Some definite integrls connected with Guss s sums Messenger of Mthemtics XLIV 95 75 85. If n is rel nd positive nd I(t where I(t is the imginry prt of t is less thn either n or we hve cos πtx coshπx e iπnx2 dx = 2 = n exp When n = the ove formul reduces to cos πtx coshπx sinπx2 dx = tn{ 8 π( t2 } cosπtxcos2πxy e iπnx2 dxdy cosh πy ( } { 4 iπ t2 cos πtx n coshπnx eiπnx2 dx. ( cos πtx coshπx cosπx2 dx. (2 if t = nd φ( = Ψ( = cosπnx 2 coshπx dx sinπnx 2 coshπx dx (3 then φ( = (2 Ψ ( +Ψ( Ψ( = (2 φ ( φ(. Similrly if 2 3 I(t is less thn either or n we hve (3 cos πtx +2cosh(2πx/ 3 eiπnx2 dx = ( } n exp { 4 iπ t2 cos πtx n +2cosh(2πnx/ 3 e iπnx2 dx. (4
76 Pper 2 If in (4 we suppose n = we otin nd if t = nd then cosπtxsinπx 2 +2cosh(2πx/ 3 dx = tn{ 8 π( t2 } φ( = Ψ( = cosπnx 2 +2cosh(2πx/ 3 dx sinπnx 2 +2cosh(2πx/ 3 dx φ( = (2 Ψ ( +Ψ( cosπtxcosπx 2 +2cosh(2πx/ 3 dx; (5 (2 ( Ψ( = φ φ(. In similr mnner we cn prove tht sin πtx tnhπx e iπnx2 dx = { } n exp 4 (+ iπ t2 sin πtx n tnhπnx eiπnx2 dx. (7 If we put n = in (7 we otin Now sin πtx tnhπx cosπx2 dx = tn{ 8 π(+t2 } sin tx lim t t tnhx eicx2 dx = lim t t = e icx e 2 x (6 (6 sin πtx tnhπx sinπx2 dx. (8 2sintx e 2x eicx2 dx+lim t sin tx e icx2 dx t dx+ i 2c. (9 Hence dividingothsidesof (7 yt ndmkingt weotin theresultcorresponding to (3 nd (6 viz.: if cos πnx φ( = e 2π x dx Ψ( = ( 2πn + sin πnx e 2π x dx
Some definite integrls connected with Guss s sums 77 then φ( = n (2 Ψ ( Ψ( Ψ( = n (2 φ ( +φ(. ( 2. I shll now shew tht the integrl ( my e expressed in finite terms for ll rtionl vlues of n. Consider the integrl J(t = If R( nd t re positive we hve J(t = 4 π r= r= costx dx cosh 2 πx 2 +x 2. ( r (2r + x 2 +(2r + 2 costx 2 +x 2 dx ( r = 2 {e (2r+t 2 (2r + 2 } (2r +e t = πe t 2cos +2 2π r= ( r e (2r+t 2 (2r + 2 ( nd it is esy to see tht this lst eqution remins true when t is complex provided R(t > nd I(t 2π. Thus the integrl J(t cn e expressed in finite terms for ll rtionl vlues of. Thus for exmple we hve costx dx cosh 2 πx +x 2 = coshtlog(2cosht tsinht (2 cos 2tx dx cosh πx+x 2 = 2cosht (e2t tn e t +e 2t tn e t nd so on. Now let Then if R( > F( = e n F(dn = cos 2tx coshπx e iπnx2 dx. (3 cos 2tx dx cosh πx+iπx2. (4
78 Pper 2 Now let Then f( = ( r exp{ (2r +t+ 4 (2r +2 iπn} r= + { ( } exp i n 4 π t2 πn r= ( r exp { (2r + t } n 4 (2r +2iπ.(5 n e n f(dn = = r= ( r e (2r+t ( π 4 (2r +2 iπ + 2 exp{ (2/π( it} (+icosh{(+i ( 2 π} cos 2tx dx cosh πx+iπx2 (6 in virtue of (; nd therefore Now it is known tht if φ( is continuous nd e n {F( f(}dn =. (7 e n φ(dn = for ll positive vlues of (or even only for n infinity of such vlues in rithmeticl progressio then φ( = for ll positive vlues of n. Hence F( = f(. (8 Equting the rel nd imginry prts in (3 nd (5 we hve cos 2tx coshπx cosπnx2 dx = { e t cos πn 4 e 3t cos 9πn 4 +e 5t cos 25πn } 4 + ( π {e t/n cos n 4 t2 πn + π ( π e 3t/n cos 4n 4 t2 πn + 9π } + (9 4n
Some definite integrls connected with Guss s sums 79 { cos 2tx coshπx sinπnx2 dx = e t sin πn 4 e 3t sin 9πn 4 +e 5t sin 25πn } 4 + ( π {e t/n sin n 4 t2 πn + π ( π e 3t/n sin 4n 4 t2 πn + 9π } +. (2 4n We cn verify the results (8 (9 nd (2 y mens of the eqution (. This eqution cn e expressed s functionl eqution in F( nd it is esy to see tht f( stisfies the sme eqution. The right-hnd side of these equtions cn e expressed in finite terms if n is ny rtionl numer. For let n = / where nd re ny two positive integers nd one of them is odd. Then the results (9 nd (2 reduce to ( cos 2tx πx 2 2cosht coshπx cos dx = [cosh{( t} cos(π/4 cosh{(3 t} cos(9π/4 +cosh{(5 t}cos(25π/4 to terms] ( [ {( + cosh } ( π t cos 4 t2 π + π 4 {( cosh 3 } ( π t cos 4 t2 π + 9π 4 + to terms ] (2 ( cos 2tx πx 2 2cosht coshπx sin dx = [cosh{( t} sin(π/4 cosh{(3 t} sin(9π/4 +cosh{(5 t}sin(25π/4 to terms] ( [ {( + cosh } ( π t sin 4 t2 π + π 4 {( cosh 3 } ( π t sin 4 t2 π + 9π ] + to terms. (22 4 Thus for exmple we hve when = nd = cosπx 2 coshπx cos2πtx dx = + 2sinπt 2 2 2coshπt (23
8 Pper 2 sinπx 2 coshπx It is esy to verify tht (23 nd (24 stisfy the reltion (2. The vlues of the integrls + 2cosπt 2 cos2πtx dx = 2. (24 2coshπt cosπnx 2 coshπx dx sinπnx 2 coshπx dx cn e otined esily from the preceding results y putting t = nd need no specil discussion. By successive differentitions of the results (9 nd (2 with respect to t nd n we cn evlute the integrls 2m sintx x cosh πx 2m costx x cosh πx cos sin πnx2 dx cos sin πnx2 dx for ll rtionl vlues of n nd ll positive integrl vlues of m. Thus for exmple we hve x 2cosπx2 coshπx dx = 8 2 4π (26 x 2 sinπx2 coshπx dx = 8 8 2. (25 3. We cn get mny interesting results y pplying the theory of Cuchy s reciprocl functions to the preceding results. It is known tht if φ(xcosknx dx = Ψ( (27 then (i 2 α{ 2 φ(+φ(α+φ(2α+φ(3α+ } = 2Ψ(+Ψ(β+Ψ(2β+Ψ(3β+ (27 with the condition αβ = 2π/k; (ii α 2{φ(α φ(3α φ(5α+φ(7α+φ(9α } = Ψ(β Ψ(3β Ψ(5β+Ψ(7β+Ψ(9β (27 with the condition αβ = π/4k;
Some definite integrls connected with Guss s sums 8 (iii α 3{φ(α φ(5α φ(7α+φ(α +φ(3α } = Ψ(β Ψ(5β Ψ(7β+Ψ(β+Ψ(3β (27 with the condition αβ = π/6k where 5 7 3... re the odd nturl numers without the multiples of 3. There re of course corresponding results for the function such s φ(xsinknx dx = Ψ( (28 α{φ(α φ(3α+φ(5α } = Ψ(β Ψ(3β+Ψ(5β with the condition αβ = π/2k. Thus from (23 nd (27 (i we otin the following results. If F(αβ = { } α 2 + cosr 2 πα 2 sinr 2 πβ 2 β cosh rπα coshrπβ (29 r= r= then F(αβ = F(βα = (2α{ 2 +e πα +e 4πα +e 9πα + } 2 provided tht αβ =. 4. If insted of strting with the integrl ( we strt with the corresponding sine integrl we cn shew tht when R( nd R(t re positive nd I(t π sintx dx sinh πx 2 +x 2 = 2 2 πe t 2sinπ + r= ( r e rt 2 r 2. (3 Hence the ove integrl cn e expressed in finite terms for ll rtionl vlues of. For exmple we hve From (3 we cn deduce tht sintx sinh 2 πx dx +x 2 = et tn e t e t tn e t. (3 sin 2tx sinhπx e iπnx2 dx = 2 e 2t+iπn +e 4t+4iπn e 6t+9iπn + {( } exp n 4 π + t2 i {e (t+ 4 iπ/n +e (3t+9 4 iπ/n + } (32 πn
82 Pper 2 R(t eing positive nd I(t 2π. The right-hnd side cn e expressed in finite terms for ll rtionl vlues of n. Thus for exmple we hve cosπx 2 sinhπx sinπx 2 sinhπx coshπt cosπt2 sin2πtx dx = (33 2sinhπt sin2πtx dx = sinπt2 2sinhπt (34 nd so on. Applying the formul (28 to (33 nd (34 we hve when αβ = 4 α ( rcos{(2r +2 πα 2 } sinh{(2r +πα} r= + β ( rcos{(2r +2 πβ 2 } sinh{(2r +πβ} r= = 2 α{ 2 +e 2πα +e 8πα +e 8πα + } 2 ; (35 α ( rsin{(2r +2 πα 2 } sinh{(2r +πα} r= = β ( rsin{(2r +2 πβ 2 } sinh{(2r +πβ}. By successive differentition of (32 with respect to t nd n we cn evlute the integrls 2m costx cos x sinh πx sin πnx2 dx 2m sintx x sinh πx r= cos sin πnx2 dx for ll rtionl vlues of n nd ll positive integrl vlues of m. Thus for exmple we hve x cosπx2 sinhπx dx = 8 x sinπx2 sinhπx dx = 4π x 3cosπx2 sinhπx dx = ( 6 4 3 π 2 x 3sinπx2 sinhπx dx = 6π nd so on. The denomintors of the integrnds in (25 nd (36 re cosh πx nd sinh πx. Similr integrls hving the denomintors of their integrnds equl to r coshπ r xsinhπ r x (36 (37
Some definite integrls connected with Guss s sums 83 cn e evluted if r nd r re rtionl y splitting up the integrnd into prtil frctions. 5. The preceding formulæ my e generlised. Thus it my e shewn tht if R( nd R(t re positive I(t π nd < R(θ < then sinπθ = π 2 costx coshπx+cosπθ dx 2 +x 2 e t sinπθ cosπ+cosπθ + r= { } e (2r+ θt 2 (2r + θ 2 e (2r++θt 2 (2r ++θ 2. (38 From (38 it cn e deduced tht if n nd R(t re positive I(t π nd < θ < then costx sinπθ coshπx+cosπθ e iπnx2 dx = {e (2r+ θt+(2r+ θ2iπn e (2r++θt+(2r++θ2iπn } r= + } exp { (π 4 i t2 ( r sinrπθe (2rt+r2iπ/4n. (39 n πn The right-hnd side cn e expressed in finite terms if n nd θ re rtionl. In prticulr when θ = 3 we hve r= costx +2cosh(2πx/ 3 e iπnx2 dx = 2 {e 3 (t 3 iπ e 3 (2t 3 4iπ +e 3 (4t 3 6iπ } + } { (π 2 n exp 4 i t2 πn {e (t 3+iπ/3n e (2t 3+4iπ/3n +e (4t 3+6iπ/3n } (4 where 245... re the nturl numers without the multiples of 3.
84 Pper 2 As n exmple when n = we hve cosπx 2 cosπtx +2cosh(2πx/ 3 dx = 2sin{(π 3πt2 /2} 8cosh(πt/ 3 4 sinπx 2 cosπtx +2cosh(2πx/ 3 dx = 3+2cos{(π 3πt 2 /2} 8cosh(πt/. 3 4 (4 6. The formul (32 ssumes net nd elegnt form when t is chnged to t + 2iπ. We hve then sintx tnhπx e iπnx2 dx (n > t > = { 2 +e t+iπn +e 2t+4iπn +e 3t+9iπn + } { ( } exp n 4 i π + t2 { πn 2 +e (t+iπ/n +e (2t+4iπ/n + }. (42 In prticulr when n = we hve cosπx 2 tnhπx sin2πtx dx = 2 tnhπt{ cos( 4 π +πt2 } sinπx 2 tnhπx sin2πtx dx = 2 tnhπtsin( 4 π +πt2. (43 We shll now consider n importnt specil cse of (42. It cn esily e seen from (9 tht the left-hnd side of (42 when divided y t tends to cos πnx e 2π x dx i 2πn + sin πnx e 2π x dx s t. But the limit of the right-hnd side of (42 divided y t cn e found when n is rtionl. Let then n = / where nd re ny two positive integers nd let (44 φ( = cos πnx e 2π x dx Ψ( = 2πn + sin πnx e 2π x dx.
Some definite integrls connected with Guss s sums 85 The reltion etween φ( nd Ψ( hs een stted lredy in (. From (42 nd (44 it cn esily e deduced tht if nd re oth odd then ( r= φ = 4 ( 2rcos r= ( r= Ψ = 4 ( 2rsin r= ( r 2 π ( r 2 π ( 4 + 4 ( r= ( 2rsin r= r= ( 2rcos r= ( 4 π + r2 π ( 4 π + r2 π (45 It cn esily e seen tht these stisfy the reltion (. Similrly when one of nd is odd nd the other even it cn e shewn tht φ ( Ψ ( σ = 4π 2 ( + 2 σ = 4π + 2 ( 2 r= r= r( r r= r= r( r r= r= r( r r= r= r( r ( cos r 2 π ( sin 4 π + r2 π ( sin r 2 π cos ( 4 π + r2 π (46 where σ = ( cos 4 π + r2 π σ = sin ( 4 π + r2 π = ( sin r 2 π = cos ( r 2 π. (47 Thus for exmple we hve φ( = 2 φ( = 2 2 8 φ(2 = φ(6 = 3 4 3 44 φ ( 2 = 4π φ( 2 5 6 φ(4 = 3 2 = 8 3 5 6 32 (48 nd so on. By differentiting (42 with respect to n we cn evlute the integrls x m e 2π x cos sin πnx dx (49
86 Pper 2 for ll rtionl vlues of n nd positive integrl vlues of m. Thus for exmple we hve nd so on. xcos 2 πx e 2π x dx = 3 4π 8π 2 xcos2πx e 2π x dx = 64 ( 2 3 π + 5 π 2 x 2 cos2πx e 2π x dx = 256 ( 5π + 5π 2 (5