EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 14, No. 4, 2021, 1249-1265 ISSN 1307-5543 – ejpam.com Published by New York Business Global Definite integral of logarithmic trigonometric functions expressed in terms of the incomplete gamma function Robert Reynolds1,∗, Allan Stauffer1 1 Department of Mathematics and Statistics, Faculty of Science, York University, Toronto, Ontario, Canada, M3J1P3 Abstract. We present a method using contour integration to derive definite integrals and their associated infinite sums which can be expressed as a special function. We give a proof of the basic equation and some examples of the method. The advantage of using special functions is their analytic continuation which widens the range of the parameters of the definite integral over which the formula is valid. We give as examples definite integrals of logarithmic functions times a trigonometric function. In various cases these generalizations evaluate to known mathematical constants such as Catalan’s constant and π. 2020 Mathematics Subject Classifications: 30-02, 30D10, 30D30, 30E20, 11M35, 11M06, 01A55 Key Words and Phrases: Entries in Gradshteyn and Rhyzik, Lerch function, Logarithm func- tion, Contour Integral, Cauchy, Infinite Integral 1. Introduction We will derive integrals as indicated in the abstract in terms of special functions. Some special cases of these integrals have been reported in Gradshteyn and Ryzhik [3]. In 1867 David Bierens de Haan [4] derived hyperbolic integrals of the form∫ ∞ 0 sinh(ax) ( e−mx(log(α)− x)k − emx(log(α) + x)k ) (cosh(ax) + cos(t))2 dx In our case the constants in the formulas are general complex numbers subject to the restrictions given below. The derivations follow the method used by us in [8]. The generalized Cauchy’s integral formula is given by xk Γ(k + 1) = 1 2πi ∫ C ewx wk+1 dw. (1) ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v14i4.4063 Email addresses: milver@my.yorku.ca (R. Reynolds), stauffer@yorku.ca (A. Stauffer) http://www.ejpam.com 1249 © 2021 EJPAM All rights reserved. R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1250 This method involves using a form of equation (1) then multiplys both sides by a function, then takes a definite integral of both sides. This yields a definite integral in terms of a contour integral. Then we multiply both sides of equation (1) by another function and take the infinite sum of both sides such that the contour integral of both equations are the same. 2. Derivation of the definite integral of the contour integral We use the method in [8]. Here the contour is similar to Figure 2 in [8]. Using a generalization of Cauchy’s integral formula we first replace x by ix+ log(a) then multiply both sides by emx for the first equation and the replace x with −x and multiplying both sides by e−mx to get the second equation. Then we subtract these two equations, followed by multiplying both sides by − sinh(ax) 2(cosh(ax)+cos(t))2 to get − sinh(ax) ( e−mx(log(α)− x)k − emx(log(α) + x)k ) 2Γ(k + 1)(cosh(ax) + cos(t))2 = 1 2πi ∫ C w−k−1αw sinh(ax) sinh(x(m+ w)) (cosh(ax) + cos(t))2 dw (2) where the logarithmic function is defined in equation (4.1.2) in [1]. We then take the definite integral over x ∈ [0,∞) of both sides to get (3) − ∫ ∞ 0 sinh(ax) ( e−mx(log(α)− x)k − emx(log(α) + x)k ) 2Γ(k + 1)(cosh(ax) + cos(t))2 dx = 1 2πi ∫ ∞ 0 ∫ C w−k−1αw sinh(ax) sinh(x(m+ w)) (cosh(ax) + cos(t))2 dwdx = 1 2πi ∫ C ∫ ∞ 0 w−k−1αw sinh(ax) sinh(x(m+ w)) (cosh(ax) + cos(t))2 dxdw = 1 2πi ∫ C πmw−k−1 csc(t)αw csc ( π(m+w) a ) sin ( t(m+w) a ) a2 dw + 1 2πi ∫ C πw−k csc(t)αw csc ( π(m+w) a ) sin ( t(m+w) a ) a2 dw from equation (2.5.48.18) in [7] and the integrals are valid for a, m, k, t and α complex and −1 < Re(w +m) < 0 and Re(α) ̸= 0. We are able to switch the order of integration over w and x using Fubini’s theorem since the integrand is of bounded measure over the space C× [0,∞). R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1251 3. Derivation of the infinite sum of the contour integral 3.1. Derivation of the first contour integral In this section we will again use the generalized Cauchy’s integral formula to derive equivalent contour integrals. First we multiply equation (1) by eimt/α/2i then replace by x by p+ it/α for the first equation and then p− it/α for the second equation to get ie− imt a (( p− it a )k − e 2imt a ( p+ it a )k) 2Γ(k + 1) = 1 2π ∫ C w−k−1ewp sin ( t(m+ w) a ) dw (4) Then we replace p with πi(2p+ 1)/a+ log(α) and multiply both sides by −2π a1 to get ie− imt a (( iπ(2p+1) a − it a + log(α) )k − e 2imt a ( iπ(2p+1) a + it a + log(α) )k) 2Γ(k + 1) = 1 2πi ∫ C w−k−1 sin ( t(m+ w) a ) e w ( log(α)+ iπ(2p+1) a ) dw (5) Then we multiply both sides by −2iπ a2 e iπm(2y+1) a and take the sum over p ∈ [0,∞) and simplify the left-hand side in terms of the Lerch function to get 2kπk+1 ( i a )k e im(π−t) a ( Φ ( e 2imπ a ,−k, −t−ia log(α)+π 2π ) − e 2imt a Φ ( e 2imπ a ,−k, t−ia log(α)+π 2π )) a2Γ(k + 1) = 1 2πi ∞∑ p=0 ∫ C w−k−1 sin ( t(m+ w) a ) e w ( log(α)+ iπ(2p+1) a ) dw = 1 2πi ∫ C ∞∑ p=0 w−k−1 sin ( t(m+ w) a ) e w ( log(α)+ iπ(2p+1) a ) dw = 1 2π ∫ C πw−k−1αw csc ( π(m+w) a ) sin ( t(m+w) a ) a2 dw (6) from equation (1.232.3) in [3] where csch(ix) = −i csc(x) from equation (4.5.10) in [1] and Im(w) > 0 for the sum to converge. The log terms cannot be combined in general. 3.2. Derivation of the second contour integral Next we will derive the second equation by using equation (6), multiplying by m csc(t) and taking the infinite sum over p ∈ [0,∞) to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1252 2kπk+1m ( i a )k csc(t)e im(π−t) a ( Φ ( e 2imπ a ,−k, −t−ia log(α)+π 2π ) − e 2imt a Φ ( e 2imπ a ,−k, t−ia log(α)+π 2π )) a2Γ(k + 1) = 1 2πi ∫ C πmw−k−1 csc(t)αw csc ( π(m+w) a ) sin ( t(m+w) a ) a2 dw (7) Then we replace k with k − 1 to get 2k−1πk ( i a )k−1 csc(t)e im(π−t) a ( Φ ( e 2imπ a , 1− k, −t−ia log(α)+π 2π ) − e 2imt a Φ ( e 2imπ a , 1− k, t−ia log(α)+π 2π )) a2(k − 1)! = 1 2πi ∫ C πw−k csc(t)αw csc ( π(m+w) a ) sin ( t(m+w) a ) a2 dw (8) from equation (1.232.3) in [3] where csch(ix) = −i csc(x) from equation (4.5.10) in [1] and Im(w) > 0 for the sum to converge. 4. The Lerch function The Lerch function see section (24.14) in [6] has a series representation given by Φ(z, s, v) = ∞∑ n=0 (v + n)−szn (9) 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 (10) where Re(v) > 0, or |z|≤ 1, z ̸= 1, Re(s) > 0, or z = 1, Re(s) > 1. 5. Definite integral in terms of the Lerch function Since the right-hand sides of equation (3), (6) and (8) are equivalent we can equate the left-hand sides simplify the factorial to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1253 (11) ∫ ∞ 0 sinh(ax) ( e−mx(log(α)− x)k − emx(log(α) + x)k ) (cosh(ax) + cos(t))2 dx = − k(2π)k ( i a )k−1 csc(t)e im(π−t) a Φ ( e 2imπ a , 1− k, −t−ia log(α)+π 2π ) a2 + k(2π)k ( i a )k−1 csc(t)e im(π−t) a + 2imt a Φ ( e 2imπ a , 1− k, t−ia log(α)+π 2π ) a2 − (2π)k+1m ( i a )k csc(t)e im(π−t) a Φ ( e 2imπ a ,−k, −t−ia log(α)+π 2π ) a2 + (2π)k+1m ( i a )k csc(t)e im(π−t) a + 2imt a Φ ( e 2imπ a ,−k, t−ia log(α)+π 2π ) a2 The integral in equation (11) can be used as an alternative method to evaluating the Lerch function. 6. Evaluation of special cases of definite Integrals 6.1. Special case 1 For this special case we will form a second equation using (11) by replacing m by −m taking the difference from the original equation and simplifying to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1254 (12) − ∫ ∞ 0 2 sinh(ax) sinh(mx) ( (log(α)− x)k + (log(α) + x)k ) (cosh(ax) + cos(t))2 dx = k(2π)k ( i a )k−1 csc(t)e− im(π−t) a Φ ( e− 2imπ a , 1− k, −t−ia log(α)+π 2π ) a2 − k(2π)k ( i a )k−1 csc(t)e− im(π−t) a − 2imt a Φ ( e− 2imπ a , 1− k, t−ia log(α)+π 2π ) a2 − k(2π)k ( i a )k−1 csc(t)e im(π−t) a Φ ( e 2imπ a , 1− k, −t−ia log(α)+π 2π ) a2 + k(2π)k ( i a )k−1 csc(t)e im(π−t) a + 2imt a Φ ( e 2imπ a , 1− k, t−ia log(α)+π 2π ) a2 − (2π)k+1m ( i a )k csc(t)e− im(π−t) a Φ ( e− 2imπ a ,−k, −t−ia log(α)+π 2π ) a2 + (2π)k+1m ( i a )k csc(t)e− im(π−t) a − 2imt a Φ ( e− 2imπ a ,−k, t−ia log(α)+π 2π ) a2 − (2π)k+1m ( i a )k csc(t)e im(π−t) a Φ ( e 2imπ a ,−k, −t−ia log(α)+π 2π ) a2 + (2π)k+1m ( i a )k csc(t)e im(π−t) a + 2imt a Φ ( e 2imπ a ,−k, t−ia log(α)+π 2π ) a2 6.2. Special case 2 For this special case we use equation (12) setting α = 1 and taking the first partial derivative with respect to m simplifying to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1255∫ ∞ 0 xk sinh(ax) cosh(mx) (cosh(ax) + cos(t))2 dx = 2k−1πk ( i a )k+1 csc(t)e− im(t+π) a a (−1 + eiπk) ( ake 2imt a Φ ( e− 2imπ a , 1 − k, π − t 2π ) − akΦ ( e− 2imπ a , 1− k, t+ π 2π ) − 2iπm ( e 2imt a Φ ( e− 2imπ a ,−k, π − t 2π ) − Φ ( e− 2imπ a ,−k, t+ π 2π )) + e 2iπm a ( akΦ ( e 2imπ a , 1− k, π − t 2π ) + 2iπmΦ ( e 2imπ a ,−k, π − t 2π )) − e 2im(t+π) a ( akΦ ( e 2imπ a , 1− k, t+ π 2π ) + 2iπmΦ ( e 2imπ a ,−k, t+ π 2π ))) (13) 7. Derivation of entry 3.514.4 in [3] Using equation (12) we proceed by setting α = 1 and simplifying to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1256 ∫ ∞ 0 xk sinh(ax) sinh(mx) (cosh(ax) + cos(t))2 dx = 2kπk+1m ( i a )k csc(t)e 2imt a − im(t+π) a Φ ( e− 2imπ a ,−k, π−t 2π ) a2 ((−1)k + 1) − 2kπk+1m ( i a )k csc(t)e− im(t+π) a Φ ( e− 2imπ a ,−k, t+π 2π ) a2 ((−1)k + 1) + 2kπk+1m ( i a )k csc(t)e 2iπm a − im(t+π) a Φ ( e 2imπ a ,−k, π−t 2π ) a2 ((−1)k + 1) − 2kπk+1m ( i a )k csc(t)e im(t+π) a Φ ( e 2imπ a ,−k, t+π 2π ) a2 ((−1)k + 1) + i2k−1kπk ( i a )k csc(t)e 2imt a − im(t+π) a Φ ( e− 2imπ a , 1− k, π−t 2π ) a ((−1)k + 1) − i2k−1kπk ( i a )k csc(t)e− im(t+π) a Φ ( e− 2imπ a , 1− k, t+π 2π ) a ((−1)k + 1) − i2k−1kπk ( i a )k csc(t)e 2iπm a − im(t+π) a Φ ( e 2imπ a , 1− k, π−t 2π ) a ((−1)k + 1) + i2k−1kπk ( i a )k csc(t)e im(t+π) a Φ ( e 2imπ a , 1− k, t+π 2π ) a ((−1)k + 1) (14) Note: When we replace k by k − 1 we get the Mellin transform. Next we set k = 0 and m = b simplify to get (15) ∫ ∞ 0 sinh(ax) sinh(bx) (cosh(ax) + cos(t))2 dx = πb csc(t) csc ( πb a ) sin ( bt a ) a2 from entry (2) in Table (64:12:7) in [5], where −π < Re(t) < π and 0 < |b|< a. 8. Derivation of entry (2.3.1.19) in [2] Using equation (13) and setting m = 0 simplifying we get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1257 (16) ∫ ∞ 0 xk sinh(ax) (cosh(ax) + cos(t))2 dx = 2k−1kπk ( 1 a )k+1 csc ( πk 2 ) csc(t) ( ζ ( 1− k, π − t 2π ) − ζ ( 1− k, t+ π 2π )) Next we set t = π/2 simplify to get (17) ∫ ∞ 0 xs−1 tanh(ax)sech(ax)dx = −2s−2πs−1(s− 1) ( 1 a )s ( ζ ( 2− s, 1 4 ) − ζ ( 2− s, 3 4 )) sec (πs 2 ) from entries (2) and (3) in Table (64:12:7) in [5]. 9. Derivation of a new entry for Table 3.514 in [3] Using equation (12) and setting k = −1, α = −1, a = 1, t = π/2,m = 1/2 and simplifying we get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1258 (18) ∫ ∞ 0 sinh ( x 2 ) tanh(x)sech(x) x2 + π2 dx = √ 2 32π2 ( −ψ(1) ( 3 8 ) + ψ(1) ( 5 8 ) + ψ(1) ( 7 8 ) − ψ(1) ( 9 8 )) + 16π (√ 2 + log ( tan (π 8 ))) from entry (3) Table (64:12:7:2) and entry (4) Table (64:12:7:3). 10. Definite integral in terms of the Hurwitz zeta function Using equation (14) and setting m = 1 and a = 2 to get (19) ∫ ∞ 0 xk sinh(x) sinh(2x) (cos(t) + cosh(2x))2 = 2k−3e iπk 2 kπk csc ( t 2 ) ζ ( 1− k, π−t 4π ) (−1)k + 1 − 2k−3e iπk 2 kπk csc ( t 2 ) ζ ( 1− k, t+π 4π ) (−1)k + 1 − 2k−3e iπk 2 kπk csc ( t 2 ) ζ ( 1− k, 34 − t 4π ) (−1)k + 1 + 2k−3e iπk 2 kπk csc ( t 2 ) ζ ( 1− k, 14 ( t π + 3 )) (−1)k + 1 + 2k−2e iπk 2 πk+1 sec ( t 2 ) ζ ( −k, π−t 4π ) (−1)k + 1 + 2k−2e iπk 2 πk+1 sec ( t 2 ) ζ ( −k, t+π 4π ) (−1)k + 1 − 2k−2e iπk 2 πk+1 sec ( t 2 ) ζ ( −k, 34 − t 4π ) (−1)k + 1 − 2k−2e iπk 2 πk+1 sec ( t 2 ) ζ ( −k, 14 ( t π + 3 )) (−1)k + 1 Next we apply L’Hôpital’s rule to the right-hand side as k → 0 to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1259 (20) ∫ ∞ 0 x sinh(x) sinh(2x) (cos(t) + cosh(2x))2 dx = 1 2 π sec ( t 2 ) ζ ( −1, π − t 4π ) + 1 2 π sec ( t 2 ) ζ ( −1, t+ π 4π ) − 1 2 π sec ( t 2 ) ζ ( −1, 3 4 − t 4π ) − 1 2 π sec ( t 2 ) ζ ( −1, 1 4 ( t π + 3 )) + 1 4 csc ( t 2 ) log ( tan ( t+ π 4 )) from entry (1) in Table (64:4:2) in [5], where −π < Re(t) < π. 11. Definite Integral in terms of the log-gamma log(Γ(x)) and Harmonic number Hk functions Using equation (19) taking the first partial derivative with respect to k and applying L’Hopitals’ rule as k → 0 and simplifying to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1260 (21) ∫ ∞ 0 log(x) sinh(x) sinh(2x) (cos(t) + cosh(2x))2 dx = 1 16 ( csc ( t 2 ) ( H− t+π 4π −H− t+3π 4π + ψ(0) ( t+ π 4π ) − ψ(0) ( 1 4 ( t π + 3 ))) + 2π sec ( t 2 ) log ( 2πΓ ( 3 4 − t 4π ) Γ ( 1 4 ( t π + 3 )) Γ ( π−t 4π ) Γ ( t+π 4π ) )) from equations (64:4:1), (64:9:2), and (64:10:2) in [5]. 11.1. Example 1 Using equation (21) and setting t = π/2 simplifying to get (22) ∫ ∞ 0 log(x) sinh(x) tanh(2x)sech(2x)dx = 1 8 ( 4 sinh−1(1) + √ 2π log ( 2πΓ ( 5 8 ) Γ ( 7 8 ) Γ ( 1 8 ) Γ ( 3 8 ) )) 11.2. Example 2 Using equation (21) and setting t = π/3 simplifying to get (23) ∫ ∞ 0 log(x) sinh(x) sinh(2x) (2 cosh(2x) + 1)2 dx = 1 288 ( 10 √ 3π log(2) + 6 log(64) + 9 √ 3π log(π) + 6 √ 3π log ( Γ ( 5 6 ) Γ ( 1 6 )2 )) R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1261 11.3. Example 3 Using equation (21) and setting t = π/4 simplifying to get (24) ∫ ∞ 0 log(x) sinh(x) sinh(2x) (2 cosh(2x) + 1)2 dx = 1 288 ( 10 √ 3π log(2) + 6 log(64) + 9 √ 3π log(π) + 6 √ 3π log ( Γ ( 5 6 ) Γ ( 1 6 )2 )) 11.4. Example 4 Using equation (21) and setting t = 2π/3 simplifying to get (25) ∫ ∞ 0 log(x) sinh(x) sinh(2x) (2 cosh(2x)− 1)2 dx = 1 16 ( 4 coth−1 (√ 3 ) + π log ( 2πΓ ( 7 12 ) Γ ( 11 12 ) Γ ( 1 12 ) Γ ( 5 12 ) )) 11.5. Example 5 Using equation (21) and setting t = 0 and applying L’Hopital’s rule as t→ 0 simplifying to get (26) ∫ ∞ 0 log(x) tanh2(x)sech(x)dx = 2C π + 1 4 π log ( 2πΓ ( 3 4 )2 Γ ( 1 4 )2 ) 12. Derivation of hyperbolic and algebraic forms 12.1. Example 1 Using equation (12) setting k = −1, t = π/2 and replacing α by eiβ simplifying we get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1262 (27) ∫ ∞ 0 x tanh(ax)sech(ax) cosh(mx) β2 + x2 dx = e− 3iπm 2a 8πa ( −2iπmΦ ( e− 2imπ a , 1, aβ 2π + 3 4 ) + e iπm a ( 2iπmΦ ( e− 2imπ a , 1, 2aβ + π 4π ) + aΦ ( e− 2imπ a , 2, 2aβ + π 4π )) − aΦ ( e− 2imπ a , 2, aβ 2π + 3 4 ) + e 2iπm a ( aΦ ( e 2imπ a , 2, 2aβ + π 4π ) − 2iπmΦ ( e 2imπ a , 1, 2aβ + π 4π )) + ie 3iπm a ( 2πmΦ ( e 2imπ a , 1, aβ 2π + 3 4 ) + iaΦ ( e 2imπ a , 2, aβ 2π + 3 4 ))) Next we take the first partial derivative with respect to m and simplifying to get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1263 (28) ∫ ∞ 0 x tanh(ax)sech(ax) cosh(mx) β2 + x2 dx = e− 3iπm 2a 8πa ( −2iπmΦ ( e− 2imπ a , 1, aβ 2π + 3 4 ) + e iπm a ( 2iπmΦ ( e− 2imπ a , 1, 2aβ + π 4π ) + aΦ ( e− 2imπ a , 2, 2aβ + π 4π )) − aΦ ( e− 2imπ a , 2, aβ 2π + 3 4 ) + e 2iπm a ( aΦ ( e 2imπ a , 2, 2aβ + π 4π ) − 2iπmΦ ( e 2imπ a , 1, 2aβ + π 4π )) + ie 3iπm a ( 2πmΦ ( e 2imπ a , 1, aβ 2π + 3 4 ) + iaΦ ( e 2imπ a , 2, aβ 2π + 3 4 ))) from equation (9.550) in [3]. Next we set m = 0 simplifying in terms of the Trigamma function ψ(1)(z) to get (29) ∫ ∞ 0 x tanh(ax)sech(ax) β2 + x2 dx = ψ(1) ( 2aβ+π 4π ) − ψ(1) ( aβ 2π + 3 4 ) 4π from equation (64:4:1) in [5]. 12.2. Example 2 Using equation (12) and setting k = −2, t = π/2 and replacing α by eiβ simplifying we get R. Reynolds, A. Stauffer / Eur. J. Pure Appl. Math, 14 (4) (2021), 1249-1265 1264 (30) ∫ ∞ 0 ( 1 (x+ iβ)2 + 1 (x− iβ)2 ) tanh(ax)sech(ax) sinh(mx)dx = e− 3iπm 2a 4π2 ( πmΦ ( e− 2imπ a , 2, aβ 2π + 3 4 ) + e iπm a ( iaΦ ( e− 2imπ a , 3, 2aβ + π 4π ) − πmΦ ( e− 2imπ a , 2, 2aβ + π 4π )) − iaΦ ( e− 2imπ a , 3, aβ 2π + 3 4 ) − e 2iπm a ( πmΦ ( e 2imπ a , 2, 2aβ + π 4π ) + iaΦ ( e 2imπ a , 3, 2aβ + π 4π )) + e 3iπm a ( πmΦ ( e 2imπ a , 2, aβ 2π + 3 4 ) + iaΦ ( e 2imπ a , 3, aβ 2π + 3 4 ))) Next we take the first partial derivative with respect tom and settingm = 0 simplifying to get (31) ∫ ∞ 0 x(x− β)(β + x) tanh(ax)sech(ax) (β2 + x2)2 dx = πψ(1) ( 2aβ+π 4π ) − πψ(1) ( aβ 2π + 3 4 ) + aβ ( ζ ( 3, aβ2π + 3 4 ) − ζ ( 3, 2aβ+π 4π )) 4π2 from equations (64:12:1) (64:13:3) and (64:4:1) in [5]. 13. Discussion In this article we derived the integrals of hyperbolic and logarithmic functions in terms of the Lerch function. Then we used these integral formula to derive known and new re- sults. We were able to produce a formal derivation for equation (27) Table 27 in Bierens de Haan [4] and equation (3.514.4) in [3] not previously published. The results presented were numerically verified for both real and imaginary values of the parameters in the in- tegrals using Mathematica by Wolfram. In this work we used Mathematica software to numerically evaluate both the definite integral and associated Special function for com- plex values of the parameters k, α, a, m and t. We considered various ranges of these REFERENCES 1265 parameters for real, integer, negative and positive values. We compared the evaluation of the definite integral to the evaluated Special function and ensured agreement. 14. Conclusion In this paper, we have derived a method for expressing definite integrals in terms of Special functions using contour integration. The contour we used was specific to solving integral representations in terms of the Hurwitz zeta function. We expect that other contours and integrals can be derived using this method. Acknowledgements This paper is fully supported by the Natural Sciences and Engineering Research Coun- cil (NSERC) Grant No. 504070. References [1] Milton Abramowitz and Irene A. Stegun. Handbook of Mathematical Functions: With Formulas, Graphs, and Mathematical Tables. Courier Corporation, 01 1965. [2] Yu A. Brychkov, O. I. Marichev, and N. V. Savischenko. Handbook of Mellin Trans- forms. CRC Press, 10 2018. [3] I. S. Gradshteyn and I. M. Ryzhik. Table of Integrals, Series, and Products. Academic Press, 05 2014. [4] David Bierens de Haan. Nouvelles tables d’intégrales définies. P. Engels, 1867. [5] Keith B. Oldham, Jan Myland, and Jerome Spanier. An Atlas of Functions: with Equator, the Atlas Function Calculator. Springer Science & Business Media, 07 2010. [6] F.W.J. Olver, A. B. Olde Daalhuis, D. W. Lozier, B. I. Schneider, R. F. Boisvert, C. W. Clark, B. R. Miller, B. V. Saunders, H. S. Cohl, and M. A. 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