𝟏 )

𝒏 =


(𝟏 + 𝟐∑𝟏

𝟐𝒌 + 𝟓

𝒏+𝟏

𝒌=𝟏

)(𝟏 + 𝟐∑

𝟏𝟐𝒌 + 𝟓

𝒏+𝟏𝒌=𝟏

𝟏 + 𝟐∑𝟏

𝟐𝒌 + 𝟓𝒏𝒌=𝟏

)

𝒏

𝟐𝒌 + 𝟓 ≤ 𝟐𝒌 + 𝟔 ⇒𝟐

𝟐𝒌 + 𝟓>

𝟏

𝟐𝒌 + 𝟑⇒ ∑

𝟏

𝟐𝒌 + 𝟓

𝒏+𝟏

𝒌=𝟏

≥ ∑𝟏

𝒌+ 𝟑

𝒏+𝟏

𝒌=𝟏

; (𝐥𝐢𝐦𝒌→∞

𝟏

𝒌 + 𝟑= ∞)

⇒ 𝐥𝐢𝐦𝒏→∞

∑𝟐

𝟐𝒌 + 𝟓

𝒏+𝟏

𝒌=𝟏

= ∞ ⇒ 𝟏 + 𝟐∑𝟐

𝟐𝒌 + 𝟓

𝒏+𝟏

𝒌=𝟏

= ∞; (𝟏)

𝒍𝒆𝒕 𝒖(𝒏) =𝟏 + 𝟐∑

𝟏𝟐𝒌 + 𝟓

𝒏+𝟏𝒌=𝟏

𝟏 + 𝟐∑𝟏

𝟐𝒌 + 𝟓𝒏𝒌=𝟏


(𝟏 + 𝟐∑

𝟏𝟐𝒌 + 𝟓

𝒏+𝟏𝒌=𝟏

𝟏 + 𝟐∑𝟏

𝟐𝒌 + 𝟓𝒏𝒌=𝟏

)

𝒏


(𝟏 + (𝒖(𝒏) − 𝟏))𝟏

𝒖(𝒏)−𝟏⋅𝒏(𝒖(𝒏)−𝟏)

=

= 𝐞𝐱𝐩 {𝐥𝐢𝐦𝒏→∞

𝟐𝒏

𝟐𝒏 + 𝟕⋅

𝟏

𝟏 + 𝟐∑𝟏

𝟐𝒌 + 𝟓𝒏𝒌=𝟏

} = 𝒆𝟎 = 𝟏; (𝟐)



𝟏

𝒏𝟐(𝟏 + 𝟐∑

𝟏

𝟐𝒌 + 𝟓

𝒏

𝒌=𝟏

)

𝒏

= ∞

1597. Find:


𝟏

𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑

𝚪′(𝒌)

𝚪(𝒌)

𝒏

𝒌=𝟐

)


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𝟏

𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑

𝚪′(𝒌)

𝚪(𝒌)

𝒏

𝒌=𝟐

) = 𝐥𝐢𝐦𝒏→∞

𝐥𝐨𝐠(𝒏!) − ∑ 𝝍(𝒌)𝒏𝒌=𝟐

𝒏(𝑯𝒏 − 𝟏)=𝑪−𝑺


𝐥𝐨𝐠(𝒏 + 𝟏) − 𝝍(𝒏 + 𝟏)

(𝒏 + 𝟏)(𝑯𝒏 +𝟏

𝒏 + 𝟏) − 𝒏𝑯𝒏 + 𝒏=


𝐥𝐨𝐠(𝒏 + 𝟏) − 𝝍(𝒏 + 𝟏)

𝒏𝑯𝒏 +𝑯𝒏 + 𝟏 − 𝒏− 𝟏 − 𝒏𝑯𝒏 + 𝒏= 𝐥𝐢𝐦𝒏→∞

𝐥𝐨𝐠(𝒏 + 𝟏) − 𝝍(𝒏 + 𝟏)

𝑯𝒏=𝑪−𝑺


𝐥𝐨𝐠 (𝒏 + 𝟐𝒏 + 𝟏) − 𝝍

(𝒏 + 𝟐) + 𝝍(𝒏 + 𝟏)

𝟏𝒏 + 𝟏


𝐥𝐨𝐠 (𝒏 + 𝟐𝒏 + 𝟏) −

𝟏𝒏 + 𝟏

𝟏𝒏 + 𝟏

=


(𝒏 + 𝟏) 𝐥𝐨𝐠 (𝒏 + 𝟐

𝒏 + 𝟏) − 𝟏 = 𝐥𝐢𝐦

𝒏→∞𝐥𝐨𝐠 (𝟏 +

𝟏

𝒏 + 𝟏)𝒏+𝟏

− 𝟏 = 𝐥𝐨𝐠 𝒆 − 𝟏 = 𝟎



𝟏

𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑

𝚪′(𝒌)

𝚪(𝒌)

𝒏

𝒌=𝟐

) =


𝟏

𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝚪(𝒏 + 𝟏)) − 𝝍(𝒏)(𝒏 − 𝟏)) =

=𝑪−𝑺


𝐥𝐨𝐠 (𝚪(𝒏 + 𝟐)𝚪(𝒏 + 𝟏)

) − 𝝍(𝒏 + 𝟏)(𝒏) + 𝝍(𝒏)(𝒏 − 𝟏)

(𝒏 + 𝟏)(𝑯𝒏+𝟏 − 𝟏) − (𝒏)(𝑯𝒏 − 𝟏)=


𝐥𝐨𝐠(𝒏 + 𝟏) − 𝝍(𝒏) − 𝟏

𝑯𝒏; (𝝍(𝒙)~ 𝐥𝐨𝐠 𝒙)


𝐥𝐨𝐠 (𝒏 + 𝟏𝒏 ) − 𝟏

𝑯𝒏= 𝐥𝐢𝐦𝒏→∞

−𝟏

𝑯𝒏= 𝟎


𝐥𝐨𝐠 𝒏! −∑𝚪′(𝒌)

𝚪(𝒌)

𝒏

𝒌=𝟐

=∑𝐥𝐨𝐠 𝒌 −∑𝝍(𝒏)

𝒏

𝒌=𝟐

𝒏

𝒌=𝟏

= ∑𝐥𝐨𝐠𝒌

𝒏

𝒌=𝟏

−∑(𝑯𝒏−𝟏 − 𝜸)

𝒏

𝒌=𝟐

⇒

𝟏

𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑

𝚪′(𝒌)

𝚪(𝒌)

𝒏

𝒌=𝟐

) =𝟏

𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠 𝒏! − (𝒏 − 𝟏)(𝑯𝒏−𝟏 − 𝜸))

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=𝟏

𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠 𝒏! − (𝒏 − 𝟏) 𝐥𝐨𝐠(𝒏 − 𝟏)); (∵ 𝑯𝒏−𝟏 − 𝜸 ≅ 𝐥𝐨𝐠(𝒏 − 𝟏))

𝐥𝐨𝐠 𝒏! = 𝐥𝐨𝐠(𝒏 + 𝟏) ≅ (𝒏 +𝟏

𝟐) 𝐥𝐨𝐠(𝒏 + 𝟏) − 𝒏 − 𝟏 +

𝟏

𝟐𝐥𝐨𝐠𝟐𝝅

(𝐥𝐨𝐠𝒏! − (𝒏 − 𝟏) 𝐥𝐨𝐠(𝒏 − 𝟏)) ≅ (𝒏 +𝟏

𝟐) 𝐥𝐨𝐠(𝒏 + 𝟏) − 𝒏 − 𝟏 +

𝟏

𝟐𝐥𝐨𝐠 𝟐𝝅 −

−(𝒏 − 𝟏) 𝐥𝐨𝐠(𝒏 − 𝟏)

(𝒏+𝟏

𝟐) 𝐥𝐨𝐠(𝒏+𝟏)

𝒏(𝑯𝒏−𝟏)≅ 𝟏 and

(𝒏−𝟏) 𝐥𝐨𝐠(𝒏−𝟏)

𝒏(𝑯𝒏−𝟏)≅ 𝟏

𝒏 − 𝟏 +𝟏𝟐𝐥𝐨𝐠 𝟐𝝅

𝒏(𝑯𝒏 − 𝟏)→ 𝟎

Therefore,


𝟏

𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑

𝚪′(𝒌)

𝚪(𝒌)

𝒏

𝒌=𝟐

)

1598.

Find a closed form:

𝛀 = ∑𝟏

𝒏∫ (𝟏 + 𝒙𝟐)−𝒏 ⋅ 𝐭𝐚𝐧−𝟏 𝒙𝟏

𝟎

𝒅𝒙

∞

𝒌=𝟏



𝛀 =∑𝟏

𝒏∫ (𝟏 + 𝒙𝟐)−𝒏 ⋅ 𝐭𝐚𝐧−𝟏 𝒙𝟏

𝟎

𝒅𝒙

∞

𝒌=𝟏

= ∫ 𝐭𝐚𝐧−𝟏 𝒙 ⋅∑(

𝟏𝟏+ 𝒙𝟐

)𝒏

𝒏

∞

𝒏=𝟏

𝒅𝒙𝟏

𝟎

∑𝒙𝒏

𝒏

∞

𝒏=𝟏

= ∫(∑𝒙𝒏−𝟏∞

𝒏=𝟏

)𝒅𝒙 = ∫𝟏

𝟏 − 𝒙𝒅𝒙 , 𝒇𝒐𝒓 |𝒙| < 𝟏 ⇒

∑𝒙𝒏

𝒏

∞

𝒏=𝟏

= − 𝐥𝐨𝐠 |𝒙 − 𝟏| ⇒ ∑(

𝟏𝒙𝟐 + 𝟏

)𝒏

𝒏

∞

𝒏=𝟏

= − 𝐥𝐨𝐠 |𝒙𝟐

𝒙𝟐 + 𝟏| = 𝐥𝐨𝐠 (

𝒙𝟐 + 𝟏

𝒙𝟐)

𝛀 = ∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏)𝟏

𝟎

𝒅𝒙 − 𝟐∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

; (𝟏)

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∫𝐥𝐨𝐠(𝒙𝟐 + 𝟏) ⋅ 𝒙′𝒅𝒙 = 𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) −∫𝟐𝒙𝟐 + 𝟐 − 𝟐

𝒙𝟐 + 𝟏𝒅𝒙

= 𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) − 𝟐𝒙 + 𝟐 𝐭𝐚𝐧−𝟏 𝒙

∫ 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) 𝐭𝐚𝐧−𝟏 𝒙𝟏

𝟎

𝒅𝒙 = (𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) − 𝟐𝒙 + 𝟐 𝐭𝐚𝐧−𝟏 𝒙)|𝟎

𝟏−

−∫ (𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏)

𝒙𝟐 + 𝟏−

𝟐𝒙

𝒙𝟐 + 𝟏+𝟐 𝐭𝐚𝐧−𝟏 𝒙

𝒙𝟐 + 𝟏)𝒅𝒙

𝟏

𝟎

=

= (𝐥𝐨𝐠𝟐 − 𝟐 +𝝅

𝟐) ⋅𝝅

𝟒− (𝐥𝐨𝐠𝟐(𝒙𝟐 + 𝟏) − 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) + (𝐭𝐚𝐧−𝟏 𝒙)𝟐)|

𝟎

𝟏=

=𝝅

𝟒𝐥𝐨𝐠 𝟐 −

𝝅

𝟐+𝝅𝟐

𝟖−𝟏

𝟒𝐥𝐨𝐠𝟐 𝟐 −

𝝅𝟐

𝟏𝟔; (𝟐)

∫ 𝐥𝐨𝐠𝒙 𝐭𝐚𝐧−𝟏 𝒙𝟏

𝟎

𝒅𝒙 = (𝒙 𝐥𝐨𝐠𝒙 − 𝒙) 𝐭𝐚𝐧−𝟏 𝒙|𝟎

𝟏−∫ (

𝒙 𝐥𝐨𝐠𝒙

𝒙𝟐 + 𝟏−

𝒙

𝒙𝟐 + 𝟏)𝒅𝒙

𝟏

𝟎

=

= −𝝅

𝟒−∫

𝒙 𝐥𝐨𝐠 𝒙

𝒙𝟐 + 𝟏

𝟏

𝟎

𝒅𝒙 +𝟏

𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝟏)|

𝟎

𝟏

= −𝝅

𝟒+𝟏

𝟐𝐥𝐨𝐠 𝟐 − ∫

𝒙 𝐥𝐨𝐠𝒙

𝒙𝟐 + 𝟏𝒅𝒙

𝟏

𝟎

𝟏

𝟐∫𝟐𝒙 𝐥𝐨𝐠 𝒙

𝒙𝟐 + 𝟏

𝟏

𝟎

𝒅𝒙 =𝟏

𝟐(−∫

𝐥𝐨𝐠(𝒙𝟐 + 𝟏)

𝒙𝒅𝒙

𝟏

𝟎

)

∫𝐥𝐨𝐠(𝒙𝟐 + 𝟏)

𝒙𝟐 + 𝟏𝒅𝒙

𝟏

𝟎

=𝒙𝟐=𝒕

∫𝐥𝐨𝐠(𝒕 + 𝟏)

√𝒕⋅𝟏

𝟐√𝒕

𝟏

𝟎

𝒅𝒕 =𝟏

𝟐∫𝐥𝐨𝐠(𝟏 + 𝒕)

𝒕

𝟏

𝟎

𝒅𝒕 =

=𝟏

𝟐∑(−𝟏)𝒏−𝟏∞

𝒏=𝟏

∫𝒕𝒏−𝟏

𝒏𝒅𝒕

𝟏

𝟎

=𝟏

𝟐∑(−𝟏)𝒏−𝟏

𝒏𝟐

∞

𝒏=𝟏

=𝝅𝟐

𝟐𝟒⇒

∫𝒙 𝐥𝐨𝐠 𝒙

𝒙𝟐 + 𝟏

𝟏

𝟎

𝒅𝒙 = −𝝅𝟐

𝟒𝟖⇒ ∫ 𝐥𝐨𝐠 𝒙 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙

𝟏

𝟎

= −𝝅

𝟒+𝟏

𝟐𝐥𝐨𝐠 𝟐 +

𝝅𝟐

𝟒𝟖; (𝟑)

From (1),(2),(3) it follows that:

𝛀 =𝟏

𝟒𝟖(𝝅𝟐 + 𝟏𝟐𝝅 𝐥𝐨𝐠 𝟐 − 𝟏𝟐 𝐥𝐨𝐠𝟐 𝟐)


𝛀 = ∫ ∑((𝟏 + 𝒙𝟐)−𝟏)𝒏

𝒏

∞

𝒏=𝟏

𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝟏

𝟎

= ∫ − 𝐥𝐨𝐠 (𝟏 −𝟏

𝟏 + 𝒙𝟐) 𝐭𝐚𝐧−𝟏 𝒙

𝟏

𝟎

𝒅𝒙 =

= ∫ 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) 𝐭𝐚𝐧−𝟏 𝒙 𝒅𝒙𝟏

𝟎

− 𝟐∫ 𝐥𝐨𝐠 𝒙 𝐭𝐚𝐧−𝟏 𝒙𝟏

𝟎

𝒅𝒙 = 𝑰𝟏 − 𝟐𝑰𝟐,

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𝑰𝟏 = ∫ 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝟏

𝟎

= 𝐥𝐨𝐠(𝟏 + 𝒙𝟐)∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙 −∫𝟐𝒙

𝟏 + 𝒙𝟐⋅ ∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝒅𝒙 =

= 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) [𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 −𝟏

𝟐𝐥𝐨𝐠(𝟏 + 𝒙𝟐)] − 𝟐∫

𝒙𝟐 𝐭𝐚𝐧−𝟏 𝒙

𝟏 + 𝒙𝟐𝒅𝒙 +∫

𝒙 𝐥𝐨𝐠(𝟏 + 𝒙𝟐)

𝟏 + 𝒙𝟐𝒅𝒙

=

= (𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − 𝟐) [𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 −𝟏

𝟐𝐥𝐨𝐠(𝟏 + 𝒙𝟐)] + (𝐭𝐚𝐧−𝟏 𝒙)𝟐 +

𝟏

𝟒𝐥𝐨𝐠𝟐(𝟏 + 𝒙𝟐)

Putting limits, we get:

𝑰𝟏 =𝟏

𝟒𝐥𝐨𝐠𝟐 𝟐 +

𝝅𝟐

𝟏𝟔+ (𝐥𝐨𝐠𝟐 − 𝟐) (

𝝅

𝟒−𝟏

𝟐𝐥𝐨𝐠𝟐)

Now,

𝑰𝟐 = ∑(−𝟏)𝒏+𝟏

𝟐𝒏 − 𝟏

∞

𝒏=𝟏

∫ 𝒙𝟐𝒏−𝟏 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

=𝐥𝐨𝐠 𝒙=−𝒖

−∑(−𝟏)𝒏+𝟏

𝟐𝒏 − 𝟏∫ 𝒖𝒆−𝟐𝒖𝒏∞

𝟎

𝒅𝒖

∞

𝒏=𝟏

=𝒕=𝟐𝒏

= −𝟏

𝟒∑

(−𝟏)𝒏+𝟏

𝒏𝟐(𝟐𝒏 − 𝟏)

∞

𝒏=𝟏

∫ 𝒕𝒆−𝒕∞

𝟎

𝒅𝒕 = −𝟏

𝟒∑

(−𝟏)𝒏+𝟏

𝒏𝟐(𝟐𝒏 − 𝟏)

∞

𝒏=𝟏

Using partial fraction decomposition, we get:

𝑰𝟐 =𝟏

𝟒∑(−𝟏)𝒏+𝟏

𝒏𝟐

∞

𝒏=𝟏

+𝟏

𝟐∑(−𝟏)𝒏+𝟏

𝒏

∞

𝒏=𝟏

−∑(−𝟏)𝒏+𝟏

𝟐𝒏 − 𝟏

∞

𝒏=𝟏

=𝟏

𝟒⋅𝝅𝟐

𝟏𝟐+𝟏

𝟐⋅ 𝐥𝐨𝐠 𝟐 −

𝝅

𝟒

𝛀 = 𝑰𝟏 − 𝟐𝑰𝟐 =𝟏

𝟒𝐥𝐨𝐠𝟐 𝟐 +

𝝅𝟐

𝟏𝟔+ (𝐥𝐨𝐠𝟐 − 𝟐) (

𝝅

𝟒−𝟏

𝟐𝐥𝐨𝐠𝟐) − 𝟐 (

𝝅𝟐

𝟒𝟖+𝟏

𝟐⋅ 𝐥𝐨𝐠 𝟐 −

𝝅

𝟒) =

=𝝅

𝟒𝐥𝐨𝐠𝟐 −

𝟏

𝟒𝐥𝐨𝐠𝟐 𝟐 +

𝝅𝟐

𝟒𝟖

Solution 3 by Katrick Chandra Betal-India

𝛀 = ∑𝟏

𝒏∫

𝐭𝐚𝐧−𝟏 𝒙

(𝟏 + 𝒙𝟐)𝒏𝒅𝒙

𝟏

𝟎

∞

𝒏=𝟏

= −∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠 (𝟏 −𝟏

𝟏 + 𝒙𝟐)𝒅𝒙

𝟏

𝟎

=

= −∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠 (𝒙𝟐

𝟏 + 𝒙𝟐)

𝟏

𝟎

𝒅𝒙 = ∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) 𝒅𝒙𝟏

𝟎

− 𝟐∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏

𝟎

=

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= [𝐥𝐨𝐠(𝟏 + 𝒙𝟐) {𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)

𝟐}]𝟎

𝟏

− 𝟐∫ (𝒙𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)

𝟐)𝒙𝒅𝒙

𝟏 + 𝒙𝟐

𝟏

𝟎

−

−𝟐 [𝐥𝐨𝐠 𝒙 {𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)

𝟐}]𝟎

𝟏

+ 𝟐∫ (𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)

𝟐)𝒅𝒙

𝒙

𝟏

𝟎

=

= 𝐥𝐨𝐠 𝟐 (𝝅

𝟒−𝐥𝐨𝐠𝟐

𝟐) − ∫

𝟐𝒙𝟐 𝐭𝐚𝐧−𝟏 𝒙 − 𝒙 𝐥𝐨𝐠(𝟏 + 𝒙𝟐)

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

+∫ (𝟐 𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)

𝒙)𝒅𝒙

𝟏

𝟎

=

=𝝅

𝟒𝐥𝐨𝐠 𝟐 −

𝐥𝐨𝐠𝟐 𝟐

𝟐− 𝟐∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙

𝟏

𝟎

+ 𝟐∫𝐭𝐚𝐧−𝟏 𝒙

𝟏 + 𝒙𝟐𝒅𝒙

𝟏

𝟎

+𝟏

𝟐∫𝐥𝐨𝐠(𝟏 + 𝒙)

𝟏 + 𝒙𝒅𝒙

𝟏

𝟎

+

+𝟐∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝟏

𝟎

−𝟏

𝟐∫𝐥𝐨𝐠(𝟏 + 𝒙)

𝒙𝒅𝒙

𝟏

𝟎

=

=𝝅



𝟐+ [(𝐭𝐚𝐧−𝟏 𝒙)𝟐]𝟎

𝟏 +𝟏

𝟐[𝐥𝐨𝐠𝟐(𝟏 + 𝒙)

𝟐]𝟎

𝟏

−𝜻(𝟐)

𝟒=

=𝝅



𝟐+𝝅𝟐

𝟏𝟔+𝐥𝐨𝐠𝟐 𝟐

𝟒−𝝅𝟐

𝟐𝟒=𝝅



𝟒+𝝅𝟐

𝟐𝟒

1599. Find:


(𝒏 − 𝟏)!∑𝟏

(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌

𝒏

𝒌=𝟎


Solution 1 by Kamel Gandouli Habib Rezgui-Tunisia

𝝎𝒏 = ∑𝟏

(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌

𝒏

𝒌=𝟎

; 𝒗𝒏(𝒌) =𝟏

(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌

𝒗𝟐𝒏(𝒌) =𝟏

(𝒌 + 𝟏)𝒌(𝟐𝒏 + 𝟏 − 𝒌)𝟐𝒏−𝒌; 𝒗𝒏(𝟐𝒏) =

𝟏

(𝟐𝒏 + 𝟏)𝟐𝒏 𝒇𝒐𝒓 𝒌 = 𝟎 𝒂𝒏𝒅

𝒏 = 𝐦𝐢𝐧 𝒗𝟐𝒏(𝒌) ⇒ 𝐦𝐚𝐱𝒗𝟐𝒏(𝒌) =𝟏

(𝒏 + 𝟏)𝒏(𝒏 + 𝟏)𝒏=

𝟏

(𝒏 + 𝟏)𝟐𝒏 𝒇𝒐𝒓 𝒌 = 𝒏.

𝟏

(𝟐𝒏 + 𝟏)𝟐𝒏≤ 𝒗𝟐𝒏(𝒌) ≤

𝟏

(𝒏 + 𝟏)𝟐𝒏, ∀𝒌 ∈ ℕ

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𝟐𝒏 + 𝟏

(𝟐𝒏 + 𝟏)𝟐𝒏≤∑𝒗𝟐𝒏(𝒌)

𝟐𝒏

𝒌=𝟎

≤𝟐𝒏 + 𝟏

(𝒏 + 𝟏)𝟐𝒏, ∀𝒌 ∈ ℕ

(𝟐𝒏 + 𝟏)(𝟐𝒏 − 𝟏)!

(𝟐𝒏 + 𝟏)𝟐𝒏≤ (𝟐𝒏 − 𝟏)!∑𝒗𝒏(𝟐𝒏)

𝟐𝒏

𝒌=𝟎

≤(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)

(𝒏 + 𝟏)𝟐𝒏

(𝟐𝒏 − 𝟏)!

(𝟐𝒏 + 𝟏)𝟐𝒏−𝟏≤ (𝟐𝒏 − 𝟏)!∑𝒗𝒏(𝟐𝒏)

𝟐𝒏

𝒌=𝟎

≤(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)

(𝒏 + 𝟏)𝟐𝒏

(𝟐𝒏 − 𝟏)!

(𝟐𝒏 + 𝟏)𝟐𝒏−𝟏≤ 𝝎𝟐𝒏 ≤

(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)

(𝒏 + 𝟏)𝟐𝒏

∵ 𝒏! = √𝟐𝒏𝝅 (𝒏

𝒆)𝒏

(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)

(𝒏 + 𝟏)𝟐𝒏≅√𝟐𝝅(𝟐𝒏 − 𝟏) (

𝟐𝒏 − 𝟏𝟐 )

𝒏

(𝒏 + 𝟏)𝟐𝒏(𝟐𝒏 + 𝟏) =

=√𝟐𝝅(𝟐𝒏 − 𝟏)(𝟐𝒏− 𝟏)𝒏

(𝒏 + 𝟏)𝟐𝒏𝒆𝒏(𝟐𝒏 + 𝟏)

(𝟐𝒏 − 𝟏)𝒏

(𝒏 + 𝟏)𝟐𝒏= (

(𝟐𝒏− 𝟏)

𝒏𝟐 + 𝟐𝒏 + 𝟏)

𝒏

→ 𝟎

√𝟐𝝅(𝟐𝒏 − 𝟏)(𝟐𝒏 + 𝟏)

𝒆𝒏≅ √𝟐𝝅(𝟐𝒏 − 𝟏)(𝟐𝒏 + 𝟏)𝒆−𝒏 ≤ (𝟐𝒏 + 𝟏)𝟐𝒆−𝒏 → 𝟎

Similarly for 𝟐𝒏 + 𝟏

𝒗𝟐𝒏+𝟏(𝒌) ≤𝟏

(𝟐𝒏 − 𝟏)𝟐𝒏(𝟐𝒏)𝟐𝒏,

(𝟐𝒏)! ∑ 𝒗𝟐𝒏+𝟏(𝒌)

𝟐𝒏+𝟏

𝒌=𝟎

≤(𝟐𝒏 + 𝟐)𝟐𝒏!

(𝟐𝒏 − 𝟏)𝟐𝒏(𝟐𝒏)𝟐𝒏→ 𝟎 ⇒ 𝝎𝟐𝒏+𝟏 → 𝟎

Therefore,


(𝒏 − 𝟏)!∑𝟏

(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌

𝒏

𝒌=𝟎

= 𝟎


Let 𝒎 ∈ ℕ− {𝟎}, 𝒇(𝒙) = (𝒙 + 𝟏)𝒙(𝒎 + 𝟏 − 𝒙)𝒎−𝒙, 𝟎 ≤ 𝒙 ≤ [𝒎

𝟐]

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𝐥𝐨𝐠 𝒇(𝒙) = 𝒙 𝐥𝐨𝐠(𝒙 + 𝟏) + (𝒎− 𝒙) 𝐥𝐨𝐠(𝒎+ 𝟏 − 𝒙)

𝒇′(𝒙)

𝒇(𝒙)= 𝐥𝐨𝐠(𝒙 + 𝟏) − 𝐥𝐨𝐠(𝒎+ 𝟏 − 𝒙) −

𝒎 − 𝒙

𝒎 + 𝟏 − 𝒙=

= 𝐥𝐨𝐠 (𝒙 + 𝟏

𝒎+ 𝟏 − 𝒙) −

𝒎 − 𝒙

𝒎 + 𝟏 − 𝒙< 𝟎,∀𝟎 < 𝒙 < [

𝒎

𝟐]

Thus, 𝒇 decreases on [𝟎, [𝒎

𝟐]] ⇒ 𝒇(𝒙) ≥ 𝒇([

𝒎

𝟐]) , ∀𝒙 ∈ [𝟎, [

𝒎

𝟐]].

Hence,

𝟏

(𝒌 + 𝟏)𝒌(𝟐𝒏 + 𝟏 − 𝒌)𝟐𝒏−𝒌≤

𝟏

(𝒏 + 𝟏)𝒏(𝒏 + 𝟏)𝒏=

𝟏

(𝒏 + 𝟏)𝟐𝒏

and

𝟏

(𝒌 + 𝟏)𝒌(𝟐𝒏 + 𝟐 − 𝒌)𝟐𝒏+𝟏−𝒌≤

𝟏

(𝒏 + 𝟏)𝟐𝒏+𝟏

(𝟐𝒏 − 𝟏)!∑𝟏

(𝒌 + 𝟏)𝒌(𝟐𝒏 − 𝒌 + 𝟏)𝟐𝒏−𝒌

𝟐𝒏

𝒌=𝟎

<(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)

(𝒏 + 𝟏)𝟐𝒏=

(𝟐𝒏 + 𝟏)!

𝟐𝒏(𝒏 + 𝟏)𝟐𝒏

Let 𝒃𝒏 =(𝟐𝒏+𝟏)!

𝟐𝒏(𝒏+𝟏)𝟐𝒏 and

(𝟐𝒏 + 𝟏 − 𝟏)! ∑𝟏

(𝒌 + 𝟏)𝒌(𝟐𝒏 + 𝟏 − 𝒌 + 𝟏)𝟐𝒏+𝟏−𝒌

𝟐𝒏+𝟏

𝒌=𝟎

<(𝟐𝒏)! (𝟐𝒏 + 𝟐)

(𝒏 + 𝟏)𝟐𝒏+𝟏

=(𝟐𝒏 + 𝟐)!

(𝟐𝒏 + 𝟏)(𝒏 + 𝟏)𝟐𝒏+𝟏

Let 𝒄𝒏 =(𝟐𝒏+𝟐)!

(𝟐𝒏+𝟏)(𝒏+𝟏)𝟐𝒏+𝟏. We prove that: 𝒃𝒏 , 𝒄𝒏 → ∞ for 𝒏 → ∞.

𝒃𝒏𝒃𝒏+𝟏

= (𝟏+𝟏

𝒏) ⋅

(𝒏 + 𝟐)𝟐

(𝟐𝒏 + 𝟐)(𝟐𝒏+ 𝟑)[(𝟏 +

𝟏

𝒏 + 𝟏)𝒏+𝟏

]

𝟐

→𝒆𝟐

𝟒

As 𝒆𝟐

𝟒> 𝟏, 𝒃𝒏 → 𝟎 as 𝒏 → ∞. Similarly, 𝒄𝒏 → 𝟎 as 𝒏 → ∞.

Now,

𝟎 <∑𝟏

(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌

𝒏

𝒌=𝟎

< 𝒃𝒏, 𝒄𝒏

As 𝒃𝒏 → 𝟎, 𝒄𝒏 → 𝟎 as 𝒏 → ∞. Therefore,

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(𝒏 − 𝟏)!∑𝟏

(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌

𝒏

𝒌=𝟎

= 𝟎

1600.

𝛀𝒏(𝒙) = ∫𝒅𝒙

𝒙(𝟏 + 𝒙𝒏), 𝒏 ∈ ℕ∗, 𝛀𝒏(𝟏) = 𝐥𝐨𝐠 𝟐

Find:

𝛀(𝒙) = 𝐥𝐢𝐦𝒏→∞

(𝒏𝛀𝒏(𝒙)) , 𝒙 > 𝟎




𝒙(𝟏 + 𝒙𝒏)= ∫

𝒙𝒏−𝟏

𝒙𝒏(𝟏 + 𝒙𝒏)𝒅𝒙 = ∫(

𝟏

𝒙𝒏−

𝟏

𝟏 + 𝒙𝒏)𝒙𝒏−𝟏 𝒅𝒙 =

=𝟏

𝒏∫(𝟏

𝒕−

𝟏

𝒕 + 𝟏)𝒅𝒕 =

𝟏

𝒏𝐥𝐨𝐠 (

𝒕

𝒕 + 𝟏) + 𝑪 =

𝟏

𝒏𝐥𝐨𝐠 (

𝒙𝒏

𝟏 + 𝒙𝒏) + 𝑪

𝛀𝒏(𝟏) =𝟏

𝒏𝐥𝐨𝐠 (

𝟏

𝟐) = −

𝟏

𝒏𝐥𝐨𝐠𝟐 + 𝑪 ⇒ 𝑪 =

𝒏 + 𝟏

𝒏𝐥𝐨𝐠 𝟐

Thus,

𝒏𝛀𝒏(𝒙) = (𝒏 + 𝟏) 𝐥𝐨𝐠 𝟐 + 𝐥𝐨𝐠 (𝒙𝒏

𝟏 + 𝒙𝒏) = 𝐥𝐨𝐠𝟐 + 𝐥𝐨𝐠 (

𝟐𝒏𝒙𝒏

𝟏 + 𝒙𝒏)

If 𝟎 < 𝒙 <𝟏

𝟐, 𝟎 < 𝟐𝒙 < 𝟏 ⇒ (𝟐𝒙)𝒏 → 𝟎


𝒏𝛀𝒏(𝒙) = −∞ 𝒊𝒇 𝟎 < 𝒙 <𝟏

𝟐.

For 𝒙 =𝟏

𝟐 we have: 𝒏𝛀𝒏(𝒙) = 𝐥𝐨𝐠 𝟐 + 𝐥𝐨𝐠 (

𝟏

𝟏+(𝟏

𝟐)𝒏) → 𝐥𝐨𝐠 𝟐 as 𝒏 → ∞.

For 𝟏

𝟐< 𝒙 < 𝟏, (𝟐𝒙)𝒏 → ∞,𝒙𝒏 → ∞ ad 𝒏𝛀𝒏(𝒙) → ∞ as 𝒏 → ∞.

For 𝒙 ≥ 𝟏,𝒏𝛀𝒏(𝒙) = 𝐥𝐨𝐠 𝟐 − 𝐥𝐨𝐠 (𝟏

𝟐𝒏+

𝟏

𝟐𝒏𝒙𝒏) → ∞ as 𝒏 → ∞.


For 𝐱 > 𝟏:

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𝛀𝒏(𝒙) = ∫𝟏

𝒙(𝒙𝒏 + 𝟏)𝒅𝒙 = 𝐥𝐨𝐠 𝒙 −

𝐥𝐨𝐠(𝒙𝒏 + 𝟏)

𝒏

Hence,


(𝒏𝛀𝒏(𝒙)) = 𝐥𝐢𝐦𝒏→∞

𝒏(𝐥𝐨𝐠𝒙 −𝐥𝐨𝐠(𝒙𝒏 + 𝟏)

𝒏) = 𝐥𝐢𝐦

𝒏→∞[𝒏 𝐥𝐨𝐠𝒙 − 𝐥𝐨𝐠(𝒙𝒏 + 𝟏)] =


𝐥𝐨𝐠 (𝒙𝒏

𝒙𝒏 + 𝟏) = 𝐥𝐢𝐦

𝒏→∞

𝟏

𝟏𝒙𝒏 + 𝟏

= 𝟎

Solution 3 by Satyam Roy-India

For 𝐱 > 𝟏:


𝒙(𝟏 + 𝒙𝒏)= ∫

𝒙−𝒏−𝟏

𝟏𝒙𝒏 + 𝟏

𝒅𝒙 =

𝟏𝒙𝒏+𝟏=𝒖

−𝟏

𝒏∫𝟏

𝒖𝒅𝒖 = −

𝟏

𝒏𝐥𝐨𝐠|𝒖| + 𝑪 =

= −𝟏

𝒏𝐥𝐨𝐠 |

𝟏

𝒙𝒏+ 𝟏| + 𝑪

Hence,


(𝒏𝛀𝒏(𝒙)) = 𝐥𝐢𝐦𝒏→∞

(𝒏 ⋅−𝟏

𝒏𝐥𝐨𝐠 |

𝟏

𝒙𝒏+ 𝟏|) = − 𝐥𝐢𝐦

𝒏→∞𝐥𝐨𝐠 |

𝟏

𝒙𝒏+ 𝟏| = 𝟎

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It’s nice to be important but more important it’s to be nice.

At this paper works a TEAM.

This is RMM TEAM.

To be continued!

Daniel Sitaru

ROMANIAN MATHEMATICAL MAGAZINE

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