Some Elliptic Function Formulae

Some Elliptic Function FormulaeAuthor(s): Thomas CraigSource: American Journal of Mathematics, Vol. 5, No. 1 (1882), pp. 62-75Published by: The Johns Hopkins University PressStable URL: http://www.jstor.org/stable/2369534 .

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Sonte Ellptic Function Formulae.

BY THOMAS CRAIG, Johns Hopkins University.

On page 102 of Prof. Cayley's Treatise on Elliptic Functions occur the formulae for the differentiation of sn u, cn u, dn u, with respect to the modulus k; these are

dsnu_ k k dk =-*2 cn u dn focn udu+ V2sn

u cn2 u,

denu k dnucn2u du- sn2u cnu, dk = Y2 sn u c2

d dnu ks 12 k dk = t2 snu cnuj0 cn u du - sn2 u dn u.

The following three forms for-these differential coefficients may be obtained from these equations, or they may be obtained directly in a slightly different manner. The formulae may have been given before, but I do not remember to have seen them; they are, however, slufficiently interesting to make a note of them here. Starting with the relation

d sn u cn ub

-log dnu=6-k dn dqb dn ub

drop the factor - Z and differentiate this a second time; we have thus on reduction,

d snfuCn 2fen C2fU - sn z6+n

du dn s dn2n

Subtract this from cn2 u and we have d snu cnu _ cn2u

cn2qb-7 dun1- dqb dn'u dn 2b

or

2 d sn u cnu k'2sn2 n, dc - dn u du



CRAIG: Some Elliptic Function Frmulae. 63

and on integration

A ~2Ssn2 -.J0cn2 u du sn d en u 02 en22uu od t dnq But

fcn2 u du- u- sn 2u du =

Assume now =p am u and write

U ,V 1--k sn2 l

then du _ k sn2 cfdcp dk Jo (1- k2Sn2 o)-g

or du k_ (sn2u du dk J dn2 U

Equation A now becomes

A' /c/~~~du E- {(k1)+ (u),sn uen u At 7CYB2 dk = (K7,2) U

+ 0 (U)t 7? dn v,

Now d snuq dub d dk =-cnu dnu du

dk-~~~d dk usn = dn u du

dk- dk

d- dn-u =ksnu cn u du dk dk

substituting from A' for dwe have the following formulae:

d snu -1 E to_()__M

dk - cnu dnu ( -

u + O(U) + k2 snuncU,

d,Nj; eni ut 1 E 2\ 0l (Uf) k 2 B dka u dia u kkU + K(- sn u cn', dk KJ2 >uO(Uflksn12 n

ddnlu k E #(U) k sn2 Ucn2 U dk -2 SnCnU K -2U () + k2 dnu



64 CRAIG: Some Elliptic Function Formulae.

By means of the formulae

cn u - e 4K k (u( + K+ uK')*(u)

dn u 6-/k E) (u6 + K) *. 19 (u6), these may be written in the forms

dsnu - *e 1 O(u+K+iK')&(u+K) E J2"\w (u) dk -e4k'k/k- 02 (f) - K Y2 +

(U))

3* i & (u + iK') 62 (u+ k + iK') -e k4 kk 68(f)

d cnu - 4 0 &(u+K) (u +iK') E 2 + of () dk kk!Kkl 1 62(U) K J ((u))

3* ,5-2 (U + iK') (u + K + iK') k' 08 (u)

d dnu - t i (u+iK') )(u+K+iK')',,E \ ('(u)| dk e k'2U &2(u) 1 KK U

((u) 5 *2 02(u + iK') 92(u + K + iK')

k'+/kl 0 (u + K) 0- (uf)

where for brevity * is written-to denote- (K' -(2)u).

Substituting in equation A the value off en2 u du we have

de n2u du (E,d2\u + et 1(U)T - d nuen Jn2u du k2k'2 (K (u)) k2 dnun

We can easily find eight other relations of this character, i. e. we can get the group of nine integrals,

f 2 cn u dn2u du, du J du, Jn s2 qh d Jsnu2 q sn2u

sn2u r 1 Pdn2 U du, I du I du,

f en2 u . cn2 ' cn2 u

sn2 u cn2 r2 J du2d, diu, dn 2U



CRAIG: Some Elliptic Function Formulae. 65

These together with

sn2 2u du, Jcn2 u du, Jdn2 u da

make a series of twelve integrals of which the differential coefficients relatively

to u are the squares of the twelve quantities

snu, cnu, dnu

1 cn u dn u sn u sn u sn u

sn u 1 dn u

Ccnn cnn cn u

snn cnw 1 dn-u dn u' dnn

whose integrals with respect to u have been given by Mr. J. W. L. Glaisher in the B. A. Report for 1881. We easily find the following values

1 2 d cnu dnu - s= i U I - - 7

sn2' du sn n

en2 tb d cn u dn u = - dn2t6-- -- 8

sn2 u d?b sn u

dn2U d cn u dn u 2 = kP sn2 u - k7 -

sn2 q dqb sn ub

sn 2 t d2 sn u dnu

en2 u 1+k2 k du cn

1 +k2 k 2 1

++2k2 d

snn dntt Cj2 U 3+ 2 + ;bn t6. + -1 + 2

2 I + k +1+kdu cnnu

dn2 U , ?d sn ul anu = le,' ssn2 t + I

en 2n dun enu

sn2 it 1 d Sn uCenU

dn2n k2 c di kZ,2 d dn u

t-,n2 6 cl sn uz en qu *fl U~sn2u+ 4 n cdn 2U du dn u

1 1 k2 P k2 d 4 t Phlb

dn sn u-- n

The integrals of these quantities are dependent then upon the integral of un u. Taking no account of signs or numerical factors we see that in the first

group of thr ee equationts, which has as the denominlator on the left-hand side tho VOL. v



66 CRAIG: Some Elliptic Function Formnulae.

d2 quantity sn2 u, the exact differential is du2 log sn u; in the second group where the

corresponding denominator is cn2 u the exact differential is d2 log cn u; and also

in the third group with denominator dn2 u the differential is d2 log dnub. The die

complete set of integrals is du _ E ' (u)) d

J -n2 (1-- u- - - logsnu,

en2 u E 6/ (u)- d fsn2bdu = -K + ( d log snu,

fdn2u, {(K; 0-() d

dn2 1 E 6' (u))1 d (5 du = k2- losnu

cn2 u 1 +k2 K 6 (u) 1 + P2 duslgcu

fdu kl 2 +k1{ E 1 0'(u) 1+2k2 d

fcn dnu ] ( E>\ 6' (u1) 1 d u d KI -- ) - - log cnnu,

CIII2d.t I +t k/ K P6 + (u(6 I + k2 2 o dut6 fun snion1 du I Ef t- i 6l(u) d log 1n ue dutain 2U \K / t(u)e duandiogn

dn2u Jo2 k2k/s + I d

in fact it cl is oi us tha the wil lol dendupn, ntga

fn 2 du, ds e n e O,O (u) d2

dn2 U Jk2 i n iog 2 u n

fdI-d,u=;{- < + - -log dniu.

Of course the whole set of initegrals might be given in terms of the e- functions onily, by suibstitUting for Ilog snl it, log c-nu, ~-log dnu their

values, but it is not worth while to write them down. It is not difficult to obtain the integrals of the quan-tities

sn 2u cn 2u, cn 2u dn u, dn' u sn2u; in fact it is obvious that they will all depend upon the integral

fSn4 u dia,

and this can be sliowni to depend uponi the initegral of sn 2 u.




The above referelnce to Mr. Glaisher's paper was inade from mnemory, but I

find on referring to it that he remarks that the integral of 1 may be deduced

from that of sn2 u by the formula

d 2 log snu - 71 sn* -

he further shows in general how the integral of snn u may be obtained. I may perhaps be permitted to quote the last few lines of Mr. Glaislher's paper, which bear directly upon this question. In connection with the formulae of reduction

for fsnn u du, Mr. Glaisher says: " We have

1. due snntu= =u(ut-1)snun-2 2(1 + 72)snnU+ U(U + 1)k72 sn??2 . ...

and by means of this formula the integral of sna u may be reduced to depend 1 1

upon the integrals of sn u sn2 u, 1 , according as u is positive and uneven,

positive and even, negative and uneven, negative and even.

The integral of sn2 u involves the function Z (u)" (i. e. as above

"and the integral of may be deduced from that of sn2 u by the formula

2. 1~~~n 2. dzUffl log sn ub = 7iV s112 ou-

Corresponding to (1) and (2) there are eleven other pairs of formulae which involve the other, eleven functions in place of sn u, and differ from one another only in the k-coefficients. It can thus be shown, that the integrals of thle nAh

powers of the twelve functions are all finitely expressible in terms of elliptic functions if u is uneven, and in terms of elliptic functions, and of the Zeta fanction if u is even; and that the twelve formulae of reduction are similar in form and deducible from any one of them."

The italics are my own. Mr. Glaisher has given in his paper the values of the integrals of the group of twelve primary functions marked (C) in the above.

It may be as well to work out the integral of sn4u, as by means .of it we can find at once the integrals of

cn4 u, dn4 u, sn2 u, cn2u, cn2u dn2 u dn2 u sn2 u &c.




Applying Mr. Glaisher's general formula for the second differential coeffi- cient of sn z u with respect to u we have

d-2 sn u - 2 4 (1 + k) sn2 u + 67k2 sn4 ut

Prom this on integration and reduction follows

fsn4 u du 3k- {1+ (1+ )(1 _K ) - ( k +(snun+dn

This taken with the formula for fsn2 u du serves at once for the reduction

of cn4u du ,dn4u du, sn2u cn 2u du, &c.

For the integral of 2 2 it is not necessary to use the above formula, sn2uz cna ff

for multiplying this by sn2 t + cn2 u = 1 it reduces to the sum of the two integrals

Jocn2 qb + Josn2 M

the values of which are given above; on substituting these values we have

Jsn2u e6 n2u qb [ K( k) (6) (1+k2

k2 d d -1 +k2 - log cn - --log sn u cn u 1 du du

or

[ K (1 + k 2) O(uf) (1 +k k2

d I + 2A0

- w log sn u (cn u) i+k2

In like manner since, 1 1 k2

sn2u dn2u q sn2u + dn2u we have N

r du r 1-2k2 E 1-2k2 __(U)

Jo sn2uqb dn2uqb L- A kl2 K k12 (9(U)

k2 d d nu + -- log dn u - d-log s



CRAIG: Some Elliptic Fanction F["ormulae. 69

or

[1l 1-2k2 E 1 -

1-2k2 6(u) k/g2 K k'2 - (tt)

dk2 -d log sn u (dn it)

anid also since

cn2u dnu q (/2 n dn2 U we have

du n~~ E1~ ~ 1 1 +k4 a 6(U) .IOen2ndn 2nb= k l4(1+ k)2) L kK - -k4 1+k2 a((U)

1 1+2k2 d 1 d -

1 +k 2 d log enu + 1

-d log dnu

or

k4(1 +k2) [k(1 + ) k4 1 ++k2 (U) 1+ 2k2

1 d (en Ub)1 +k2

--2 -logf 2 U dnu

A further set not mentioned in Mr. Glaisher's paper, nor in Prof. Cayley's Treatise, are the following three integrals

f -sun fd f cn' en f dnn d Clllbudnn o'lubu 'J sn dn uc

or

ti f du du d- log sn u, d log cnu, d- log dn u

These are very easily found; take for example the first one d__u du:

write sn u = x, then the integral is

X1-2.1 k2x2

Make x A+ J3+ + D

1X2.1- k2x2 1-x 1 +x 1 -kx 1 + kx

and we find at once 1, B, D k' 1 -k k

A, B, C, D ~2k,~ 21'21'2'




Substituting these values-we have f xdx 1 1 -k2x2

J1 - x'.1 -kx220 log 1 x2

Replacing x by its value gives

sn u 1 dn u .Jocnn dunqzdu = d /2 log

and similarly rcnnb sun

d dtdu = log du

dsn udu= oe n dun ucun n t du=-log

vOsunzcunq sunq

From any two of these we obtain the third by addition. The similar integrals involving the squares of sn u, cn u, dn u will involve the e-functions. We have, viz:

f(cu2 q n q du f /2J(c~n2 hdIn )dU s n2K zl F I ~~-du -=)du

e? sn2u dn2 ./OU \ ksn2n -dn2nu e n 2u du_ = 1 + k2 )du

JOsu2cn2 ndu = + du

Substituting from the above values of these quantities, we find readily

cn2u dnu =- (1 + k2)(l -k2)2II+k+2(1 [) K ]+ 1+2k2 1

2 __ (_) 1 d r 1+k2 dk2] (1+k2)(1 k2)2 19(U) 1 -2dogu dn

(cn2du - E [P O(n)1 d snnu Josn2n dn2n Lu=- -K u+ O(n) + d log dn

I' dn2u 1d4 (2 + 5(E) + V)-2 (I k ) Jsn2nudn2n u = l+k 2 (4 5 +k2(+k K

2 (1 + k2) (n) -d log s un 9() du g 1k2-2k4

(cnnu) 1+k2

Of course, one could go on and extend these integrals indefinitely, but all subsequent forms would only depend upon those given in the above and in




Mr. Glaisher's paper. It may be worth while, however, to give the formulae for the integration of the quantities

snau cn Cu

cna u dnP u dna u snP u

in the several cases where a and ,3 are either both even, one even and the other odd. or both odd, i. e. four cases in all for each of the three products. Suppose a and i3 both even, say a = 2n, i3_ 2m:

Then sn2n6 cn2nu= sn12n sn' u)m

Again make a =2n + I, 3 2m2n: theni

sna2f+1 u en2m en u sn2+l un (1 sn2 u)m.

Again make a =2n, , =2m + 1: then

sn 2n cn 2m+1 u=cn2m+lu^6 (u1. cn2u)n.

For these three cases we have

h = 0 h-O

2. fsn2n u cn2m u dt= )hm(mln 1). .h( + l) fsn 2(h + n ) u d h=O hm

3. fsn~ 2n1 en2m ul du= E h )mf(m1) . (m- h+l)f l 2 (It?m?' ) it du

and for the case of both odd wre have

h=m h = m

4. fsn?n+ u cn2m+ u du = n ()hn m- (hn-i ) e. ( hn-h + l )

4fsn2 +n?uen cnu dd

Each term in (4) is of the form

fsn 2 ?1 q' cn t dci;



72 CRAIG : Some Elliptic Function Formulae.

this is

-1f2 ,snu d(dnu) -(- k(?1+)J(1 dn2'u)i d dnu

1 =0

t =j

k2~ ~~I (> + 1) .

(-) -- I

2 j +-1) dn21 + lu.6 - k2 (zi?1) (1! (21+1)dn 1

1=0

Giving j the value that it his in (4), viz.

j=2h+2n+ 1

we have for the integral in the general term of (4) I=2(h+rn )

fsn2(h??k)+ud) k(=+1 (_)1 (2h + 2n + 1) (2h + 2n) ... (2h + 2n-1+ 2) du21+ 1 k2(j+') ~~~1! (21 +1)

1=0

and so

4. Jsn2 + 1u cn2mn + Iu dat

h = m

_____ _h_ (m(m- h +1)

I = 2h 4- 2n + 1

(-1 (2h + 2n + 1) (2h + 2n). . . (2h + 2n-1 + 2) dn21 + t __ ~~~~~1! (21+1)dn1-u

1=0

by 'aid of the relations /d2 (1 + 2cn2 u) = dn2 u

1 (~dn 2U en2` __

T K' k12 )-CfU P2

wve readily find the integrals of cnau drl u

for the various odd and even values of a and T These are h=m

if. fenlu dn2mu du /d2mE (n-1) ..u (d7-h +1) e2fn2( h+f)u du, h =O

h=c2? m

2 e n2 i lt6 dn2 m t6 dJtu = 74,t en m, du, m1) - +1)2




h=n 3t PcXn6dnm+t dt=(_)n 1 _nn(n-1) ... (n-h + 1) 'h 3'. Jcnnu dn2m+l-u du - (-

2n- ~() Jdnh u dun, h=O ~ h.

A= 0

4'. fen2n +?u dn2m?+ u udu=

h=n I = 2h+2n+1

_72 (n-1)* (n-h + 1)T2h _)(2h + 2n.+ 1)....*(2h+ 2n-1+ 2)sn2l+1U A ~hI 1! (21 +1) n+u

h=0 1=0

Again since

sn2u=+u (1-dn2 u),

we have h=m

1". fdn2n u sn2mU du = 1m ()h (in h+l) I dn2(h+n)u dun, h=O

h=m I n' 1__ hm(m-1). ... (m-h +1)Pd'(+n ud 2".Jdn2n2+l mu s2mt du = k2 ()h h J0dn2(h+n liudu,

h=O

3 "itd2~ sn2m+u du=E )snl h 9s u du, h ='

h=m

4"/ jdn2n+1u sn2m+1u du = k-2 ()hm(m-1) *. (m-h+1)(_ J2h+2n+1)

h=O

I =2h + 2n + 1 \' (2h + 2n+ 1) . .. (2h + 2n- l+ 2) e21 21

1=0 11(21+1)C

The formnulae 2, 3, 2/' 3/' 2// 3" might have been given in forms similar to 4, 4', 4". For the integrations in these sets of equations we need only to employ the formulae

d' - sn~ n = nI dU2 sn $=n(n-) sn- tbn' (1 + kl) snn u-+ n (n + snn n+I u2

d 2 cn n = u (n -1) I'2 cn-IU + n2 (k2 _ k2) cn n U n (n + 1) k2 cnn+?u,

din2 dnnu =- n (n- 1)kY2 dn 2u + n2 (1 + 7'2) dnt _-n (n + 1) dnn+ 2u.

VOL. V.




I have tried to find reasonably simple and symmetrical forms for the expansions of the right-hand members of the above equations, by grouping together all terms containing only elliptic functions, and all containing the 0- function, but, though I think such forms should exist, I have n1ot been able to hit upon them. Another set of formulae might be found for the integrals of

1 1 1 snau cnn u cn?u u dng u dn 'u sng u sna Ucnnu cna a Udn u dna Usn qu

dnuY u sny u cn'u u

dnezu sn'y u cny u sna Ucngu' cna U dng u dnau sng u

but they are easily deducible from the preceding and following forms.

The formulae of integration for the several cases involved in the expression

fsna/ u cnn' u dnY'yM du

are very easily found; there are in all 8 cases, viz.

1. 2a, 2g3, 2y, 5. 2a+1, 23+1, 2y, 2. 2a.+1, 2g3, 2y, 6. 2a, 23 + 1, 2y + 1, 3. 2a, 2g3+1, 2y, 7. 2a+1, 23, 2y+1, 4. 2a,, 2g3 2y+1, 8. 2a+l, 2/3+1, 2y+1.

For brevity write

[N g_ g(g 1). * ;(g ]+1A

It is very easy now to find the integral of sn a'U cn 'u dny'u for the above 8 cases by aid of the relations

1-sn2u=cn2 u, 1 -k2sn2u=dn2u, 1 -cn2usn2u

-4 (1- >dt2d ) =cn2u, 2 1d2(1 + e2cn2 u) = dn2u, 1 (1 -dn2u) = sn2u

where as above 9 - -2 . The integrals are h=9 1=9

1. f(sn2au cn2 u dn2yudu= (_)h +1[g, I]E[y, 1]7?zfsn2h?z+audu h=O 1=0




h=P I=y

2. fsn 2a+lu en2 3u dn2"Yu= u [gB , ][ 1] ]k2lsn2(h+l+ a+ ) u du h=O 1=0

h=a l =y

3.fsn2au en29+l u dn2y u du = 2V (_)h+ l[a, h][, l}zr21 h=O 1=0

Jcnn2(h+l+?+?u du

h=a I-=, ,la U la,____h] [P,I( , 4.Jnsu2a d C n1 + u u da = (-)P 2P2k2a > [a, /h1 1lf dn2 (???u du

h=O 1=0

5fsn2la+lu cn21+lu dn2ru du

h=a 1=P

()+1 < (_)h+1 [a, h][P, 1] dn2(h+l+y+ D u

k.2 (afi+ ) z = > k'21 2(h+l+r+ 1) h=O 1=0

h=P 1=h

6.fsn2au en23+1 u dn2y+luduE h][y, l]k2_2(h+I+ a)- h=O 1=0

h=a l =y

7.sn2a+lU cn2P u dn2y+ 'u du - - y_ (_)h [a, h] _en L' 2 (h ++ I + +) h=O - 1=0

h=P I =y

0.sl2af+1lt c2P+1t d2y+ 1 6ds=E (_)h+I 1[z, 71] [y ]21 s2 (h + I

1

U1 fSi+u en+lu dn+lu di()+, 11h 2(h+l+a+1)

h=0 1=0--



Some Elliptic Function Formulae

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