EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 4, 2021, 1337-1349 ISSN 1307-5543 – ejpam.com Published by New York Business Global Double integral involving logarithmic and quotient function with powers expressed in terms of the Lerch function Robert Reynolds1,∗, Allan Stauffer1 1 Department of Mathematics and Statistics, Faculty of Science, York University, Toronto, Ontario, Canada, M3J1P3 Abstract. In this work the authors use their contour integral method to derive the double integral given by ∫∞ 0 ∫∞ 0 xm−1ym+ q 2 −1 logk(axy) (xq+1)2(yq+1)2 dxdy in terms of the Lerch function. This integral formula is then used to derive closed solutions in terms of fundamental constants and special functions. There are some useful results relating double integrals of certain kinds of functions to ordinary integrals for which we know no general reference. Thus a table of integral pairs is given for interested readers. All the results in this work are new. 2020 Mathematics Subject Classifications: 30E20, 33-01, 33-03, 33-04, 33-33B, 33E20 Key Words and Phrases: Catalan’s constant, Double integral, Apéry’s constant, Lerch function, Contour integral 1. Introduction The double integral in terms of the Lerch function derived in this work is used to provide formal derivations and new formulae in the form of a summary table of integrals. The Lerch function being a special function has the fundamental property of analytic con- tinuation, which enables us to widen the range of evaluation for the parameters involved in our definite integral. The definite integral derived in this manuscript is given by∫ ∞ 0 ∫ ∞ 0 xm−1ym+ q 2 −1 logk(axy) (xq + 1)2 (yq + 1)2 dxdy (1) where the parameters k, a are general complex numbers and 0 < Re(m) < 1. This work is important because the authors were unable to find similar derivations in current literature. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4085 Email addresses: milver@my.yorku.ca (R. Reynolds), stauffer@yorku.ca (A. Stauffer) http://www.ejpam.com 1337 © 2021 EJPAM All rights reserved. R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1338 The derivation of the definite integral follows the method used by us in [3] which involves Cauchy’s integral formula. The generalized Cauchy’s integral formula is given by yk Γ(k + 1) = 1 2πi ∫ C ewy wk+1 dw. (2) where C is in general an open contour in the complex plane where the bilinear concomi- tant has the same value at the end points of the contour. This method involves using a form of equation (2) then multiply both sides by a function, then take a definite integral of both sides. This yields a definite integral in terms of a contour integral. A second contour integral is derived by multiplying equation (2) by a function and performing some substitutions so that the contour integrals are the same. 2. Definite integral of the contour integral We use the method in [3]. The variable of integration in the contour integral is t = m + w. The cut and contour are in the first quadrant of the complex z-plane. The cut approaches the origin from the interior of the first quadrant and the contour goes round the origin with zero radius and is on opposite sides of the cut. Using equation (2) we replace y by log(axy) then multiply by xm−1ym+ q 2−1 (xq+1)2(yq+1)2 . Next we take the double infinite integral over x ∈ (0,∞) and y ∈ (0,∞) to get (3) 1 Γ(k + 1) ∫ ∞ 0 ∫ ∞ 0 xm−1ym+ q 2 −1 logk(axy) (xq + 1)2 (yq + 1)2 dxdy = 1 2πi ∫ ∞ 0 ∫ ∞ 0 ∫ C aww−k−1xm+w−1ym+ q 2 +w−1 (xq + 1)2 (yq + 1)2 dwdxdy = 1 2πi ∫ C ∫ ∞ 0 ∫ ∞ 0 aww−k−1xm+w−1ym+ q 2 +w−1 (xq + 1)2 (yq + 1)2 dxdydw = 1 2πi ∫ C π2aww−k−1(m− q + w)(2m− q + 2w) csc ( 2π(m+w) q ) q4 dw from equation (3.241.5) in [1] where Re(w + m) < 2q. We are able to switch the order of integration over z, x and y using Fubini’s theorem since the integrand is of bounded measure over the space C × R× R. 3. The Lerch function We use (9.550) and (9.556) in [1] where Φ(z, s, v) is the Lerch function which is a generalization of the Hurwitz zeta ζ(s, v) and Polylogarithm functions Lin(z). The Lerch function has a series representation given by R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1339 Φ(z, s, v) = ∞∑ n=0 (v + n)−szn (4) where |z|< 1, v ̸= 0,−1, .. and is continued analytically by its integral representation given by Φ(z, s, v) = 1 Γ(s) ∫ ∞ 0 ts−1e−vt 1− ze−t dt = 1 Γ(s) ∫ ∞ 0 ts−1e−(v−1)t et − z dt (5) where Re(v) > 0, and either |z|≤ 1, z ̸= 1, Re(s) > 0, or z = 1, Re(s) > 1. 4. Infinite sum of the contour integral 4.1. Derivation of the first contour integral In this section we will again use Cauchy’s integral formula (2) and taking the in- finite sum to derive equivalent sum representations for the contour integrals. We pro- ceed using equation (2) and replace y by log(a) + 2iπ(2y+1) q and multiply both sides by −4iπ2m2e 2iπm(2y+1) q q4 and take the infinite sum over y ∈ [0,∞) simplifying in terms of the Lerch function to get (6) 4k+1πk+2m2 ( i q )k−1 e 2iπm q Φ ( e 4imπ q ,−k, 12 − iq log(a) 4π ) q5Γ(k + 1) = − 1 2πi ∞∑ y=0 ∫ C 4iπ2m2aww−k−1e 2iπ(2y+1)(m+w) q q4 dw = − 1 2πi ∫ C ∞∑ y=0 4iπ2m2aww−k−1e 2iπ(2y+1)(m+w) q q4 dw = 1 2πi ∫ C 2π2m2aww−k−1 csc ( 2π(m+w) q ) q4 dw from equation (1.232.3) in [1] where Im(m+ w) > 0 for convergence of the sum. 4.2. Derivation of the second contour integral In this section we will again use Cauchy’s integral formula (2) and taking the infinite sum to derive equivalent sum representations for the contour integrals. We proceed using equation (2) and replace y by log(a)+ 2iπ(2y+1) q and multiply both sides by 6iπ2me 2iπm(2y+1) q q3 and take the infinite sum over y ∈ [0,∞) simplifying in terms of the Lerch function to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1340 (7) 3i22k+1πk+2m ( i q )k e 2iπm q Φ ( e 4imπ q ,−k, 12 − iq log(a) 4π ) q3Γ(k + 1) = 1 2πi ∞∑ y=0 ∫ C 6iπ2maww−k−1e 2iπ(2y+1)(m+w) q q3 dw = 1 2πi ∫ C ∞∑ y=0 6iπ2maww−k−1e 2iπ(2y+1)(m+w) q q3 dw = − 1 2πi ∫ C 3π2maww−k−1 csc ( 2π(m+w) q ) q3 dw from equation (1.232.3) in [1] where Im(m+ w) > 0 for convergence of the sum. 4.3. Derivation of the third contour integral In this section we will again use Cauchy’s integral formula (2) and taking the infinite sum to derive equivalent sum representations for the contour integrals. We proceed using equation (2) and replace y by log(a)+ 2iπ(2y+1) q and multiply both sides by −2iπ2e 2iπm(2y+1) q q2 and take the infinite sum over y ∈ [0,∞) simplifying in terms of the Lerch function to get (8) 22k+1πk+2 ( i q )k−1 e 2iπm q Φ ( e 4imπ q ,−k, 12 − iq log(a) 4π ) q3Γ(k + 1) = − 1 2πi ∞∑ y=0 ∫ C 2iπ2aww−k−1e 2iπ(2y+1)(m+w) q q2 dw = − 1 2πi ∫ C ∞∑ y=0 2iπ2aww−k−1e 2iπ(2y+1)(m+w) q q2 dw = 1 2πi ∫ C π2aww−k−1 csc ( 2π(m+w) q ) q2 dw from equation (1.232.3) in [1] where Im(m+ w) > 0 for convergence of the sum. 4.4. Derivation of the fourth contour integral In this section we will again use Cauchy’s integral formula (2) and taking the infinite sum to derive equivalent sum representations for the contour integrals. We proceed using equation (2) and replace y by log(a)+ 2iπ(2y+1) q and multiply both sides by −4iπ2e 2iπm(2y+1) q q4 and replace k → k − 2 and take the infinite sum over y ∈ [0,∞) simplifying in terms of the Lerch function to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1341 (9) 4k−1πk ( i q )k−3 e 2iπm q Φ ( e 4imπ q , 2− k, 12 − iq log(a) 4π ) q5Γ(k − 1) = − 1 2πi ∞∑ y=0 ∫ C 4iπ2aww1−ke 2iπ(2y+1)(m+w) q q4 dw = − 1 2πi ∫ C ∞∑ y=0 4iπ2aww1−ke 2iπ(2y+1)(m+w) q q4 dw = 1 2πi ∫ C 2π2aww1−k csc ( 2π(m+w) q ) q4 dw from equation (1.232.3) in [1] where Im(m+ w) > 0 for convergence of the sum. 4.5. Derivation of the fifth contour integral In this section we will again use Cauchy’s integral formula (2) and taking the infinite sum to derive equivalent sum representations for the contour integrals. We proceed using equation (2) and replace y by log(a)+ 2iπ(2y+1) q and multiply both sides by −4iπ2e 2iπm(2y+1) q q4 and replace k → k − 1 and take the infinite sum over y ∈ [0,∞) simplifying in terms of the Lerch function to get (10) 22(k−1)+3πk+1m ( i q )k−2 e 2iπm q Φ ( e 4imπ q , 1− k, 12 − iq log(a) 4π ) q5Γ(k) = − 1 2πi ∞∑ y=0 ∫ C 8iπ2maww−ke 2iπ(2y+1)(m+w) q q4 dw = − 1 2πi ∫ C ∞∑ y=0 8iπ2maww−ke 2iπ(2y+1)(m+w) q q4 dw = 1 2πi ∫ C 4π2maww−k csc ( 2π(m+w) q ) q4 dw from equation (1.232.3) in [1] where Im(m+ w) > 0 for convergence of the sum. 4.6. Derivation of the sixth contour integral In this section we will again use Cauchy’s integral formula (2) and taking the infinite sum to derive equivalent sum representations for the contour integrals. We proceed using equation (2) and replace y by log(a) + 2iπ(2y+1) q and multiply both sides by 6iπ2e 2iπm(2y+1) q q3 R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1342 and replace k → k − 1 and take the infinite sum over y ∈ [0,∞) simplifying in terms of the Lerch function to get (11) 3i22(k−1)+1πk+1 ( i q )k−1 e 2iπm q Φ ( e 4imπ q , 1− k, 12 − iq log(a) 4π ) q3Γ(k) = 1 2πi ∞∑ y=0 ∫ C 6iπ2aww−ke 2iπ(2y+1)(m+w) q q3 dw = 1 2πi ∫ C ∞∑ y=0 6iπ2aww−ke 2iπ(2y+1)(m+w) q q3 dw = − 1 2πi ∫ C 3π2aww−k csc ( 2π(m+w) q ) q3 dw from equation (1.232.3) in [1] where Im(m+ w) > 0 for convergence of the sum. Main results 5. Definite integral in terms of the Lerch function Theorem 1. For a, k ∈ C, 0 < Re(m) < 1, Re(q) > 0,∫ ∞ 0 ∫ ∞ 0 xm−1ym+ q 2 −1 logk(axy) (xq + 1)2 (yq + 1)2 dxdy = 4k−1πk q4 ( i q )k e 2iπm q ( i ( (k−1)kq2Φ ( e 4imπ q , 2−k, 1 2 − iq log(a) 4π ) −8π2(q−2m)(q−m)Φ ( e 4imπ q ,−k, 1 2 − iq log(a) 4π )) + 2πkq(3q − 4m)Φ ( e 4imπ q , 1− k, 1 2 − iq log(a) 4π )) (12 ) Proof. Since the addition of the right-hand sides of equations (6), (7), (8), (9), (10) and (11) is equal to the right-hand side of equation (3) we can equate the left-hand sides and reduce the factorial to get the stated result. R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1343 6. Integral representation for Catalan’s constant G, Apéry’s constant ζ(3) and π Proposition 1. (13 ) ∫ ∞ 0 ∫ ∞ 0 4 √ x (√ x √ y − 1 ) (x+ 1)2 4 √ y(y + 1)2 log(xy) dxdy = G Proof. Use (12) and form a second equation by replacing m → p and take their difference. Then set k = −1, a = 1, q = −1,m = −3/4, p = −1/4 and simplify using entries (1), (2), (4) in Table below (64:12:7) in [2]. Proposition 2. (14 ) ∫ ∞ 0 ∫ ∞ 0 4 √ x 4 √ y − 1 x3/4(x+ 1)2 4 √ y(y + 1)2 log(xy) dxdy = G− 7ζ(3) 8π Proof. Use (12) and form a second equation by replacing m → p and take their difference. Then set k = −1, a = 1,m = 1/4, p = 1/2. Then apply L’Hopital’s rule to the right-hand side as q → 1 and simplify using entries (1), (2), (4) in Table below (64:12:7), equation (64:12:1) and entry (2) in Table below (64:7) in [2]. Proposition 3. (15 ) ∫ ∞ 0 ∫ ∞ 0 y2 ( 1− x2y2 ) log(xy) (x4 + 1)2 (y4 + 1)2 ( log2(xy) + π2 )dxdy = 1 4 (G− 1) and (16 ) ∫ ∞ 0 ∫ ∞ 0 y2 ( x2y2 − 1 ) (x4 + 1)2 (y4 + 1)2 ( log2(xy) + π2 )dxdy = 1 2π2 − 1 16 Proof. Use (12) and form a second equation by replacing m → p and take their difference. Then set k = −1, a = −1, q = 4,m = 3, p = 1 rationalize the denominator and equate real and imaginary parts and simplify using entries (1), (2), (4) in Table below (64:12:7) in [2]. R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1344 7. Definite integral in terms of the polylogarithm function Theorem 2. For k ∈ C, 0 < Re(m) < 1, Re(q) > 0, (17 ) ∫ ∞ 0 ∫ ∞ 0 xm−1ym+ q 2 −1 logk ( e 2iπ q xy ) (xq + 1)2 (yq + 1)2 dxdy = 4k−1π q4 k ( i q )k e − 2iπm q ( i ( (k − 1)kq2Li2−k ( e 4imπ q ) − 8π2(q − 2m)(q −m)Li−k ( e 4imπ q )) + 2πkq(3q − 4m)Li1−k ( e 4imπ q )) Proof. Use (12) and replace a → e 2πi q and simplify using equation (64:12:2) in [2]. 8. Definite integral in terms of the Hurwitz zeta function Theorem 3. For k ∈ C, Re(q) > 0, (18 ) ∫ ∞ 0 ∫ ∞ 0 x q 4 −1y 3q 4 −1 logk(xy) (xq + 1)2 (yq + 1)2 dxdy = 23k−4πk q2 ( 8iπkζ ( 1− k, 1 4 ) − 8iπkζ ( 1− k, 3 4 ) − (k − 1)k ( ζ ( 2− k, 1 4 ) − ζ ( 2− k, 3 4 )) + 12π2 ( ζ ( −k, 1 4 ) − ζ ( −k, 3 4 ))) ( i q )k Proof. Use (12) set a = 1 and replace m → q/4 and simplify using equation (64:12:1) and entry (4) in Table below (64:12:7) in [2]. R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1345 9. Derivation of entry 3.241.5 in [1] Proposition 4. For 0 < Re(m) < 1, 0 < Re(p) < 1, Re(p) < 2q, (19 ) ∫ ∞ 0 ∫ ∞ 0 y q 2 −1 (xpyp − xmym) x (xq + 1)2 (yq + 1)2 dxdy = π2 ( (2m− q)(q −m) csc ( 2πm q ) + (q − 2p)(q − p) csc ( 2πp q )) q4 Proof. Use equation (12) and set k = 0 and simplify using entry (2) in Table below (64:12:7) in [2]. Proposition 5. (20 ) ∫ ∞ 0 ∫ ∞ 0 √ y log(log(xy)) √ x (x2 + 1)2 (y2 + 1)2 dxdy = 1 64 ( 16G+ π2 ( 8i+ 3iπ + 2 log ( 64π3Γ ( −1 4 )6 729Γ ( −3 4 )6 ))) Proof. Use equation (18) and take the first partial derivative with respect to k and set k = 0, q = 2 and simplify using equation (64:10:2) in [2]. Proposition 6. (21 ) ∫ ∞ 0 ∫ ∞ 0 √ y log(xy) log(log(xy)) √ x (x2 + 1)2 (y2 + 1)2 dxdy = 1 16 π2 ( −6iG+ (−4− i)− 2iπ + log ( 6561Γ ( −3 4 )8 256π4Γ ( −1 4 )8 )) Proof. Use equation (18) and apply L’Hopital’s rule as k → 1 and set q = 2 and simplify using equation (64:10:2) in [2]. 10. Definite integral in terms of the Zeta function of Riemann Lemma 1. (22 ) ∫ ∞ 0 ∫ ∞ 0 √ y √ x (x2 + 1)2 (y2 + 1)2 ( log2(xy) + π2 )dxdy = 3 32 ( 2ζ ′(−2) + log(2) ) R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1346 Proof. Use equation (12) and set m = 1/2, a = −1, q = 2 and simplify in terms of the Riemann zeta function using entry (2) in Tbale below (64:7) and entry (4) in table below (64:12:7) in [2] to get (23) ∫ ∞ 0 ∫ ∞ 0 √ y logk(−xy) √ x (x2 + 1)2 (y2 + 1)2 dxdy = 2k−5(iπ)k ( −8iπ ( 2k − 1 ) kζ(1− k) + ( 2k − 2 ) (k − 1)kζ(2− k)− 6π2 ( 2k+1 − 1 ) ζ(−k) ) Next apply L’Hopital’s rule as k → −1 and simplify. Proposition 7. (24 ) ∫ ∞ 0 ∫ ∞ 0 √ y √ x (x2 + 1)2 (y2 + 1)2 ( log2(xy) + π2 )dxdy = 3 32 ( 2ζ ′(−2) + log(2) ) and (25 ) ∫ ∞ 0 ∫ ∞ 0 √ y log(xy) √ x (x2 + 1)2 (y2 + 1)2 ( log2(xy) + π2 )dxdy = −π2 96 Proof. Use equation (23) and apply L’Hopitals’ rule as k → −1 rationalize the denom- inator and equate real and imaginary parts to get stated result. Theorem 4. For k ∈ C, (26 ) ∫ ∞ 0 ∫ ∞ 0 y2 logk(ixy) (x4 + 1)2 (y4 + 1)2 dxdy = 1 128 (iπ)k ( −8iπ ( 2k − 1 ) kζ(1− k) + ( 2k − 2 ) (k − 1)kζ(2− k)− 6π2 ( 2k+1 − 1 ) ζ(−k) ) Proof. Use (12) and set m = 1, q = 4, a = i and simplify using entry (4) in Table below (64:12:7) in [2]. R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1347 Proposition 8. (27 ) ∫ ∞ 0 ∫ ∞ 0 y2 (x4 + 1)2 (y4 + 1)2 ( 4 log2(xy) + π2 )dxdy = 3 128 ( 2ζ ′(−2) + log(2) ) and (28 ) ∫ ∞ 0 ∫ ∞ 0 y2 log(xy) (x4 + 1)2 (y4 + 1)2 ( 4 log2(xy) + π2 )dxdy = − π2 768 Proof. Use equation (26) and apply L’Hopitals’ rule as k → −1 rationalize the denom- inator and equate real and imaginary parts to get stated result. R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1337-1349 1348 11. Summary table of results f(x, y) ∫∞ 0 ∫∞ 0 f(x, y)dxdy 4 √ x( √ x √ y−1) (x+1)2 4 √ y(y+1)2 log(xy) G 4 √ x 4 √ y−1 x3/4(x+1)2 4 √ y(y+1)2 log(xy) G− 7ζ(3) 8π y2(x2y2−1) (x4+1)2(y4+1)2(log2(xy)+π2) 1 2π2 − 1 16 y2(1−x2y2) log(xy) (x4+1)2(y4+1)2(log2(xy)+π2) 1 4(G− 1) y q 2−1(xpyp−xmym) x(xq+1)2(yq+1)2 π2 ( (2m−q)(q−m) csc ( 2πm q ) +(q−2p)(q−p) csc ( 2πp q )) q4 √ y log(log(xy)) √ x(x2+1)2(y2+1)2 1 64 ( 16G+ π2 ( 8i+ 3iπ + 2 log ( 64π3Γ(− 1 4) 6 729Γ(− 3 4) 6 ))) √ y log(xy) log(log(xy)) √ x(x2+1)2(y2+1)2 1 16π 2 ( −6iG+ (−4− i)− 2iπ + log ( 6561Γ(− 3 4) 8 256π4Γ(− 1 4) 8 )) √ y √ x(x2+1)2(y2+1)2(log2(xy)+π2) 3 32 (2ζ ′(−2) + log(2)) √ y log(xy) √ x(x2+1)2(y2+1)2(log2(xy)+π2) −π2 96 y2 (x4+1)2(y4+1)2(4 log2(xy)+π2) 3 128 (2ζ ′(−2) + log(2)) y2 log(xy) (x4+1)2(y4+1)2(4 log2(xy)+π2) − π2 768 12. Discussion In this work the authors derived a double integral formula in terms of the Lerch func- tion. This integral formula was then used to derive special cases in terms of fundamental constants and special functions. A table of integrals featuring some of the integral results was presented for the benefit of interested readers. We used Wolfram Mathematica to REFERENCES 1349 numerically verify the formulas for various ranges of the parameters for real and imagi- nary values. We will use our contour integral method to derive other double integrals and produce more tables of integrals in our future work. References [1] I. S. Gradshteyn and I. M. Ryzhik. Table of Integrals, Series, and Products. Academic Press, 05 2014. [2] Keith B. Oldham, Jan Myland, and Jerome Spanier. An Atlas of Functions: with Equator, the Atlas Function Calculator. Springer Science & Business Media, 07 2010. [3] Robert Reynolds and Allan Stauffer. A method for evaluating definite integrals in terms of special functions with examples. International Mathematical Forum, 15:235– 244, 2020.