ROMANIAN MATHEMATICAL MAGAZINE · R M M ROMANIAN MATHEMATICAL MAGAZINE Founding Editor DANIEL...
Transcript of ROMANIAN MATHEMATICAL MAGAZINE · R M M ROMANIAN MATHEMATICAL MAGAZINE Founding Editor DANIEL...
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 801 - 900RMM - Calculus Marathon 801 - 900
www.ssmrmh.ro
2
Proposed by
Daniel Sitaru – Romania, Srinivasa Raghava-AIRMC-India
Mokhtar Khassani-Mostaganem-Algerie
Naren Bhandari-Bajura-Nepal,Rahim Shahbazov-Baku-Azerbaijan
Mohammed Bouras-Morocco,Abdul Mukhtar-Nigeria
Vasile Mircea Popa-Romania,Ekpo Samuel-Nigeria
Jalil Hajimir-Toronto-Canada,Florică Anastase-Romania
Ahmed Hegazi-Cairo-Egypt,Radu Diaconu-Romania
Igor Soposki-Skopje-Macedonia, Dinu Șerbănescu-Romania
Abdul Hafeez Ayinde-Nigeria, Prem Kumar-India
www.ssmrmh.ro
3
Solutions by
Daniel Sitaru – Romania,Nassim Nicholas Taleb-USA,Soumava Chakraborty-Kolkata-India
Abdul Hafeez Ayinde-Nigeria, Mokhtar Khassani-Mostaganem-Algerie, Ahmed Salama
Hegazy-Cairo-Egypt, Sergio Esteban-Argentina,Gabriel Ruddy Cruz Mendez-Lima-Peru
Gerald Bernabel-Lima-Peru, Dawid Bialek-Poland, Kamel Benaicha-Algiers-Algerie, Pedro
Nagasava-Brazil,Ravi Prakash-New Delhi-India, Timson Azeez Folorunsho-Nigeria
Surjeet Singhania-India, Remus Florin Stanca-Romania,Igor Soposki-Skopje-Macedonia,
Kartick Chandra Betal-India,Yen Tung Chung-Taichung-Taiwan, Kartick Chandra Betal-
India,Florentin Vișescu-Romania , George Florin Șerban-Romania,Florică Anastase-
Romania, Avishek Mitra-West Bengal-India,Ovwie Edafe-Nigeria, Samir HajAli-Damascus-
Syria, Zaharia Burghelea-Romania, Tobi Joshua-Nigeria,Jalil Hajimir-Canada, Sanong
Huayrerai-Nakon Pathom-Thailand, Ali Jaffal-Lebanon, Lazaros Zachariadis-Thessaloniki-
Greece,Yuval Peres-USA, Chris Kyriazis-Athens-Greece, Adrian Popa – Romania, Khaled Abd
Imouti-Damascus-Syria, Soumitra Mandal-Chandar Nagore-India,Tran Hong-Dong Thap-
Vietnam, Sujit Bhowmick-North Bengal-India, Nelson Javier Villaherrera Lopez-El
Salvador,Marian Ursărescu – Romania, Naren Bhandari-Bajura-Nepal, Artan Ajredini-
Presheva-Serbie
www.ssmrmh.ro
4
801. Prove that:
(
*
( )
G: Catalan’s constant
Proposed by Mokhtar Khassani-Mostaganem-Algerie
Solution by Abdul Hafeez Ayinde-Nigeria
Let .
/
. / .
/ .
/
. / .
/ .
/∑ ( ) .
/ .
/ .
/
. / .
/ .
/
( )
∑ ( ) .
/ .
/ .
/
. / .
/ .
/
( )
∑ ( ) .
/ .
/ .
/
. / .
/ .
/
( )
∑
. / .
/ .
/
. / .
/ .
/
( )
∑
. /
. / .
/ .
/
( )
∑
( )
. / .
/.
/ .
/ .
/
∑( )
( )( )( )( )( )
∑( )
( ) ( )( )( )
www.ssmrmh.ro
5
(
∑
( )
∑
( )
( )
∑
( )
∑
( )
∑
( )
+
{
(∑
( )
( )
+
∑
( )
(∑
( )
+
(∑
( )
+}
{
(
*}
{
}
(
*
(
*
Catalan’s constant.
802. Let’s define the function
( ) ∫
( )
Prove that:
∫( (
* ( )* ( )
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Mokhtar Khassani-Mostaganem-Algerie
( ) ∫
∫
∫
( ) ∫ ∑
www.ssmrmh.ro
6
( ) ∑∫ ( )
( ) ∑8 ( )
( ) ( )
9
( ) ∑(
( )
( )
( ) *
(
* (
* (
* ( ) (
*
(
* ( ) ( )
( ) ( ) (
* ( ) ( )
Now:
∫( (
* * ( )
∫( (
* * ( )
∫( ( ) ( ) ( )
( ) ( )
( )
( ) ( )+
( )
∫ ( ) ( )
∫ ( )∑( )
∑
∫( )
∑
( )
∑(
*
∑(
( )
( )
*
www.ssmrmh.ro
7
∑(
*
( ) ( )
∫ ( )
∫ ( )
⏞
{ ( )}
∫ ( )
( ) ( )
( ) ( ) ( )
∫ ( ) ( )
⏞
{ ( )( )}
⏞
∫(
( )+
∑∫ ( )
∑(
( )
( )
( ) *
( ) ( ) ( ) ( )
(Answer)
803.
(√ √ √ √ √ +
:√ √√ (√ √ ) √ √ ;
Proposed by Naren Bhandari-Bajura-Nepal
Solution 1 by Ahmed Salama Hegazy-Cairo-Egypt
( ) ( ) ( ) ( ) ( )
Now we find ( ), let
( ) ( ) ( )
( )
www.ssmrmh.ro
8
( ) ( )
( ( ) ) ( )
√
√
, and √
√
since
√ √
√ √ , put
√
√
√
√ , by substitution we have
√ √ √ √ √
Solution 2 by Sergio Esteban-Argentina
Calculo del geometricamente. Por triangulos notables sabemos
i) Por Angulo inscrito sabemos que
www.ssmrmh.ro
9
y
ii) Sabemos que:
√ √ √ √ √ √ √ y
Usando el teorema de Ptolomeo
(√ √ )(√ ) (√ √ ) 4√ √ 5
(√ √ )(√ ) (√ √ ).√ √ /
Entonces:
√ √ √ √ √
Preliminares
i) Sabemos que el triangulo equilatero
www.ssmrmh.ro
10
Aplicando Pitagoras: ( ) √
ii) Trabajando i)
Prolongamos hasta tai que . Si damos (√ √ )
Entonces
iii) en el decagon regular
www.ssmrmh.ro
11
medida de
Se traza la bisectriz interior
Por el teorema de la bisectriz interior
. Operando .
√
/. Entonces:
: 4√
5;
√ √ si damos
www.ssmrmh.ro
12
804. Find:
Proposed by Rahim Shahbazov-Baku-Azerbaijan
Solution 1 by Gabriel Ruddy Cruz Mendez-Lima-Peru
Sabemos: ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) }
Sumando verticalmente: ∑ ( ( )) ∑ ( ( ))
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
∑ , ( )-
4 ⏟
5
∑ , ( )-
∑ ( ( ))
∑ ( ( ))
www.ssmrmh.ro
13
Solution 2 by Gerald Bernabel-Lima-Peru
∑ ( )
∑ ( )
( ) . / ( ) .
/ ( ) .
/ ( ) .
/ ( ) .
/ ( )
. / ( ) .
/ ( ) .
/ ( )
Si: ( )
( ) 4. / ( ) .
/ ( ) .
/ ( ) .
/5
. / .
/ ( ) .
/ ( ) .
/ ( ) .
/ ( )
Si: ( ) entonces la ecuacion: . / .
/ .
/ .
/
Presenta 22 raices de la forma ( ) donde * +
∑ ( )
. /
. /
805. Prove that:
∑ ( )
( )( )( )
4 ( )
5 .
/√ :
. /
√ ; 4
( )
5
4 ( )
5 ( )
Euler-Mascheroni constant; exponential integral
imaginary error function
Proposed by Mokhtar Khassani-Mostaganem-Algerie
www.ssmrmh.ro
14
Solution by Dawid Bialek-Poland
∑ ( )
( )( )( )
( )
∑(
*
( )( )
∑(
*
⏟
∑.
/
( )
⏟
∑.
/
( )
⏟
(1)
∑.
/
( ) .
/ .
/ .
( )
/ .
( )
/ (2)
Where (1) ( ) ( ) ∑
– exponential integral
∑.
/
( )
∑
.
/
( )
∑
.
/
( )
( ) (
*
. ( )
/ ( ) (3)
Where (II)
∑
∑
∑
( )
∑(
*
( )
∑
4√
5
( )
√
∑
4√
5
( )
√
[
∑
4√
5
( )
√
]
√ ∑
4√
5
( )
( )
√ √
4√
5
.
/ √ 4
. /
√ 5 .
/√ 4
.
/
√ 5 (4)
Where (III) ( )
√ ∑
( )
( )
( ) - imaginary function
Rewriting (1) with (2), (3) and (4) we get:
∑ ( )
( )( )( )
4 ( )
5 4
( )
5
4 ( )
5 ( ) .
/√ :
. /
√ ;
www.ssmrmh.ro
15
4 ( )
5 .
/√ :
. /
√ ; 4
( )
5
4 ( )
5 ( )
806.
∫√ √
√ √ √ √ √ √ √ √
⏟
– fibonacci number;
Prove that
is faster to give golden number then
Proposed by Mohammed Bouras-Morocco
Solution by Kamel Benaicha-Algiers-Algerie
∫√ √
√ √ √ √ √ √ √
Put: ( ) √ √ √ √ (1)
and ( ) √ √ √ √ √ √ √ √ √
So, ( ) ( ) ( ) (2)
We have: ( ) √ ( ) (a)
( ) ( ) √ ( ) ( ) ( )
( )
√ ( ) ( ) ( ( ))4
( )5
√ ( ) ( ) (3)
Put ( ) √ √ ( )
increasing and we have: , - ( )
www.ssmrmh.ro
16
√ √ ( ) √ √ √ √ √ ( √ )
( ) ( ) and then ( ( )) is increasing, ( ) ( ) (4)
( ) , by using (3) and (4), then
So, ( ( )) and ( )
is solution of equation (result (a)) with condion
So, ( ) (b)
∫ ( )
( )
∫ ( )
(using result (2))
, - ( ) then by using second Middle value theory:
- , such that ∫ ( )
( ) ( ) ∫ ( )
( )∫ ( )
∫ ( )
( ) (using result (b))
. We have
( ) - ,⁄
Put
where ( ) is the Fibonacci sequence.
Put ( )
is decreasing ( ) is fluctuate sequence.
By comparison of fluctuate sequence ( ) and increasing sequence ( ( ))
then
is faster than
to . denote the golden ration:
√
807.
∫
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Pedro Nagasava-Brazil
∫
( )
( )
www.ssmrmh.ro
17
∫
( )
( ) ( ) ∫
∫
( )( )( )
( )( )
∫
6. /
7
6. /
. /
7
(
*
∫
∫
< .
/
=
∫
808. Define for all ( ) √ √ √ √ ( )
⏟
then prove that
∫ ( )
( )
√ √ √ √ √
√ √
Proposed by Naren Bhandari-Bajura-Nepal
Solution by Sergio Esteban-Argentina
www.ssmrmh.ro
18
i) Probamos que si , -, se cumpie que
√ √ √ √ (
*
Demonstration:
√ √ ( ) √
| (
*|
Lado que
.
/
√ √ √ (
*
Reemplarando lo intenso:
√ √ √ √ (
* (
*
Analogamente
√ √ √ √
⏟
(
*
en el problema ( ) (1)
ii) Probamos que (2)
Es facel probar que
Entonces: ⏟
⏟
⏟
Nota:
www.ssmrmh.ro
19
Usando (1) y (2) en la integral
∫
∫
∫ ( )
(
∫
∫
)
( |
|
,
(
(
*
*
(
*
Resolvemos .
/. Consideramos el triangulo notable de y
i)
ii)
√[ (√ √ )] (√ √ )
www.ssmrmh.ro
20
√ (√ √ ) (√ √ ) (√ √ )
√ √ √
SABEMOS que √ √ √ √ esto se deduce de (√ √ )
. Entonces
(
* (
*
(√ √ )
√ √ √ √ √ √ √
√ √
Por lo tanto
.
/
√ √ √ √ √
√ √
(√ √ √ √ √
√ √ +
809. Prove that Catalan’s constant:
∑ ∑( )
( )
∑∑( )
( ) ∑ ∑
( )
( )
Proposed by Naren Bhandari-Bajura-Nepal
Solution by Sergio Esteban-Argentina
∑∑( ) ( ) ( )
( )
∑ ∑( )
( )
We calculate
∑∑( ) ( ( ) )∫ ( )( )
∫∑∑( ) ( ( ) ) ( )
∫ ( )
∑( ) [∑ ∑( )
]
∫ ( )
4
5 [
]
www.ssmrmh.ro
21
∫ ( )
( )( )( )
∫ ( ) [
( )
( )]
∫ ( )
∫ ( )
∫ ( )
Now, we calculate
∑∑( )
( )
∑∑( )
∫ ( )
∫ ( )
∑
∑( )
∫ ( )
.
/ (
*
∫ ( ) [
( )
( )]
∫ ( )
∫ ( )
∫ ( )
810. Prove that:
∫ .
/ ( )
( )
Proposed by Abdul Mukhtar-Nigeria
Solution by Dawid Bialek-Poland
∫ .
/ ( )
∫ ( )
( )
∫ ( )
( )
( )
( ) .
/ ( )
( )
www.ssmrmh.ro
22
∫ ( )
( )
∑
∫ ( )
⏞
∑
:[
( )] |
∫
; ∑
[
] |
∑
⏞
( ) ( )
∫ ( )
( )
⏞
∫ ( )
( )
⏞( )
∫
( )
∑
∫ ( )
⏞
∑
:6
( )7 |
∫ ( )
;
∑
∫ ( )
⏞
∑
:6
( )7 |
∫
;
∑
( ) 46
( )7 |
5
∑ ( )
⏞
∑ ( )
∑
( )
∑
∑
⏞
( )
( ) ( )
( ) ( ) Where:
( )∑
( ) ∑
( )
( ) ∑
( )
Rewriting (1) with (2) and (3), we get
∫ .
/ ( )
( )
www.ssmrmh.ro
23
811. Solve for :
∫. ( )/
Proposed by Daniel Sitaru – Romania
Solution 1 by Kamel Benaicha-Algiers-Algerie
∫ ( )( ( ))
(E); for
( ) ∫ ( )( ( ))
∫ ( )( ( ))
∫ ( ) ( )
( ( ))
∫
( )
( )
|
( )
( ) ( )
(E) ( ) ( )
( ) ( ) ( )
bijection function
So: (E) ( ) ( ) ( ) ( )
Solution 2 by Ravi Prakash-New Delhi-India
Let ∫
( )dt ∫
.
/ ∫
Let ( )( )
( )
∫
( )
∫ . Thus,
]
∫
∫
Thus, the given equation becomes
or or
Solution 3 by Timson Azeez Folorunsho-Nigeria
∫ . ( ) ( )/
put
www.ssmrmh.ro
24
∫ . ( )/
∫ ( ( ))
∫ ( ( ) )
∫ ( ( ) )
∫ ( ( ) )
Put
∫ ( ( ))
∫
|
|
( )|
( )
( ) is also a solution
812. Solve for real numbers:
∫(
*
(
*
Proposed by Daniel Sitaru – Romania
Solution 1 by Kamel Benaicha-Algiers-Algerie
∫ ( )
( )
∫ ( )
( ( ) *
Put: ( )
, so
( )
∫
( )
|
( )
( ) |
Then, if ∫ ( )
( )
.
/ (E)
(E) ( )
( )
( )
( )
( ) ( )
www.ssmrmh.ro
25
Put: ( ) ( )
- , ( )
( )
. So: ( )
( )
for and for: . So: ( )
- ,
is the maximum of
( )
; ( ) unique value for real numbers
( ) * +
Solution 2 by Surjeet Singhania-India
∫ ( )
( )
(
* ∫
( )
( )
Put ( )
( )
∫
( )
∫
( )( )
(
*
(
*
(
*
Comparing with other side we get
Solution 3 by Remus Florin Stanca-Romania
∫ ( )
( ) ∫
( )
( ) 4(
( )*
5
∫ ( )
( )
(
( )*
Let
( )
( )
( ) ∫
( )
( ) ∫
∫ ( )
( )( )
∫(
*
(|
|*
(|
( )
( )
|,
As we know that ( )
∫ ( )
( )
4
( )
( )5
4
( )
( )5
(
*
( )
( )
( ) ( ) ( ) ( )
www.ssmrmh.ro
26
( ) ( )
Let ( ) ( )
( )
( )
and we have that for ( ) and for
( ) and ( )
on ( ) the function is increasing ( )
and for , ) the function is
decreasing with
.
813. ( )
Find:
( ) ∫4( ) ( ) ( ) ( )
( ( ))( ( ))5
Proposed by Daniel Sitaru – Romania
Solution by Igor Soposki-Skopje-Macedonia
( ) ∫( ) ( )
( )( )
( )( ) ( )( )
( )
( )
( ) ( ) ( )
( )
( ) ∫
⏟
( ) ( )
( ) ⏟
( ) ∫
|
|
( ) |
| |
|
www.ssmrmh.ro
27
( )
( ) ( )
( )
2
( )
( ) | ( ) ( )
( ) ( ) |
814. Find:
∫
(
( )√( )( ) 4 4√
5 5
)
Proposed by Daniel Sitaru – Romania
Solution by Igor Soposki-Skopje-Macedonia
:√
; ∫
( )√( )( ) 4 √ 5
4√ 5
√
( )
( )
www.ssmrmh.ro
28
√
( )
( )
√
( )
( )
√( ) . /
( )
√( )( )
( )√( )( ) ∫
( ) ∫
( )
2
3 ∫
∫
∫
( )
4 √ 5
815. Given that:
( ) ∫4( )√
( )√ 5
Prove or disprove that:
( ) ∫ ( ) √ 4 4√
√ 55
Proposed by Ekpo Samuel-Nigeria
Solution by Mokhtar Khassani-Mostaganem-Algerie
( ) ∫
( )√ √ ∫
√
√ √
√
(√ )
√
√
( ) ∫ ( ) ∫√
√
⏞ √ √
√ ∫ √
√
( )
√ 4√
√ 5 √ 4
√
√ 5
Note: ( )
√ ∫
and ( ) ( )
www.ssmrmh.ro
29
816. Prove that for:
∫ .
/
√
( )
Proposed by Srinivasa Raghava-AIRMC-India
Solution 1 by Kartick Chandra Betal-India
∫ .
/
√ ( )
( ) ∫ .
/
√ ( )
( ) ∫
√ ( )
(
( ) *
( )
∫
√
( )
∫( )
( ) ∫
( ) ∫( )
( )
∫ ( )
( )
Solution 2 by Yen Tung Chung-Taichung-Taiwan
∫ .
/
√
⏟
∫ .
/
√
∫ (
*
∫ ( )
⏟ ( )
∫ ( )
∫
( )
www.ssmrmh.ro
30
817. Find:
( ) ∫ (
( )( )*
Proposed by Vasile Mircea Popa-Romania
Solution by Mokhtar Khassani-Mostaganem-Algerie
∫
( )( )
∫4 ( )
( )( )
( )
( )( )
( )( )5
( )
( )
( )
∫ ( )
∫:
;
( ) ∫
. /
(
*
( ) (
√ * (
√ *
∫ ( )
∫:
( )
;
( ) ∫
. /
(
*
( ) (
√ * (
√ *
{
( )
( )( ( ) (
√ * (
√ **
( )( ( ) (
√ * (
√ **
}
www.ssmrmh.ro
31
( )
√ ( )
( )
√ ( )
(
√ *4
( )
√ ( )
( )
√ ( )5
818. Show that:
∫ ( ) ( ( )) ( ( ( )) ( ))
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Mokhtar Khassani-Mostaganem-Algerie
∫ ( ) ( ( ) )
∫ ( ) ( )
∫
∫
∫
now let:
So: ∫
| |
( (
*+ (
( )
*
819. Find:
∫4 ( )
√ 5
Proposed by Ekpo Samuel-Nigeria
Solution 1 by Abdul Hafeez Ayinde-Nigeria
∫ ( )
√ ∫ ( ) ( √ ) √ ( )
√ ∫√
√ ( )
√ ∑( )
∫√
www.ssmrmh.ro
32
√ ( )
√ ∑( )
∫
√ ( ) √
√ ∑
( )
. /
√ ( ) √
√ ∑
( )
. /
√ ( ) √
√ ∑( ) .
/
. /
√ ( ) √
√ . /
. /∑( ) .
/
. /
√ ( ) √
√ ∑( ) .
/
. /
√ ( ) √
√ (
*
Where ( ) is the classic hypergeometric function.
Solution 2 by Mokhtar Khassani-Mostaganem-Algerie
∫ ( )
√
⏞
{ √ ( )}
√ ∫√
⏞
√ ( )
√ ∫
√ √ ( )
√ (
* √ ( )
√ (
*
820. Find a closed form:
∫4 ( )√
√ 5
Proposed by Ekpo Samuel-Nigeria
www.ssmrmh.ro
33
Solution 1 by Kamel Benaicha-Algiers-Algerie
∫ ( )√
√
. Let , then:
∫
( )
( )
∫
( )
( )
( ) ∫
( )
( )
Integration by parts gives:
∫
( )
∫
( )
(
*
∫
∫
(
*
∫(
* ( )
∫
∫
: ∫
∫
;
(
*
(
*
(
*
( (
* (
*+
By using Gauss theorem:
(
* (
* √ 4
√
√ 5
:√ √
√ √
√
√ ;
(
* (
* √ (√ )
:√ √
√ √
√
√ ;
www.ssmrmh.ro
34
√
(√ )
:√ √
√ √
√
√ ;
√ (√ )
√
So: ∫ ( )√
√
√ (√ )
√
Solution 2 by Kartick Chandra Betal-India
∫ ( )√
√
∫( )
√
∫ ( )
( )
∫ (
*
∫(
*
{
( ) } ∫
( )
( )
∑ ( )
( ) ∫(
*
∑ ( )
( ) <
=
∑( )
( )
. / .
/
∑( ) ( )
( ) ( )
∑( ) ( )
∑( )
8
9
∑( )
∑( )
∑( )
( )( )
∑( )
>
?
∑( )
∑( )
www.ssmrmh.ro
35
∑( )
∑( )
∑( )
∫
∫
∑( )
∫
( )
∫
∫
∫>
.
/
?
∫
.
/
. /
(√ )
∫
. /
. /
(√ )
√ 2
3
√ 6 |
√
√ |7
√
√ | √
√ |
√
√ (√ )
821. Find without softs:
∫∫
( )
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Daniel Sitaru-Romania
∫(
( ) *
⏞
∫4( )
( ) 5
∫4( )
( ) 5
∫4( )
( ) 5
www.ssmrmh.ro
36
∫4 ( )
( ) 5 ∫
∫:∫(
( ) *
;
∫.
/
822. For , we have:
∫
( ) √ ( )
Proposed by Srinivasa Raghava-AIRMC-India
Solution 1 by Florentin Vișescu-Romania
∫
√
∫
√
( ) ∫
√
∫
√
∫
√
√
∫ ( √ )
( √ )
∫ ( √ )
√
∫
|
Solution 2 by Tobi Joshua-Nigeria
∫
√
www.ssmrmh.ro
37
∫ ( √ )
( √ )( √ )
∫ ( √ )
∫
∫
∫ √
(odd fxn) ∫
(odd fxn)
∫
(even function);
1
823. Prove that:
∫∫(( ) ( ) ( ) ( ))
( ( ))
( )
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Kamel Benaicha-Algiers-Algerie
∫∫( ) ( ) ( ) ( )
( ( ))
∫∫( ) ( ) ( )( ( ) ( ))
( ( ) ( ))
∫∫
( ) ( ( ) ( ))
∫∫( ) ( )
( )
( ( ) ( ))
∫
( )4 ( ) ( )
5 ∫∫
( ) ( )
( )
( ( ) ( ))
www.ssmrmh.ro
38
∫ ∫( ) ( )
( )
( ( ) ( ))
∫
( ) 4 ( ) ( )
5
∫∫( ) ( )
( )
( ( ) ( ))
∫ ∫( ) ( )
( )
( ( ) ( ))
(1)
( {( ) ⁄
}
*( ) + {( )
}
,
∫ ( ) ∫
( )
, ( ) ( )-
∫ ( )
[
( )
( )]
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫
( )
∫
( )
( ) (2)
∫ ∫( ) ( )
( )
( ( ) ( ))
∫ ∫ ( )
( )
( ( ) ( ))
www.ssmrmh.ro
39
∫ ( )
∫( ( ) ( ( )))
∫ ( )
( )
( ( ( ) ) ( ))=
∫ ( )
[ ( )
( ( ))]
∫ : ( )
( ) ( )
( )
∫
( )
( )
;
∫ ( )
( )
∫
( )
( )
(
* ∫
( )
∫ ( )
( ) (3)
(1), (2) and (3)
( )
∫∫( ) ( ) ( ) ( )
( ( ))
( )
824. Evaluate the integral
If ∫ ∫ 4
. /
. /5
Proposed by Abdul Hafeez Ayinde-Nigeria
Solution by Pedro Nagasava-Brazil
1. Separating variables:
∫ .
/
∫
2. For the variable, let’s apply Frullani integral:
www.ssmrmh.ro
40
∫
.
/ .
/ ( )
3. For the variable, let :
∫ .
/
Now, let’s apply Differentiation Under the Integral Sign:
( ) ∫ (
*
( )
∫ .
/
(1)
Let
:
( )
∫
.
/
( ) ∫
.
/
(2)
With (1) (2):
( ) ∫ (
* .
/
√ ( ) √
( )
∫
4
5
√
6
7
∫ 4
5
√
6
7
Hence: ∫ .
/
√
0
1
√
Finally:
∫ ∫
4
. /
. /5
√
( )
825. If , then 0
1 such that:
www.ssmrmh.ro
41
∫
( )( )
( )(
*
Proposed by Florică Anastase-Romania
Solution 1 by George Florin Șerban-Romania
From Osiann-Bonet formula we have:
, - , - , -
∫ ( ) ( ) ( )∫ ( ) ( )∫ ( )
∫
( )( )
∫
( )( )
, - ( )
( )
( )
, -
, - ( )
, -
( )∫ ( ) ( ) ∫ ( )
∫ ( ) ∫
∫
. /
(
*
www.ssmrmh.ro
42
∫ ( )
∫ ( )
(
*
( )(
*
Solution 2 by proposer
Theorem (Bonnet-Weierstrass):
If , - decreasing function of class and , - continuous function,
then , -
∫ ( ) ( ) ( )∫ ( ) ( )∫ ( )
Demonstration:
Let , - ( ) ( ) ( ) ( ) , -
From theorem 2 of means , - such that:
∫ ( ) ( ) ( )∫ ( )
∫ ( )( ( ) ( )) ( ( ) ( ))∫ ( )
∫ ( ) ( ) ( )∫ ( ) ( ( ) ( ))∫ ( )
( )∫ ( ) ( )∫ ( )
q.e.d. Let 0
1 ( )
( )
( )
( )
( ) ∫
∫
∫
. /
0
1
www.ssmrmh.ro
43
∫
( )( ) ( )( ( ) ( )) .
/ ( ( ) ( ))
(
*
( )(
*
826. Find:
∫ ( ( ))
Proposed by Mokhtar Khassani-Mostaganem-Algerie
Solution by Avishek Mitra-West Bengal-India
∫
( ) ∫ [ ∑ ( )
]
∫
∑
∫
( )
∫
( ) 6 ( )
7
∫
( )
∫
( )
[
8 ( )
9
∫
( )
]
www.ssmrmh.ro
44
[
( )
∫
( )
]
( )
[
8 ( )
9
∫
( )
]
( )
[
∫ ( )
]
( )
[
8 ( )
9
∫ ( )
]
( )
6
( )
* ( )+
7
( )
( )
6
7
∑( )
∑( )
( )
( )
( ) ( )
( ) ( )
( )
( )
( )
( )
827. Evaluate the following integral:
www.ssmrmh.ro
45
∫ ( ) ( )
( )
Proposed by Ahmed Hegazi-Cairo-Egypt
Solution by Avishek Mitra-West Bengal-India
∫ ( )
( )
[
]
∫ ( )
( )
<∫ ( )
( )
∫ ( )
=
∫ ( )
, ( ) ( ) ( )- ( ) ( ) ( ) (Answer)
828. Prove that:
∫ (( )
( ) *
Proposed by Mohammed Bouras-Morocco
Solution by Kamel Benaicha-Algiers-Algerie
∫ (( ) ( ) *
∫ ( )
( )∫
( )
Put:
∑
∫
( )
( )
∑
( )
( ( ) )|
www.ssmrmh.ro
46
( )
( )
∫ (( )
( ) *
829. Let: ( ) ∑
Prove that:
∫ 4
√( ( ) ( ))5
Proposed by Mohammed Bouras-Morocco
Solution by Mokhtar Khassani-Mostaganem-Algerie
∑
∑ . /
(
*
( ) | |
Now: ∫ .
√ ( ) ( )/
∫ 4√
5
⏞
8 4√
59
√
∫
√ ( )
√
:∫
√
∫
√ ( )
;
8√ 4√
59
830. Find:
∫ ( ) ( )
Proposed by Abdul Mukhtar-Nigeria
Solution 1 by Avishek Mitra-West Bengal-India
∫ ( )
www.ssmrmh.ro
47
6 ( ) ∫
7
∫ ( )
( )
6∫
7
[
( ) ]
∫ ( )
( )
∫ ( )
( )
∑
∫
∑
( ) ( )
( )
∑
( )
∑ ( )
( )
( )
[∑ ( )( )
∑
( )
]
0∑
∑
1 , ( ) ( ) ( ) ( )- ( ) ( ) ( ) (Answer)
Solution 2 by Ovwie Edafe-Nigeria
∫ ( )
6 ( )
7
∫ ( )
∑
∫
∫
( )
( )
∑
( )
∑
∑
∑
, ( ) ( ) ( )- ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
∑
.
/ ( )
∑ ( )
( )
www.ssmrmh.ro
48
831. Prove that:
∫ ( ) ( )
(
* ( ) ( ) ( )
( )
Proposed by Mokhtar Khassani-Mostaganem-Algerie
Solution by Dawid Bialek-Poland
∫ ( ) ( )
∫ ( ) ( )
( ) ∫
( )
⏟
∫
( ) .
/
⏟
(1)
( ) ∫
( )
( ) (2)
∫ ( ) .
/
∑
∫ ( )
∑
< ( )
∫ ( )
=
( )∑
∑
∫ ( )
( ) ( ) ∑
< ( )
∫
=
( ) ( ) ( )∑
∑
[
]
( ) ( ) ( ) ( ) ∑
∑
( ) ( ) ( ) ( ) ( ) ∑
.
/
( ) ( ) ( ) ( ) ( ) .
/ (3)
Rewriting (1) with (2) and (3), we get:
www.ssmrmh.ro
49
( )∫
( )
∫ ( ) .
/
( )
( ) ( ) ( ) ( ) ( ) (
*
Note: ( ) ∑
( ) ∑
832. ( )
∏ ( √
√ *
Prove that:
∫
( )(
*
Proposed by Mohammed Bouras-Morocco
Solution by Samir HajAli-Damascus-Syria
( )
∏ 4 √
√
5
∏ : ( )
( )
;
∏ 4 ( )
5
( ( ))
( )
( )
∫
( )
(
* ∫
( )
(
*
∫ ( )
∫ ( )
∫ ( )
∑∫ ( )
∑
( )
∑
( )
∫ ( )
∑∫ ( )
∑
( )
𝛀 𝝅𝟐
𝟐𝟒 𝟏
𝟑 𝝅𝟐
𝟖 𝝅𝟐
𝟏𝟐
www.ssmrmh.ro
50
833. For any complex numbers with Re( ), Re( )
If
( ) ∫( )
( )
then show that
:∫ ( )
;
:∫ ( )
;
4
( )
( )5
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Nassim Nicholas Taleb-USA
∫ (
*
, -
We substitute , so ∫4
5 , -
We have
∫
, -
, -
, ( )- .
/, where is the
Laplace transform.
From tables, , ( )-( )
, so
, and reversing the order of
integration – deriv: ∫
[ ] |
[ ] |
[ ]
[ ]
6
7
Repeating for and summing gets the required result.
834. Find:
∫4
( ) ( )5
Proposed by Radu Diaconu-Romania
www.ssmrmh.ro
51
Solution by Zaharia Burghelea-Romania
Let
( )
∫
( ) ( )
∫ .
/
( ) . /
. /
( )
∫( ) .
/
, -,( ) ( )-
∫
( ) ( )
∫
( ) ( )
∫
( ) ( )
√ ( ) 4
( )
√ ( )5|
√ ( ) (
√ ( )
+
835. Prove that:
∫
4
5
( )
( )
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Zaharia Burghelea-Romania
∫
4
5
⏞
∫ (
*
∫ 4
5
∫ (
*
∫ 4
5
⏞
∫
∫
∫
( )
∫ (
*
⏞
∫ ( )
√
⏞
∫ ( )
( )
www.ssmrmh.ro
52
( )
( )
836. Find:
∫ ( ) ( ) , n
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Avishek Mitra-West Bengal-India
( )
∫
∫
∫
∫
( )
837.
∫
Proposed by Igor Soposki-Skopje-Macedonia
Solution by Tobi Joshua-Nigeria
∫
∫
∫
∫
∫
∫
( )
( ) ( )
∫ ( )
( ) ( )
6
∫
( )
∫
∫( )
( )7
www.ssmrmh.ro
53
[
∫
( )
. /
4√ 5
∫
. /
4√ 5
∫(
*
( *
]
[
√ (
√ *
√ (
√ *
∫
. /
. /
(√ )
]
6
√ (
√ *
√ (
√ *
√ 4
√
√ 5 7
6
√ (
√ *
√ (
√ *
√ 4
√
√ 5 7
6
√ (
√ *
√ (
√ *
√ 4
√
√ 5 7
838. Find:
( ) ∫ (
( )( )*
Proposed by Vasile Mircea Popa-Romania
Solution by Kamel Benaicha-Algeirs-Algerie
( ) ∫ (
( )( )*
∫ (
( )( )*
( ):∫
∫ 4
5
;
( )4
4
√ 5 | 5
| |
( )
( ) ∫
∫
( ) {
| |
( )
www.ssmrmh.ro
54
839. Let be positive and continuous when . Prove:
:∫ ( )
;
∫( ( ))
Proposed by Jalil Hajimir-Canada
Solution by Soumitra Mandal-Chandar Nagore-India
is a positive continuous function on , -
( ) for all , - hence ∫ ( )
Now ∫ ( )
.∫ ( )
/
. Let ∫ ( )
We have to prove, ( ) ( )
( )( ) ( ) ( )
Hence .∫ ( )
/
.∫ ( )
/
∫ ( )
:∫ ( )
;
:∫ ( )
;
∫ ( )
:∫ ( )
;
840. If
then:
∫∫( ( )
( )*
√
( )
Proposed by Jalil Hajimir-Canada
Solution by Daniel Sitaru-Romania
Denote
( )
( ) ( )
www.ssmrmh.ro
55
⏞
.
/ ⏞
.
/
. /
.
/
√
∫∫4 ( )
( )5
∫∫4 √
5
√
:∫
;
√
( )
Equality holds for .
841. If
then:
∫∫4 ( )
( )5
√ ( )
Proposed by Daniel Sitaru – Romania
Solution 1 by Jalil Hajimir-Canada
( )
√
√
if
∫∫
( )
√
∫∫
√
( )
( ) ( )
Solution 2 by Sanong Huayrerai-Nakon Pathom-Thailand
For and
, we have as following:
1. since ( )
Hence ( ( )) √
and since
( ( ))
( )
Hence
( ( ))
( ) ( √ )
( )
www.ssmrmh.ro
56
Hence √
( ( ))
( )
( √ )
Hence ( )
( ) √
Hence ∫ ∫
( )
( ) ∫ ∫
√
√
|
√
( )( ). Therefore, it is true.
Solution 3 by Ali Jaffal-Lebanon
Let ( ) 8
1
1
we have ( ) 0
1
( )
Let ( ) 0
1 ( )
then ( )
and
( )
( )
So, ( ) 0
1 then ( ) 0
1 so, is concave on 0
1
Let 0
1 where we have .
/
( ) ( ) ( )
so,
.
/
√
by GM-AM inequality we have:
(
*
(
*
4 √
5
√
So, ∫ ∫
( )
∫ ∫
√
( ) √
www.ssmrmh.ro
57
842. If then:
∫:∫:∫(√( )( )( )
( )( )( )+
;
;
(
*
Proposed by Daniel Sitaru – Romania
Solution 1 by Lazaros Zachariadis-Thessaloniki-Greece
√∏( )
∏( )
∏( )
( )
so, ∫ ∫ ∫√∏( )
∏( )
∫ ∫ ∫
∫∫
( )
( )
( ) ( )
( )
. . //
Solution 2 by Sanong Hauyrerai-Nakon Pathom-Thailand
For , we have ( ) ( ) ( )
and ( )
and since ( )
( )
( )
Hence ( ) ( ) ( ) ( )
( )
( )
( )
( )
( )
√( )
( )
√ ( )
. Hence for
√( )( )( )
( )( )( )
√
(( ) )
Hence for
∫:∫:∫√( )( )( )
( ) ( )( )
;
;
www.ssmrmh.ro
58
∫:∫:∫
;
;
|
( )( )( )
.
/
. Therefore, it is true.
Solution 3 by Jalil Hajimir-Canada
Let √( )( )( )
( )( )( ) since √
√
√
,( )( )( )-
∫∫∫
∫∫∫
( )
843. If (, -) | ( )| , - then:
∫ ( )
:∫ ( )
;
Proposed by Jalil Hajimir-Canada, Dinu Șerbănescu-Romania
Solution by Daniel Sitaru-Romania
( ) ( ) ⏞
( )( )
( ( )) ( ) ( ( )) ( ) ( )
( ) ( ( ) ( )) ( )
∫∫( )
∫∫( ( ) ( ))
∫∫ ( ) ∫∫ ( ) ( ) ∫∫ ( )
∫ ( ) ∫ ( ) ∫ ( ) ∫ ( )
www.ssmrmh.ro
59
∫ ( ) :∫ ( )
;
∫ ( ) :∫ ( )
;
844. If then:
∫∫ (
*
( )∫
∫∫ (
*
( )∫
Proposed by Daniel Sitaru – Romania
Solution 1 by Ali Jaffal-Lebanon
Let ( ) , where - ,
( )
( )
( )
Let ( )
( ) ( ) has the same sign as since for all
. We have 0
1 [.
/
] . So,
( )
( )
then ( ) for all then ( ) and is convex.
By Jensen’s inequality: .
/
( ) ( )
∫∫ (
*
( )
∫( )
( )
∫ ( )
www.ssmrmh.ro
60
4 ( )
( )
5:∫( )
;
( )∫
( )∫
Therefore: ∫ ∫ .
/
( ) ∫
( )∫
∫∫ (
*
Solution 2 by Soumitra Mandal-Chandar Nagore-India
Let ( ) for all
( )
( )
( )
Let ( ) for all .
Then ( ) {.
/
} for all . Hence is an increasing function
( ) ( ) then ( ) for all .
Hence is a convex function, so by definition:
( )
( ) .
/
( )
( ) (
* (
*
.
/ .
/ [generalizing by ]
(
* (
*
( )∫
∫∫ (
*
( ) ∫
∫ ∫ .
/
. Proved.
845. If then:
www.ssmrmh.ro
61
∫
∫
∫
√ ( )
Proposed by Daniel Sitaru – Romania
Solution 1 by Lazaros Zachariadis-Thessaloniki-Greece
( ) ( )
( ) convex, so:
∫ ( )
(
*
( )
Thus ∑ ( )
√ ( )
( )
( )
√ ( )
( )
( )
√ ( )
√ ( )
Solution 2 by Ali Jaffal-Lebanon
Let ( ) , , ( )
( )
( )
then is convex on , , so,
∫ ( )
( ) .
/ for all then
∫ ( )
∫ ( )
∫ ( )
(
* (
* (
*
√ .
/ .
/ .
/
by GM-AM inequality
√ ( ) ( ) ( )
√ ( ) ( ) ( )
but ( ) ( ) ( )
(( ) ( ) ( ))
since ( ) is convex on , , then
√ ( ) ( ) ( )
√ ( )
Therefore:
∫
∫
∫
√ ( )
www.ssmrmh.ro
62
846. If then:
∫ ∫ ∫ :√
√
√
;
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Daniel Sitaru-Romania
√
√
√
√
( )
√
( )
√
⏞
( ) √
(
√
)
(
√
)
(
√
)
( ) √
( ) ( )
( ) √
( ) ( )
⏞
( ) ( ) ( ) ( )
( )
( )
√
√
√
∫ ∫ ∫ :√
√
√
;
∫ ∫ ∫
( )( )( )
847. If then:
∫ ∫ ∫ :√ √
√ √ √ ; √
Proposed by Jalil Hajimir-Toronto-Canada
www.ssmrmh.ro
63
Solution by Daniel Sitaru-Romania
( ) ( ) ( ) . By Popoviciu’s inequality:
∑ ( )
(
*
∑ (
*
∑
∑
√
(√ √ √ )
√
√ √ √ √
√
√ √ √ √
∫ ∫ ∫ :√ √
√ √ √ ; ∫ ∫ ∫ √
√
Equality holds for .
848. If
then:
( ) ( ) :∫
; ∫
. /
:∫
;:∫
;
Proposed by Daniel Sitaru – Romania
Solution 1 by Ali Jaffal-Lebanon
We have ( ) is continuous and convex function on . Since ( ) for
all then
:
∫ (
*
;
∫ ( (
**
But . .
//
. So, .
∫ .
/
/
∫
We have ( ) ( ) .∫
/
∫ .
/
( ) ( ) :∫
;(
*∫
www.ssmrmh.ro
64
( ) :∫
;:∫
;
But ( ) .∫
/
4∫ .
/
.
/
5
By Cauchy-Schwarz .∫
/.∫
/
Therefore: ( ) .∫
/ .∫
/ .∫
/ .∫
/
Therefore ( ) ( ) .∫
/
∫ .
/
∫
∫
Solution 2 by Soumava Chakraborty-Kolkata-India
( ) ( ):∫
; ∫
. /
( )
:∫
;:∫
;
The continuous analogue of AM-GM inequality:
∫
( )
( ) ∫
. /
and
.
/ ∫
∫
( )
(1).(2) .∫
/ .∫
/ ( )
( ) .∫
/
∫ .
/
Again, continuous analogue of reverse CBS inequality
:∫
;:∫
; ( )
:∫√
√
;
( )
(3) (4) RHS of (a) LHS of (a) (a) is true (Proved)
849. If , - ( ), continuous, then:
∫∫:√ ( ) ( )
√ ( ) ( );
( )∫ ( )
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
65
Solution 1 by Yuval Peres-USA
It suffices to verify that , -
√ ( ) ( )
√ ( ) ( )
( ) ( ) (*)
and then integrate both sides. Squaring, (*) is equivalent to:
( ) ( )
( ) ( ) √
( ) ( )
( ) ( )
( ) ( ) ( ) ( )
that is: √, ( ) ( )- ( ) ( ) [ ( ) ( )] ( ) ( )
This is just the AM-GM inequality √
Solution 2 by Chris Kyriazis-Athens-Greece
It holds that √ ( ) ( )
( ) ( )
and √ ( ) ( )
( ) ( )
So, adding those inequalities, we have:
√ ( ) ( )
√ ( ) ( ) ( ) ( )
Integrating from to , we have:
∫∫<√ ( ) ( )
√ ( ) ( )=
∫∫, ( ) ( )-
( )∫ ( )
(*)
(*) ∫ 2∫ , ( ) ( )-
3
∫ 0∫ ( )
( )( )1
∫ ( )
( ) ∫ ( )
( ) ( )∫ ( )
Solution 3 by Jalil Hajimir-Canada
We know for any two positive real numbers and
( ) ( )
www.ssmrmh.ro
66
∫∫[ ( ( ) ( )) ( ( ) ( )) ]
∫∫( ( ) ( ))
∫∫√ ( ) ( )
√ ( ) ( ) ( )∫ ( )
850. If then:
∫∫∫ √ (
*
( ) ∫ √
Proposed by Daniel Sitaru – Romania
Solution by Ali Jaffal-Lebanon
Let ( ) ( )
( )
( )
( )
( )
(
* ( )
we have ( ) for all , - then is concave on , -
Let , -, so, by Jensen’s inequality we have:
(
*
( ) ( ) ( )
∫∫∫ (
*
∫∫∫ ( ) ( ) ( )
∫∫∫ ( ) ( ) ( )
∫
∫
∫ ( )
∫
∫
∫ ( )
∫
∫
∫ ( )
www.ssmrmh.ro
67
( ) ∫ ( )
( ) ∫ ( )
( )
∫ ( )
( )( )
∫ ( )
( ) ∫ ( )
. So,
∫∫∫ (
*
( ) ∫ ( )
Therefore: ∫ ∫ ∫ √ .
/
( ) ∫ √
851. If – continuous, ( ) ( ) ( ) then:
∫∫∫ ( )
∫ ( )
Proposed by Daniel Sitaru – Romania
Solution by Adrian Popa – Romania
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
Similarly: ( ) ( ) ( ) ( )
( )
( ) ( ) ( )
So, ( ) ( ( ) ( ) ( ))
∫∫∫ ( )
∫∫∫( ( ) ( ) ( ))
∫ ( )
∫ ( )
∫ ( )
852. ( ) ∫ (
*
( ) ∫ (
*
www.ssmrmh.ro
68
Prove that:
( ) ( )
0
1
Proposed by Daniel Sitaru – Romania
Solution by Șerban George Florin – Romania
√
( )(
)
√
( )(
) √
( )( ) √
√
because: ( ) , - and
( ) , -
( ) ( ) ∫(
+
∫√
∫
|
|
0
1 , - and
because , - ( ) ( )
, true, because
853. Prove without softs:
∫∫∫ √
( )( )
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Soumava Chakraborty-Kolkata-India
√ √
www.ssmrmh.ro
69
∫∫∫ √
( )( )
⏞( )
∫∫∫
( )( )
∫
.
/
√
∫
√
∫√
⏞( )
∫√
∫√ ∫√
∫(√ √ ) √ ∫( )
√
√ ∫ ( )
√ ( ) √ ∫
√ ( )
⏞( )
√ ( )
∫(√ √ ) √ ∫( )
√
√ ∫( )
√( ) √ ∫
√ ( )
⏞( )
√ | √ |
( ) ( ) ∫√
√ ( ) √ | √ |
∫√
√ .
/
√ |
√
| √ ( ) √ | √ |
√ (√ ) √ .
/
∫√
⏞( )√
4
(√ )5
www.ssmrmh.ro
70
( ) ( ) ∫
⏞( )√
4
(√ )5
∫
∫( )
∫
∫
⏞( )
( ) ( ) ∫∫∫ √
( )( )
4√
54
(√ )5∫∫
√ 4
(√ )5 (
*∫
√
4
(√ )5 ( )
( )
854. Let be continuous on , -. If ( ) ( ) , -, prove:
∫ ( )
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Khaled Abd Imouti-Damascus-Syria
Suppose: √ . So: (
√ *
(
√ *
So:
√
√ ( ) ( )
√
For: , -. Then: ( ) ( )
∫ ( )
∫ ( )
2
8
∫ ( )
∫ ( )
{
{
www.ssmrmh.ro
71
∫ ( )
∫ ( )
∫ ( )
∫
, -
∫ ( )
855. If ( ) then:
∫∫(( ( ) ( )) 4 ( ) ( )
5+
( )∫( ( ) ( ( )))
Proposed by Daniel Sitaru – Romania
Solution by Florentin Vișescu-Romania
( ) ( )
( )
( )
( )
( )
( ) convexe .
/
( ) ( )
( )
( )
( ) ( )
∫∫, ( ) ( )-
( ) ( )
∫∫ ( )
( ) ( ) ( )
∫∫( ( ) ( ))
( ) ( )
( )∫ ( )
( )
856. Prove without softs:
∫ ( )
www.ssmrmh.ro
72
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Daniel Sitaru-Romania
, - ( ) ⏞
(
*
( ( )) (
*
∫ ( ) ∫(
*
4(
*
(
*
5
∫ ( )
(
*
857. If .
/ continuous then:
∫ ( ( ))
∫ 4
( )5
( )
Proposed by Daniel Sitaru – Romania
Solution by Khaled Abd Imouti-Damascus-Syria
∫( ( ( )) 4
( )5
( )
Let be the function: ( ) ( ) .
/ 1
0
( )
.
/ 1
0
, ( )-
, ( )-
( )
.
/ ( )
.
/
( )
( )
So: ∫ ( ( ( )))
(
( )* ∫
( )
www.ssmrmh.ro
73
858. Prove without softs:
∫√ ( )
( )
√
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Daniel Sitaru-Romania
∫4 ( )
( ) 5
⏞
∫4( ) ( )
( ) 5 ( )
∫4( ) ( )
( ) 5
∫4( ) ( )
( ) 5
∫4( ( ) ) ( )
( ) 5
∫( ( ))
∫:√ ( )
( ) ;
⏞
√ ∫
√
859. Prove without softs:
∫ (( ) ( )
)
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Daniel Sitaru-Romania
[
] [
]
√
∫ (( ) ( )
)
⏞
www.ssmrmh.ro
74
∫ (( √ )
( √ )
* ⏞
∫ 4√( √ )
5
⏞
∫ 4√( √ )
5
∫ 4√(√ ) 5
∫ (√ )
√ √
(
*
√
√
860. If then:
:∫
;:∫
; :∫
;:∫
;
Proposed by Daniel Sitaru-Romania
Solution 1 by proposer
:∫
;:∫
; :∫( )
;:∫(
)
;
⏞
:∫( )
;:∫( )
; :∫
;:∫
;
Equality holds for .
Solution 2 by Sergio Esteban-Argentina
Let ( ) , - since is positive and nonincreasing, it follows that
∫<∫ ( )
( )( (
) ( )
) =
we would just have to prove that:
(
) ( )
www.ssmrmh.ro
75
The function ( ) is a convex on , so according to Karamata’s inequality it
suffices to prove that without loss of generality such that ( )
majorizes ( )
i) and ( ) ( )
ii) ( ) ( ) true! Expanding , we obtain
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
∫ ( )
the proof is now complete.
861. Let , - be continuos and increasing. Prove that:
∫∫∫4 ( ) ( ) ( )
5 ( ) ∫ ( )
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Daniel Sitaru-Romania
Let , -. WLOG: .
increasing ( ) ( ) ( ) ( ) ( ) ( )
By Cebyshev’s inequality: ( ) ( ) ( )
( )( ( ) ( ) ( ))
( ) ( ) ( )
( ( ) ( ) ( ))
∫∫∫4 ( ) ( ) ( )
5
∫∫∫4
( ( ) ( ) ( ))5
www.ssmrmh.ro
76
∫∫∫ ( ) ( ) ∫ ( )
862. If ( ) – continuous, then:
∫∫
(
(
( ( ))( ( ))
( ( ) ( )
*
)
)
( )∫ ( )
:∫ ( )
;
Proposed by Daniel Sitaru – Romania
Solution 1 by Adrian Popa-Romania
( ( ))( ( ))
4 ( ) ( )
5
4 ( ) ( )
5
( ( ))( ( ))
( ( ) ( )
*
( ( ))( ( ))
( ( ) ( )
*
∫ ∫ ( ( ))( ( ))
. ( ) ( )
/
(1)
:∫( ( ) )
;
∫ ( )
∫
|
∫ ( )
( )∫ ( )
( )∫ ( )
.∫ ( )
/
(2)
From (1) and (2) ∫ ∫ ( ( ))( ( ))
. ( ) ( )
/
( )∫ ( )
.∫ ( )
/
Solution 2 by Soumitra Mandal-Chandar Nagore-India
Let ( ) ( ) for all then ( )
( )
( )
hence is convex function ( ) ( ( )) ( ) ( ( ))
4 ( ) ( )
5
4 ( ) ( )
5
( ) ( )
4 ( ) ( )
5
( ( ))( ( ))
4
( ) ( )
5
www.ssmrmh.ro
77
( ) ( ) ( ) ( )
( ( ))( ( ))
( ( ) ( )
*
∫∫ ( )
∫∫ ( )
:∫ ( )
;:∫ ( )
;
∫∫ ( ( ))( ( ))
( ( ) ( )
*
( )∫ ( )
:∫ ( )
;
∫∫ ( ( ))( ( ))
( ( ) ( )
*
863. Prove:
∫√
( )
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Soumitra Mandal-Chandar Nagore-India
∫√ ( )
∫√
(
*
∫√
(
*
<
√ ( )=
<
√ ( )=
. Again,
√ ( ) for all
Let ( ) for all
( ) ( ) ( ) ( )
for ( )
( )
www.ssmrmh.ro
78
.
/ √ hence has local maximum at
So,
( )
√
√
√ √
√ √
( )
√ ( )
√ √
√ ∫
∫√ ( )
√ √ ∫√ ( )
√ ∫ √
( )
√ √
. Hence proved
864. , - ( ) – continuous. Find: such that:
∫∫∫.( ( ) ( )) ( )( ( ) ( ))
( )( ( ) ( ))
( )/
∫ ( )
Proposed by Daniel Sitaru – Romania
Solution by Tran Hong-Dong Thap-Vietnam
Let ( ) ( ) ( )
We choose such that . Using Jensen’s Inequality:
( ) ( ) ( ) ( , -)
( ) ( ) ( ) ( , -) ( )
( )
( )
∫∫∫( ( ) ( )) ( )( ( ) ( ))
( )( ( ) ( ))
( )
∫∫∫, ( ) ( ) ( )-
<>∫ ( )
∫
∫
? >∫ ( )
∫
∫
? >∫ ( )
∫
∫
?=
www.ssmrmh.ro
79
<∫ ( )
∫ ( )
∫ ( )
=
∫ ( )
∫ ( )
865. If * + then:
( ) ∫(∏ ( )
+
∏∫ ( )
Proposed by Daniel Sitaru – Romania
Solution by Avishek Mitra-West Bengal-India
( )
√ ∫
( )
√
[for all ]
( ) in , - * +
Also, we know for ( ) ( ) for any in the interval
, - * + if ( ) ( ) , -
( ) ( ) is monotonically increasing in , - * +
also ∫ ( )
∫ ( )
[ ( )
√ ]
[ ( ) ( )
√ ]
( ) ( ) is the integrable between , - * +
( ) ∫ ( )
( ) ( )
∫ ( )
∫ ( )
∫ ( )
[Let us denote ( ) ( ) ( ) ( ) ( ) ( )]
( ) ∫ ( )
( ) ( ) ∫ ( )
∫ ( )
∫ ( )
www.ssmrmh.ro
80
( ) ∫∏ ( )
∏∫ ( )
866. If , - ( ) – continuous then:
:∫√ ( )
;
∫ ( )
:∫ √ ( )
;
Proposed by Daniel Sitaru – Romania
Solution by Tran Hong-Dong Thap-Vietnam
Using Cauchy – Schwarz inequality:
:∫√ ( )
;
:∫ √ ( )
;
∫
∫ ( )
∫ ( )
:∫√ ( )
;
:∫ √ ( )
√ ( )
√ ( )
;
∫ 0√ ( )
1
∫ 0√ ( )
1
∫ 0√ ( )
1
:∫ √ ( )
;
:∫√ ( )
;
∫ ( )
:∫ √ ( )
;
867. If then:
∫ (
)
∫
:∫
;
Proposed by Daniel Sitaru – Romania
Solution by Sergio Esteban-Argentina
We can notice that: ( ) ( ) and ( )
www.ssmrmh.ro
81
as ( )
( ) is strictly convex. By integral form of Jensen’s inequality
∫
:∫ ( )
; ∫
:∫
;
868. If
then:
( ) ∫
( )
Proposed by Daniel Sitaru – Romania
Solution by Florentin Vișescu-Romania
.
/ ( )
( )
( )
( )
( ) .
/
( )
( )
( ) ( )
( ) ( )
( ) ( )
( )
( )
. So, has a root in .
/
( )
( )
. So ( ) and has a root in
.
/
www.ssmrmh.ro
82
( )
( )
So, ( ) and has a root in ( ). It doesn’t matter where is towards
( )
( )
So, ( ) .
/. Then:
∫
( )
( )
( ) ∫
( )
869. If , - has a continuous second derivative and ( ) ( )
then:
( ( )) ∫( ( ))
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Daniel Sitaru-Romania
:∫( ( ))
;:∫
; ⏞
:∫ ( )
;
:∫( ( ))
;
: ( ) ( ) ∫ ( )
;
( ( ) : ( ) ( ) ∫ ( )
;,
: ( ) ( ) ∫ ( )
;
( ( ) ( ) ( ) ( ( ) ( ))) ( ( ))
www.ssmrmh.ro
83
:∫( ( ))
; ( ( ))
870. If then:
∫∫
√ ∫∫
√
(
*
Proposed by Daniel Sitaru-Romania
Solution by Avishek Mitra-West Bengal-India
√ √
( ) ( )
√ ( )
( )
( )
( ) ( )
√ √
∫∫
( )
√ ∫∫
√
( )
∫∫(
*
∫∫
∫∫
6
7 , -
6
7 , -
( )( ) (
*( )
(
*( )
( )
871. If
then:
www.ssmrmh.ro
84
∫∫( ( )
( )*
√ ( )
Proposed by Daniel Sitaru-Romania
Solution by Tran Hong-Dong Thap-Vietnam
( ) √ ( )
( ) √ ( ) √
4
√
5 ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) √
Hence f .
/
√
√
( )
( ) , ( )-
( ) √
. .
/ ( ) /. Hence
∫∫( ( )
( )*
∫∫4 √
5
√ ( )
Proved. Equality for a=b.
872. Prove without softs:
∫( ( ))
Proposed by Jalil Hajimir-Toronto-Canada
Solution by Tran Hong-Dong Thap-Vietnam
( ) ( ) ( )
( )
( ) ⏞
We compute: .
/ .
/ ( )
www.ssmrmh.ro
85
( ) ( ) ( )
( ) ( ) .
/
( )
∫
∫ ( )
∫
∫ ( )
873. If then:
∫
( )
( )( )
Proposed by Daniel Sitaru-Romania
Solution 1 by Sujit Bhowmick-North Bengal-India
( )
∫
( )
( )( )
∫
( ) ( )
⁄
( )
:
( ) ( ) ( ) ( ( ))
;
( ) ( ( ))
∫
||
( )
( )
( )
( )
www.ssmrmh.ro
86
Solution 2 by Ravi Prakash-New Delhi-India
( )
( )( )
( ( ))
( )
4 ( ( ))( ) ( ( ))
( )5
( )
4 ( )
5
∫ ( )
( )( )
∫
4 ( )
5
( )
| ( )
⁄
∫
( )
( )( )
( )
( )
874. If , - continuous then:
∫∫∫ √( ( ) ( ) ( ) ( ) ( ) ( ))
∫ ( )
Proposed by Daniel Sitaru-Romania
Solution by Florentin Vişescu-Romania
( )( )
( ) ( ) ( )
We must show: ( ) ( ) ( )
( ) ( ( ) ( ) ) (( ) ( ))
( ) ( )
( ) ( )
( ) ( ) ( )
www.ssmrmh.ro
87
( ) ( ) true from:
⇒ {
( ) ( ) ( ). So,
( ) ( )
∫∫∫ √( ( ) ( ) ( ) ( ) ( ) ( ))
∫∫∫( ( ) ( ) ( ))
∫ ( )
875. Evaluate the following sum:
∑(
( )
( )
( )
( ) *
Proposed by Prem Kumar-India
Solution 1 by Avishek Mitra-West Bengal-India
∑
( )
∑
. /
(
*
( )
∑
( )
∑
( )
( )
( ) ( )
∑
( )
∑
. /
(
*
( )
∑
( )
∑
( )
www.ssmrmh.ro
88
Solution 2 by Nelson Javier Villaherrera Lopez-El Salvador
∑[
( )
( )
( )
( ) ]
∑[
( )
( )
( )
( ) ]
∑{
, ( ) -
, ( ) -
, ( ) -
( ) }
∑[
( )
( )
( )
]
∑[
( )
( ) ]
[∑
( )
∑
]
∑( )
( )
[∑
∑
( )
∑
]
(∑
∑
+
∑
( )
( ) ∑
876. Find:
:∑∑∑∑
( )( )( )( )
;
Proposed by Daniel Sitaru – Romania
Solution 1 by Mokhtar Khassani-Mostaganem-Algerie
Let ∑ ∑ ∑ ∑
( )( )( )( )
∑∑∑
( )( )( )
www.ssmrmh.ro
89
∑∑
( ) ( )
∑∑
( )( )
∑
( )
We will prove that:
In first note that there are of permutations of
( )
( )
( )
( ) let’s
discuss the 04 cases:
Case 01: if all 4! combinations are then contained with in
Case 02: if and are distinct we need 6 of because some possibilities are
contained in
Case 03: if we need 3 of because some possibilities are contained in
and
Case 03: if we need 8 of because some possibilities are contained in
and
Case 03: this appears exactly once in every sum, which thus us 12
copies of
Since: ∑
( )
∑
( )
∑
∑
( )
( ) ( )
( ( ) )
∑
( )
∑(
( ) *
∑(
( ) *
∑
( )
∑(
( ) *
∑(
( ) *
∑(
( ) *
( ) ( ) ( ) ( )
Solution 2 by Kartick Chandra Betal-India
∑∑∑∑
( )( )( )( )
www.ssmrmh.ro
90
∑∑∑∑
( )( )( )
{
}
∑∑∑
( )( )( )
(
*
∑∑
( )( )
. ( ) /
∑∑
( )
{ ( )
( )
} ∑∑
( )
{
}
∑
( )
∑{( ( )
( )
+
( ) }
∑
( )
(
*
∑
( )
< ( )
∑{
( )}
2 ( ) 3= ∑
( )
∑
( )
[ ( )
( ) ] ( )
∑
( )
[
( )
( ) ]
∑
( )
∑ ( )
( )
∑ ( )
( )
, * ( ) +- ∑{
( ) }
( ) ∑{
( )
( )
}
∑{
( )
( )
}
∑ ( )
( )
∑[{ ( )
( )
}
( ) ]
∑ ( )
www.ssmrmh.ro
91
{ ∑(
*
∑
( )
} ∑ ( )
∑ ( )
* ( ) + ∑ ( )
877. Find:
:∫4
5
;
Proposed by Daniel Sitaru – Romania
Solution 1 by Naren Bhandari-Bajura-Nepal
Here
( ) ∫
∫(∑
4
5
+
∑
( )
and hence we have the limit: ( ( )) . .∑
( ) //
(
( ∑
( )
++
:
(( ) ∑
( )
+
;
(
(
** ( ) √
Solution 2 by Abdul Hafeez Ayinde-Nigeria
Let ∫
and ( )
∫
∑
∫
www.ssmrmh.ro
92
(∑
( )
+
(∑
( )
+
( ∑
( )
+
∑( )
(∑
( )
+
{∑( )
(
*
}
8
(
*
9
{
( )}
{
( )}
878. Find:
(
:∑√
∑
√
∑(√ √ )
√
;,
Proposed by Daniel Sitaru – Romania
Solution by Remus Florin Stanca – Romania
∑√
∑
√
∑(
√ √ *
√ (√ √ )
√ (√ √ )
√ √ √ (
√
√ * √ (
√
√ * √
√
∑ √
∑ √
www.ssmrmh.ro
93
∑ :√
√
;
∑ 4 √
√ 5
∑(√ √ )
√
( )
∑
(√ √ )
√
∑(√ √ )
√
∑√
∑
√
∑(√ √ )
√
∑√
∑
√
∑(√ √ )
√
∑(√ √ )
√
∑(√ √ )
√
:∑√
∑
√
∑(√ √ )
√
;
879. Find:
(
∫ 4
5
√ ∑ .
/
∫ 4
5
√ ∑ .
/
)
Proposed by Daniel Sitaru – Romania
Solution 1 by Remus Florin Stanca-Romania
Let’s compute ∫
and ∫
∫
∫
∫( )( )
∫( )( )
∫( )
So,
www.ssmrmh.ro
94
“ ”
√ ∑
(
√ ∑
* (
√ ∑
*
√ ∑
.
√ ∑
/ .
√ ∑
/ (1)
Let’s compute
√ ∑
∑
(√ )
∑
(√ )
(√ )
( ) ( ) ( )
( )
(√ )
4
√ 5
(
.
/
√ )
( )
.
/ .
/
Solution 2 by Ali Jaffal-Lebanon
We have: ∑
∑
∑
. So,
∑
( )
( )
( )
Where ( )
So,
√ ∑
√
√
( )
( )
Then,
√ ∑
tends to
when
and
√ ∑
√
( )
Let
then
www.ssmrmh.ro
95
∫
∫
∫
∫
∫
∫
∫( )( )
∫
∫( )
Let ∫
and ∫ ( )
∫
We have
∫
∫( )( )
∫( )
So,
then
Therefore
880. Find:
(. /
:
∑:
. /
.
/;
;,
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
96
Solution 1 by Mokhtar Khassani-Mostaganem-Algerie
Let ∑. /
.
/
∑ .
/
( ) ( )
( )
( )∑. /
( ) ( )
( )
( )∑. /
∫
( )
( )∫ ( )
∑. /
( )
( )∫ ( )
( ) ( ) ( )
( )
( )
( ) ( )( )
( )
( ) ( )( )
( ) ( )( )
Now: (. /
*
( )
( ) ( )( )
Solution 2 by Ali Jaffal-Lebanon
Let ( )
∑
∑
( ) ( )
( )
∑
( ) ( )
( ) ( )
∑
∫
( )
∫∑ ( )
( )
∫( )
( )
∫ ( )
www.ssmrmh.ro
97
( ) ∫ ( )
Let ∫ ( )
( ) ( )
( )
( ) ( )
( )
( )
So, . Then
881.
(
(
∑
,
(
∑
+
)
(
(
∑
+
(
∑
,
)
Find:
(∏
+
Proposed by Daniel Sitaru – Romania
Solution by Ali Jaffal-Lebanon
Let
∑
∑
and √∏
By GM-AM inequality: then: ( ) ( )
( )
We have ( ) ( )
. So,
( ) ( ) (*)
( ) ( ) (**)
where ( ) ( )
by (*) and (**) we have: ( ) ( )
So: ( ) then ( )
And also, by (*) and (**) we have:
www.ssmrmh.ro
98
( ) ( )( ) ( ) ( ) ( )
then ( ) ( ) ( ) ( ) ( )
( )
( )
So, then ( ) but ( )
then
( ) . we have ( )
( ) ( )
then by Sandwich theorem: ( ) ( )
( ) then
( ) and ∏
882. If , different in pairs, then:
( ) ( ) ( ) :∏:√
√
;
;
( ) ∑(
4
5
+
Proposed by Daniel Sitaru – Romania
Solution 1 by Soumitra Mandal-Chandar Nagore-India
( ) ∑
4
5
∑
4
5
4
5
(
*
4
√ 5 :√
√
;
∑ ( )
∑ :√
√
;
.
/
∑ :√
√
;
( )
www.ssmrmh.ro
99
∑ ( )
∑ :√
√
;
Solution 2 by Adrian Popa-Romania
( )
?
|∫
( ) ∑
( )
( )
∑
4(
*
5
4
( )
( ) 5
( )
( )
4
√ 5
√ 4
√
√ √
√ 5
We must prove that: ∑ (√
√
√
√ * ∑ (√
√
*
⏟
(
:√
√
;, (
:√
√
;,
:√
√
;
(√
√
* (√
√
*
(True) because (1)
Similarly: (
(√
√
*+ (√
√
* (2) and
(
4√
√
5+ 4√
√
5 (3)
(1) (2) (3) ( ) ( ) ( ) (∏(√
√
**
Solution 3 by Remus Florin Stanca-Romania
( ) ∑ (
.
/
* , let
, we also know that
www.ssmrmh.ro
100
because , so (1)
( ) ∑4
5
∑4∫4
5
5
∑(∫ *
∫(∑
+ ∫
4 ( )
5 ∫
∫
(| |)
( )
4
5
.
( ) / (
√
* because
( ) 4 √
5 ( ) ( ) ( )
∑ 4 √
5
(
∏ 4√
√ 5
)
( ) (∏ (√
√
* * (2)
we know that ( ) ( )
:∏:√
√
;
; ( ) :∏:√
√
;
; ( )
( ) ( ) ( ) :∏:√
√
;
;
883. Find:
(
(( )
( ) ( )
*
( ) ( ) ( ),
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
101
Solution 1 by Ali Jaffal-Lebanon
We have by GM-AM inequality:
4( )
( ) ( )
5
.√( ) ( ) ( )
/
(
*
, ( )( ) ( )-
.
/ , ( ) ( ) ( )-
(*)
And ( ) ( ) .
/ , ( ) ( )-
(**)
Let
(( ) ( )
( ) )
( ) ( ) ( )
( ) 4
( ) ( )
5
( ) , ( ) ( ) ( )-
( ) ( )
.
/
.
/ . So,
but then so,
Solution 2 by Remus Florin Stanca-Romania
(
( (
( ) ( )
**
( ( ))
( ( ))
( ) ( )
)
4 4( )
( )
55
( ) ( ), let
4 4( )
( )
55
( ) ( )
:
4( )
( )
5
4( )
( )
5
;
4
( ) ( )
5
( ) (1)
(( )
( )
*
( )
(( )
( )
*
( )
www.ssmrmh.ro
102
(( )
( )
( ) ( )
*
.
/
(( )
( )
( ) ( )
*
.
/
(( )
( )
( ) ( )
*
( ) ( )
( ) ( )
(( )
( ) ) ( )(( )
( ) )
( )(( ) ( ) )
( )( )
(
( )( ) *
( )( )
( ) ( )
( )
(( )
( ) ) ( )(( )
( ) )
( )(( ) ( ) )
( ) ( )
( ) ( )
( ) ( )
( )
( ) ( )
( ) ( ( )
( ) ( )
)
( ( ) (
( ( )
( )* *
( ) ,
4 ( )
( ) ( )
( )5
(4 ( )
5
+
( (
4 ( )
( )5
++
4 ( )
( ) ( )
( )5
www.ssmrmh.ro
103
( )
4 ( )
( ) ( )
( )5
( )( ) ( )
( )
( ) ( ) ( )
( ) .
/
( ) .
( )
( )/
( )
(2)
4( )
( )
( ) ( )
5
(( )
( )
*
4( )
( )
( ) ( )
5 ( )
(( )
( )
*
( )
( )
4( )
( )
( ) ( )
5
.
/ .
/
( ) .
/
( )
(
( )( )
*
( )( )
( )( )
(3) ( ) ( ) ( )
Because : 4
( ) ( )
5
4( )
( )
5
;
:( )
( )
( ) ( )
;
:( )
( )
;
884. Find:
( (∑4 . /5
+ ∑
+
Proposed by Florică Anastase – Romania
Solution 1 by Marian Ursărescu – Romania
Let ∑
. We have:
www.ssmrmh.ro
104
∑
∑
∑
∑
∑
(1)
But ∑
(
*
(2)
From (1) (2)
and (geometric progress). We must find:
( ∑
+
( ∑
+
∑
(3)
Now, using Cesaro-Stolz for
:
∑
a) strictly decreasing
b)
∑
∫
c)
( )
( )( )
( )
( )
( )( )
(4)
From (3) (4)
Solution 2 by Mokhtar Khassani-Mostaganem-Algerie
Let:
www.ssmrmh.ro
105
∑. /
∑.
/
∑.
/
.
/
∑.
/
⏞
(
.
/ ∑.
/
⏞
)
.
/ . /
.
/ 4.
/ .
/5
.
/ .
/
. Also:
∑
∑
(∑
∑
+ ( )
Now: ( ( ) ) ( )
.
/
( ) (
*
( )
(answer)
885. Dedicated to teacher Mehmet Șahin
( ) ( ) ( ) ( )
Find:
:
∑ 4 (
* (
*5
;
Proposed by Daniel Sitaru – Romania
www.ssmrmh.ro
106
Solution by Florentin Vișescu – Romania
( ) ( ) ( ) ( )
( ( ) ( ) ) , ( ) ( )-
( ) ( ) ( ) ( )
( ) not verify; ( )
( ) ( )
( )
( )
( ( )
( )* ( ( )
( )*
( )
( ) ( ) ( ( ) ) ( ) ( ) ( )
( )
⟨
( ) [
]
( )
( ) [
]
( )
( ) ( )
:∫
;
. | /
∑ 4 (
* (
*5
.∑ .
/
/
∑ . /
: ∑ 4 (
* (
*5
;
[
∑ (
*
]
∑ (
*
:∫ ( )
;
∫ ( )
:∫ ( )
;
886.
www.ssmrmh.ro
107
∫ .
/
( )
Proposed by Jalil Hajimir-Canada
Solution 1 by Ali Jaffal-Lebanon
Let ∫ .
/
( )
∫ .
/
( )
∫ .
/
( )
We have | .
/
( )|
( )
Since | | | |
|∫ .
/
( )
| ∫
∫
then ∫ .
/
( )
we have:
|∫ .
/
( )
| ∫
∫
∫
(
√ *
√ [
(
√ *]
√
then ∫ .
/
( )
therefore
Solution 2 by Naren Bhandari-Bajura-Nepal
Here ∫ .
/
( )
Consider ( ) .
/
( ). We note that at
( ) and we have ( )
( )
www.ssmrmh.ro
108
( )
( .
/
*
( )
( )
| ( )| ( ) . Then, by dominating convergence, we have:
∫ ( )
∫ ( )
∫
887. Find:
∑4
(∑ )(∑
)
5
Proposed by Daniel Sitaru – Romania
Solution 1 by Adrian Popa – Romania
∑4
(∑ )(∑
)
5
∑(
*
We know that
Solution 2 by Naren Bhandari-Bajura-Nepal
We write the sum without notation
( )
( )( )
( )( )
Now, by partial fraction decomposition we have:
www.ssmrmh.ro
109
(
* (
*
Since is a telescoping series with nth partial sum as
6
∑
7
Solution 3 by Naren Bhandari-Bajura-Nepal
Here ∑ (
∑ ∑
*
∑ ( ( )
∑ ( ( ) ) ∑ ( ( ) )
*
( )
( ( ) )( ( ) ( ) )
( )
( ( ) ( ) )( ( ) ( ) ( ) )
( )
( ( ) ( ) ( ) )( ( ) ( ) ( ) )
( )
( ) ( )
( ) ( )
( ) ( ) ( )
We have telescoping sum give us
( ) (∑(
*
+
888. Find:
(
( )∑4 4
55
+
Proposed by Daniel Sitaru – Romania
Solution 1 by Ali Jaffal-Lebanon
Let ( )
, ,
www.ssmrmh.ro
110
( )
( ) then is increasing on , ,
Let and , so, ( ) ( ), then
then
then
( )
( ) ( )
( ) (*) but
( )
So,
(**). Then by (*); (**) we have:
( )
( ) ( )
( )
but is increasing, so: .
/ .
( )
( )/ (.
/
*
So,
( )∑ .
/
( )∑ ( ) (.
/
*
Where
( )∑ .
( )
( )/
We have
∑
(.
/
* ∫
( )
Let ( )
then ∫
( )
∫
, ( )-
We have
( )∑ (.
/
*
and
( )∑ ( ) (.
/
*
( )∑
4(
*
5
Then
Solution 2 by Remus Florin Stanca-Romania
We know that ∑ ( ) ( ) ∫ ( )
continuous and
, - and || || where || ||
( ), and and
, let ( )
( )
( )( )
( ) ( )
( )
( )
( ) ( )
( )
and ( )
|| || and ( )
( ) and ( ) ( ) continuous
www.ssmrmh.ro
111
∑ ( )
( )
4
5 ∫ ( )
∑
( )
.
/
∫ ( )
∑
( )
.
/
(1)
( ) increasing .
/ .
/
( )∑ .
/
( )∑
and
( )∑ 4
5
( )
( )
.
( )∑ . .
//
/
∫ ( )
(2)
∫ ( ) ∫ ( ) ( ) ∫
( )
∫
( )
( )
( )
( )
(
( )∑4 4
55
+
4
( )
5
( )
( )
( )
Solution 3 by Marian Ursărescu-Romania
(another approach) Let , - ( ) Riemann integrability
Let .
( )
( )
( )
( )
( ) /
|| ||
( )
( ( ) ( )( ))
( )
( )
|| || . Let ,
-, so that:
( )
( ) (
) ∑ 4 ( )
( )5
( )
( )∑
www.ssmrmh.ro
112
( )∑
∫
∫
|
∫
( ) |
889. Find:
:
∫. .
//
;
Proposed by Vasile Mircea Popa-Romania
Solution 1 by Ali Jaffal-Lebanon
Let
∫ .
/
∫
∫ .
/
∫ .
/
∫ 4
5
then| |
∫ .
/
We have ( ) and then
so, .
/
then | |
∫
so | | then
Solution 2 by Naren Bhandari-Bajura-Nepal
We consider ( ) (
*
. We observe that at , the limit of ( )
respectively when . Limit of ( )
when . Hence by Dominated
Convergence Theorem, we have:
∫ ( )
∫
890. Find the closed form:
www.ssmrmh.ro
113
∑ ( )( ) ( )
( )( )( ) ( )
Proposed by Florică Anastase-Romania
Solution 1 by Kamel Benaicha-Algiers-Algerie
∑ ( )( ) ( )
( )( ) ( )
Consider the series ( ∑ ) ⁄ ( )( ) ( )
( )( ) ( ), so:
( ) ∑( )( ) ( )
( )( ) ( )
⁄
( )
( ) ( )∑ ( ) ( )
( )
( )
( ) ( )∑∫ ( )
( )
( ) ( )∫∑
( )
( )
( ) ( )∫ ( )
( )
( ) ( ) ( ) ( )
( )
( )
( ) (1)
Consider the function ( ) ( )
(- ,) and ( )
( )
So, for ( ) ( ) ( )
( )
Let
So, , using the result (1):
www.ssmrmh.ro
114
∑ ( )( ) ( )
( )( ) ( )
Solution 2 by Samir HajAli-Damascus-Syria
Let put: ( ) ∑ ∏
We have: ∏
( )
( ) ( )
( ) ( ) ( )
( )
( )
( ) ( )
( ) ( )
Where ( ) is Beta function. Now: ( )
( ) ∑ ( )
But: ( ) ∫
( ) .
Since: ∑ uniformly converge
( ) ( ) ∫∑
( ) ∫ ( )
∫ ( )
( )( ) ( )
So, ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( )
Now, for:
Where:
We find: ∑ ( ) ( )
( )( ) ( )
Solution 3 by Mokhtar Khassani-Mostaganem-Algerie
∑ ( )( ) ( )
( )( ) ( )
( )
( ) ( )∑
( )
( )
( ) ( )∫
( )
∑
www.ssmrmh.ro
115
( )
( ) ( )∫
( )
( )
( ) ( ) ( ) ( )
( )
Solution 4 by Mokhtar Khassani-Mostaganem-Algerie
∑ ( )( ) ( )
( )( ) ( )
∑ ( )( ) ( )( ) ( )( ) ( )( )
( )( ) ( )
∑4
( )( ) ( )
( )( ) ( ) ( )( ) ( )
( )( ) ( )5
4
( )( ) ( )
( )( ) ( )5
891. ( ) .∑ .
/
/
Find:
:
:
;
;
Proposed by Florică Anastase – Romania
Solution by Mokhtar Khassani-Mostaganem-Algerie
Since: ∏ .
/
( .
/*
∑ . .
//
( ( .
/*
+
Now, by differentiable both side with respect to we get:
∑
( (
**
(
( . /*
,
www.ssmrmh.ro
116
∑ (
*
4 .
/5
∑ .
/
( ) (( ). /*
(( ) )
Let:
4
.
/5
4
(( ) ) 4
.
/55
(
:
.
/;
(( ) )
)
(
( ) ( )
( ) (( ) )* 4
.
/
5. Now:
: .
/
; (
(
**
√
892. Find:
4√( ) √( )
√( ) √
5
Proposed by Daniel Sitaru – Romania
Solution by Naren Bhandari-Bajura-Nepal
4√( ) √( )
√( ) √
5
:( ) 4√ ( ) (
*
5
( )
( ) (√ .
/
*
;
√
:(
*
√ ( )
( ) (
*
√
;
www.ssmrmh.ro
117
√
:( ) (
( )*
( )
( ) (
*
;
√
( 4
( ) (
( ) *5 4
(
*5+
√
(
* √
893. If: ( )
( )
( ) ( )
Find: ( √
*
Proposed by Florică Anastase – Romania
Solution 1 by proposer
( ) .
/ .
/ .
/ ( ) .
/
∫( ) ( ) |
∫( )
∫( )( ) ∫( ) ∫( )
( )
( )
√
√ ( )
( )
⏞
( )(( ) )
( ) ( )
( )
4 √
5
√
Solution 2 by Naren Bhandari-Bajura-Nepal
www.ssmrmh.ro
118
We write: ∑ ( ) .
/
∑ ( )
. /∫
∫(∑( )
. / +
∫( )
∫( )
< ( )
. / (
*=
[
√
( ) √
]
( ) ( )
( )
Now, we solve for
4 √
5
4
√
5 :
√( )
√( )
( )
;
4
√
√
5 4
√
5
Note for all inductively we can show that:
Set and hence by Squeeze theorem
894. Find:
(
√ ∑ (
√ *
+
Proposed by Vasile Mircea Popa-Romania
Solution 1 by Florentin Vișescu-Romania
We know that:
√
(
√ *
√
√
(
√ *
(
√ *
www.ssmrmh.ro
119
∑
√
∑
∑
√
√ ∑
∑
√ ∑
( )
√
( )
∑
√
( )
√
( )
√ ( )
( )
⏟
√ ( )
⏟
√ ∑
√
( )
⏟
( ) grade 6 polynomial in .
Solution 2 by Naren Bhandari-Bajura-Nepal
(
√ ∑ (
√ *
+
∑(
√ (
√ **
∑4
√ 4
√
√ 5 5
∑
∑ ( )
( )
( )
( )
We note that:
( )
∑(∑ ( )
( ) √
+
∑
( )
4
( )
( ) √( )
√
5
∑ ( )
( )
( )
895. Find:
www.ssmrmh.ro
120
( ) ∏4
( ) 5
Proposed by Florică Anastase – Romania
Solution by Kamel Benaicha-Algiers-Algerie
( ) ∏ . ( )
( ( )) /
. Let ( ) ( ( ))
( ) ∏
( )
( ) ( )
( )
If then: ( ) . If then ( )
If then ( ) ( )
( ). Let ( )
( )
then ( )
( )
So, for: and for: and
so: ( ) ( ) ( ) and ( ) ( ( ))
we know: ( ) . For
( ( ) ( ) ( )
( ) ( ) ( ) *
for: ( ) ( ) so: ( ) , then ( )
( ( ))
( )
( ) ( ) ( ( )) ( ) ( ( )) . So: ( )
∏4 ( )
( ( )) 5
896. Evaluate the limit:
:∫( ( ) ( ))
;
Proposed by Srinivasa Raghava-AIRMC-India
Solution 1 by Kamel Benaicha-Algiers-Algerie
(∫ ( ( ) ( ))
)
. Let ( ) ∫ ( ( ) ( ))
www.ssmrmh.ro
121
∫
( ( ) ( ))
. /∫
. /∫ . /
(
*
- ,
. /
(
4 .
/5
. /
)
. /:
. /
. /;
. .
/ .
//
.
/ .
/
. So: . .
/ .
// .
/
[ (
* . .
/ .
//]
4 4 .
/5 . .
/ .
//5
. ( ( )) . .
/ .
///
( ( ( ))
( ( ))*
Then, we have: ( ( ))
( ) and,
( ( ))
So:
(∫ ( ( ) ( ))
)
( denote Euler – Mascheroni constant)
Solution 2 by Mokhtar Khassani-Mostaganem-Algerie
Since: ∫ ( )
.
/ .
/
√ and ∫ ( )
.
/ .
/
√ where: and
we obtain:
:∫( ( ) ( ))
;
( . /
. .
/ .
//,
(
. . .
/ .
// ( ( ))/*
www.ssmrmh.ro
122
(
. / .
/
. / .
/
( ), .
/
897. Find:
: ( ∑( )
( ) ( )
+;
Proposed by Daniel Sitaru – Romania
Solution by Remus Florin Stanca – Romania
(∑( )
( ) ( )
∑
+
(∑( ) ( ) ( )
( ( ) ( ))( )
+
(∑ ( )
( ( ) ( ))( )
+
We know that ( )
( ( ) ( ))( )
( ( ) )( )
( ( ) )( )
∑ ( )
( ( ) ( ))( )
( )
( )( ( ) )
( )
( )( ( ) ) .∑
( )
( ) ( ) ∑
/ (*)
∑
∑
(1)
( )
continuous, let
|| ||
∑ ( ) ( ) ∫ ( )
where , - and
and and let
∑
∫
( ) ( )
www.ssmrmh.ro
123
( )
( )
∑( )
( ) ( )
( )
and and let
∑
∫
( ) ( )
∑
( )
( )
∑( )
( ) ( )
( )
( ( ) ∑( )
( ) ( )
+
( ) ∑( )
( ) ( )
(
( )
( ) ( )
( )
( ) ( )
( )
( ) ( ) )
( )
(
( )
( ) ( )
( )
( ) ( )
( )
( ) ( ) )
4 ( )
( ) ( )
5
4 ( )
( ) ( )5
4 ( )
( ) ( )5
898. Find ( – golden ratio):
www.ssmrmh.ro
124
:
√
√
√
;
Proposed by Daniel Sitaru – Romania
Solution by Ali Jaffal-Lebanon
We have by Bernoulli’s inequality:
( ) for all and then .
/
Let ∑ .
/
. So,
∑
but
then
∑
we know that ∑ .
/
.
/
and .
/
since
then therefore
899. Find:
(
√∑( ) .
/
)
Proposed by Daniel Sitaru – Romania
Solution 1 by Ali Jaffal-Lebanon
∑( )
. / .
/ ( ) .
/ ( ) .
/
( ) . / . ( )/
.
/
. / ( ) .
/ ( ) .
/ ( ) .
/
( ( )) .
/
www.ssmrmh.ro
125
0. / ( ) .
/ ( ) .
/ ( ) .
/ ( ) .
/1
6.
/ ,( ) ( ) ( ) -7
√∑( ) .
/
√.
/ ( )( )
Degree of .
/ ( )( )
is equal to
So: degree of
√.
/ ( )( )
is
4
√∑ ( ) .
/
5 4
5 (*)
But 4
5 4
( )
5 4
( )
5
So: ∑ ( ) .
/
Solution 2 by Khaled Abd Imouti-Damascus-Syria
We know that ∑
then for
Let
√∑ ( ) .
/
we have
√∑ ( )
√∑ ( )
. We know that:
∑ ( )
∑ ( )
∑
( )( )
Then
. ( )( )
/
But . ( )( )
/
(
( )( )
*
then
Solution 3 by Artan Ajredini-Presheva-Serbie
www.ssmrmh.ro
126
Let ∑ ( ) .
/ and ( ) (
) . /
Since ( ) we have that: . Now,
( )
. Hence, √ . Definitely,
√
Solution 4 by Ali Jaffal-Lebanon
Let
√∑ ( )
we know that ∑
then ∑
So, for all then
: ∑ ( )
;
: ∑ ( )
;
: ∑ ( )
;
:∑
;
:∑
;
but
Solution 5 by Naren Bhandari-Bajura-Nepal
Let be the limit and we write ∑ ( ) . /
. Now from
Bernoulli inequality we have: √
√
and hence we have
.
/ and by Stolz Cesaro theorem we have
( ) (∑( ) .
/
+
( ) (∑( ) .
/
+
( ) . /
( )
( ) . /
( )
and hence by Squeeze theorem we have .
www.ssmrmh.ro
127
900. ( ) .∑ ( )( ) ( )
( ) / * +
Find:
∑
( )
Proposed by Daniel Sitaru – Romania
Solution by Samir HajAli-Damascus-Syria
We have: ∑
( )
∑
( ) because the series converge uniformly when | |
∑ ( )
( ) (Similarly)
We find: ∑ ( )
( )
( )
( )
( )
Then: ∑ ( )( )
( ) | |
Therefore: ∑ ( )( )
( )
( )
Similarly step by step we find:
∑ ( )( ) ( )
( ) | |
Hence: ( ) ∑ ∏ ( )
( )
( )
(
*
. /
and: ∑
( ) ∑
∑