Table in Gradshteyn and Ryzhik: Derivation of definite integrals of a Hyperbolic Function
aa r X i v : . [ m a t h . C A ] F e b Table in Gradshteyn and Ryzhik: Derivation ofdefinite integrals of a Hyperbolic Function
Robert Reynolds and Allan Stauffer Faculty of Science, York University Department of Mathematics, York [email protected], stauff[email protected] 26, 2021
Abstract
We present a method using contour integration to derive definite inte-grals and their associated infinite sums which can be expressed as a specialfunction. We give a proof of the basic equation and some examples of themethod. The advantage of using special functions is their analytic contin-uation which widens the range of the parameters of the definite integralover which the formula is valid. We give as examples definite integralsof logarithmic functions times a trigonometric function. In various casesthese generalizations evaluate to known mathematical constants such asCatalan’s constant and π . 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 andRyzhik [1]. In 1867 David Bierens de Haan [9] derived hyperbolic integrals ofthe form Z ∞ sinh( ax ) (cid:0) e − mx (log( α ) − x ) k − e mx (log( α ) + x ) k (cid:1) (cosh( ax ) + cos( t )) dx In our case the constants in the formulas are general complex numbers sub-ject to the restrictions given below. The derivations follow the method used byus in [8]. The generalized Cauchy’s integral formula is given by x k k ! = 12 πi Z C e wx w k +1 dw. (1)This method involves using a form of equation (1) then multiplys both sidesby a function, then takes a definite integral of both sides. This yields a definite1ntegral 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 thecontour integral of both equations are the same. We use the method in [8]. Here the contour is similar to Figure 2 in [8]. Using ageneralization of Cauchy’s integral formula we first replace x by ix + log( a ) thenmultiply both sides by e mx for the first equation and the replace x with − x andmultiplying both sides by e − mx to get the second equation. Then we subtractthese two equations, followed by multiplying both sides by − sinh( ax )2(cosh( ax )+cos( t )) to get − sinh( ax ) (cid:0) e − mx (log( α ) − x ) k − e mx (log( α ) + x ) k (cid:1) k ! (cosh( ax ) + cos( t )) = 12 πi Z C w − k − α w sinh( ax ) sinh( x ( m + w ))(cosh( ax ) + cos( t )) dw (2)where the logarithmic function is defined in equation (4.1.2) in [15]. We thentake the definite integral over x ∈ [0 , ∞ ) of both sides to get (3) − Z ∞ sinh( ax ) (cid:0) e − mx (log( α ) − x ) k − e mx (log( α ) + x ) k (cid:1) k ! (cosh( ax ) + cos( t )) dx = 12 πi Z ∞ Z C w − k − α w sinh( ax ) sinh( x ( m + w ))(cosh( ax ) + cos( t )) dwdx = 12 πi Z C Z ∞ w − k − α w sinh( ax ) sinh( x ( m + w ))(cosh( ax ) + cos( t )) dxdw = 12 πi Z C πmw − k − csc( t ) α w csc (cid:16) π ( m + w ) a (cid:17) sin (cid:16) t ( m + w ) a (cid:17) a dw + 12 πi Z C πw − k csc( t ) α w csc (cid:16) π ( m + w ) a (cid:17) sin (cid:16) t ( m + w ) a (cid:17) a dw from equation (2.5.48.18) in [14] and the integrals are valid for a , m , k , t and α complex and − < Re ( w + m ) < Re ( α ) = 0. We are able to switch theorder of integration over w and x using Fubini’s theorem since the integrand isof bounded measure over the space C × [0 , ∞ ).2 Derivation of the infinite sum of the contourintegral
In this section we will again use the generalized Cauchy’s integral formula toderive equivalent contour integrals. First we multiply equation (1) by e imt/α / i then replace by x by p + it/α for the first equation and then p − it/α for thesecond equation to get ie − imta (cid:16)(cid:0) p − ita (cid:1) k − e imta (cid:0) p + ita (cid:1) k (cid:17) k ! = 12 π Z C w − k − e wp sin (cid:18) t ( m + w ) a (cid:19) dw (4)Then we replace p with πi (2 p + 1) /a + log( α ) and multiply both sides by − πa to get ie − imta (cid:18)(cid:16) iπ (2 p +1) a − ita + log( α ) (cid:17) k − e imta (cid:16) iπ (2 p +1) a + ita + log( α ) (cid:17) k (cid:19) k != 12 πi Z C w − k − sin (cid:18) t ( m + w ) a (cid:19) e w ( log( α )+ iπ (2 p +1) a ) dw (5)Then we multiply both sides by − iπa e iπm (2 y +1) a and take the sum over p ∈ [0 , ∞ ) and simplify the left-hand side in terms of the Lerch function to get2 k π k +1 (cid:0) ia (cid:1) k e im ( π − t ) a (cid:16) Φ (cid:16) e imπa , − k, − t − ia log( α )+ π π (cid:17) − e imta Φ (cid:16) e imπa , − k, t − ia log( α )+ π π (cid:17)(cid:17) a k != 12 πi ∞ X p =0 Z C w − k − sin (cid:18) t ( m + w ) a (cid:19) e w ( log( α )+ iπ (2 p +1) a ) dw = 12 πi Z C ∞ X p =0 w − k − sin (cid:18) t ( m + w ) a (cid:19) e w ( log( α )+ iπ (2 p +1) a ) dw = 12 π Z C πw − k − α w csc (cid:16) π ( m + w ) a (cid:17) sin (cid:16) t ( m + w ) a (cid:17) a dw (6)from equation (1.232.3) in [1] where csch( ix ) = − i csc( x ) from equation(4.5.10) in [15] and Im ( w ) > 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 get3 k π k +1 m (cid:0) ia (cid:1) k csc( t ) e im ( π − t ) a (cid:16) Φ (cid:16) e imπa , − k, − t − ia log( α )+ π π (cid:17) − e imta Φ (cid:16) e imπa , − k, t − ia log( α )+ π π (cid:17)(cid:17) a k != 12 πi Z C πmw − k − csc( t ) α w csc (cid:16) π ( m + w ) a (cid:17) sin (cid:16) t ( m + w ) a (cid:17) a dw (7)Then we replace k with k − k − π k (cid:0) ia (cid:1) k − csc( t ) e im ( π − t ) a (cid:16) Φ (cid:16) e imπa , − k, − t − ia log( α )+ π π (cid:17) − e imta Φ (cid:16) e imπa , − k, t − ia log( α )+ π π (cid:17)(cid:17) a ( k − πi Z C πw − k csc( t ) α w csc (cid:16) π ( m + w ) a (cid:17) sin (cid:16) t ( m + w ) a (cid:17) a dw (8)from equation (1.232.3) in [1] where csch( ix ) = − i csc( x ) from equation(4.5.10) in [15] and Im ( w ) > The Lerch function has a series representation given byΦ( z, s, v ) = ∞ X n =0 ( v + n ) − s z n (9)where | z | < , v = 0 , − , .. and is continued analytically by its integral repre-sentation given byΦ( z, s, v ) = 1Γ( s ) Z ∞ t s − e − vt − ze − t dt = 1Γ( s ) Z ∞ t s − e − ( v − t e t − z dt (10)where Re ( v ) >
0, or | z |≤ , z = 1 , Re ( s ) >
0, or z = 1 , Re ( s ) > Since the right-hand sides of equation (3), (6) and (8) are equivalent we canequate the left-hand sides simplify the factorial to get411) Z ∞ sinh( ax ) (cid:0) e − mx (log( α ) − x ) k − e mx (log( α ) + x ) k (cid:1) (cosh( ax ) + cos( t )) dx = − k (2 π ) k (cid:0) ia (cid:1) k − csc( t ) e im ( π − t ) a Φ (cid:16) e imπa , − k, − t − ia log( α )+ π π (cid:17) a + k (2 π ) k (cid:0) ia (cid:1) k − csc( t ) e im ( π − t ) a + imta Φ (cid:16) e imπa , − k, t − ia log( α )+ π π (cid:17) a − (2 π ) k +1 m (cid:0) ia (cid:1) k csc( t ) e im ( π − t ) a Φ (cid:16) e imπa , − k, − t − ia log( α )+ π π (cid:17) a + (2 π ) k +1 m (cid:0) ia (cid:1) k csc( t ) e im ( π − t ) a + imta Φ (cid:16) e imπa , − k, t − ia log( α )+ π π (cid:17) a The integral in equation (11) can be used as an alternative method to eval-uating the Lerch function.
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 get5 Z ∞ ax ) sinh( mx ) (cid:0) (log( α ) − x ) k + (log( α ) + x ) k (cid:1) (cosh( ax ) + cos( t )) dx = k (2 π ) k (cid:0) ia (cid:1) k − csc( t ) e − im ( π − t ) a Φ (cid:16) e − imπa , − k, − t − ia log( α )+ π π (cid:17) a − k (2 π ) k (cid:0) ia (cid:1) k − csc( t ) e − im ( π − t ) a − imta Φ (cid:16) e − imπa , − k, t − ia log( α )+ π π (cid:17) a − k (2 π ) k (cid:0) ia (cid:1) k − csc( t ) e im ( π − t ) a Φ (cid:16) e imπa , − k, − t − ia log( α )+ π π (cid:17) a + k (2 π ) k (cid:0) ia (cid:1) k − csc( t ) e im ( π − t ) a + imta Φ (cid:16) e imπa , − k, t − ia log( α )+ π π (cid:17) a − (2 π ) k +1 m (cid:0) ia (cid:1) k csc( t ) e − im ( π − t ) a Φ (cid:16) e − imπa , − k, − t − ia log( α )+ π π (cid:17) a + (2 π ) k +1 m (cid:0) ia (cid:1) k csc( t ) e − im ( π − t ) a − imta Φ (cid:16) e − imπa , − k, t − ia log( α )+ π π (cid:17) a − (2 π ) k +1 m (cid:0) ia (cid:1) k csc( t ) e im ( π − t ) a Φ (cid:16) e imπa , − k, − t − ia log( α )+ π π (cid:17) a + (2 π ) k +1 m (cid:0) ia (cid:1) k csc( t ) e im ( π − t ) a + imta Φ (cid:16) e imπa , − k, t − ia log( α )+ π π (cid:17) a (12) For this special case we use equation (12) setting α = 1 and taking the firstpartial derivative with respect to m simplifying to get Z ∞ x k sinh( ax ) cosh( mx )(cosh( ax ) + cos( t )) dx = 2 k − π k (cid:0) ia (cid:1) k +1 csc( t ) e − im ( t + π ) a a ( − e iπk ) (cid:18) ake imta Φ (cid:18) e − imπa , − k, π − t π (cid:19) − ak Φ (cid:18) e − imπa , − k, t + π π (cid:19) − iπm (cid:18) e imta Φ (cid:18) e − imπa , − k, π − t π (cid:19) − Φ (cid:18) e − imπa , − k, t + π π (cid:19)(cid:19) + e iπma (cid:18) ak Φ (cid:18) e imπa , − k, π − t π (cid:19) + 2 iπm Φ (cid:18) e imπa , − k, π − t π (cid:19)(cid:19) − e im ( t + π ) a (cid:18) ak Φ (cid:18) e imπa , − k, t + π π (cid:19) + 2 iπm Φ (cid:18) e imπa , − k, t + π π (cid:19)(cid:19)(cid:19) (13)6 Derivation of entry 3.514.4 in [1]
Using equation (12) we proceed by setting α = 1 and simplifying to get (14) Z ∞ x k sinh( ax ) sinh( mx )(cosh( ax ) + cos( t )) dx = 2 k π k +1 m (cid:0) ia (cid:1) k csc( t ) e imta − im ( t + π ) a Φ (cid:16) e − imπa , − k, π − t π (cid:17) a (( − k + 1) − k π k +1 m (cid:0) ia (cid:1) k csc( t ) e − im ( t + π ) a Φ (cid:16) e − imπa , − k, t + π π (cid:17) a (( − k + 1)+ 2 k π k +1 m (cid:0) ia (cid:1) k csc( t ) e iπma − im ( t + π ) a Φ (cid:16) e imπa , − k, π − t π (cid:17) a (( − k + 1) − k π k +1 m (cid:0) ia (cid:1) k csc( t ) e im ( t + π ) a Φ (cid:16) e imπa , − k, t + π π (cid:17) a (( − k + 1)+ i k − kπ k (cid:0) ia (cid:1) k csc( t ) e imta − im ( t + π ) a Φ (cid:16) e − imπa , − k, π − t π (cid:17) a (( − k + 1) − i k − kπ k (cid:0) ia (cid:1) k csc( t ) e − im ( t + π ) a Φ (cid:16) e − imπa , − k, t + π π (cid:17) a (( − k + 1) − i k − kπ k (cid:0) ia (cid:1) k csc( t ) e iπma − im ( t + π ) a Φ (cid:16) e imπa , − k, π − t π (cid:17) a (( − k + 1)+ i k − kπ k (cid:0) ia (cid:1) k csc( t ) e im ( t + π ) a Φ (cid:16) e imπa , − k, t + π π (cid:17) a (( − k + 1)Note: When we replace k by k − k = 0 and m = b simplify to get (15) Z ∞ sinh( ax ) sinh( bx )(cosh( ax ) + cos( t )) dx = πb csc( t ) csc (cid:0) πba (cid:1) sin (cid:0) bta (cid:1) a from entry (2) in Table (64:12:7) in [11], where − π < Re ( t ) < π and 0 < | b | < a . Using equation (13) and setting m = 0 simplifying we get716) Z ∞ x k sinh( ax )(cosh( ax ) + cos( t )) dx = 2 k − kπ k (cid:18) a (cid:19) k +1 csc (cid:18) πk (cid:19) csc( t ) (cid:18) ζ (cid:18) − k, π − t π (cid:19) − ζ (cid:18) − k, t + π π (cid:19)(cid:19) Next we set t = π/ Z ∞ x s − tanh( ax )sech( ax ) dx = − s − π s − ( s − (cid:18) a (cid:19) s (cid:18) ζ (cid:18) − s, (cid:19) − ζ (cid:18) − s, (cid:19)(cid:19) sec (cid:16) πs (cid:17) from entries (2) and (3) in Table (64:12:7) in [11]. Using equation (12) and setting k = − , α = − , a = 1 , t = π/ , m = 1 / Z ∞ sinh (cid:0) x (cid:1) tanh( x )sech( x ) x + π dx = √ π (cid:18) − ψ (1) (cid:18) (cid:19) + ψ (1) (cid:18) (cid:19) + ψ (1) (cid:18) (cid:19) − ψ (1) (cid:18) (cid:19)(cid:19) + 16 π (cid:16) √ (cid:16) tan (cid:16) π (cid:17)(cid:17)(cid:17) (18)from entry (3) Table (64:12:7:2) and entry (4) Table (64:12:7:3).
10 Definite integral in terms of the Hurwitz zetafunction
Using equation (14) and setting m = 1 and a = 2 to get819) Z ∞ x k sinh( x ) sinh(2 x )(cos( t ) + cosh(2 x )) = 2 k − e iπk kπ k csc (cid:0) t (cid:1) ζ (cid:0) − k, π − t π (cid:1) ( − k + 1 − k − e iπk kπ k csc (cid:0) t (cid:1) ζ (cid:0) − k, t + π π (cid:1) ( − k + 1 − k − e iπk kπ k csc (cid:0) t (cid:1) ζ (cid:0) − k, − t π (cid:1) ( − k + 1+ 2 k − e iπk kπ k csc (cid:0) t (cid:1) ζ (cid:0) − k, (cid:0) tπ + 3 (cid:1)(cid:1) ( − k + 1+ 2 k − e iπk π k +1 sec (cid:0) t (cid:1) ζ (cid:0) − k, π − t π (cid:1) ( − k + 1+ 2 k − e iπk π k +1 sec (cid:0) t (cid:1) ζ (cid:0) − k, t + π π (cid:1) ( − k + 1 − k − e iπk π k +1 sec (cid:0) t (cid:1) ζ (cid:0) − k, − t π (cid:1) ( − k + 1 − k − e iπk π k +1 sec (cid:0) t (cid:1) ζ (cid:0) − k, (cid:0) tπ + 3 (cid:1)(cid:1) ( − k + 1Next we apply L’Hˆopital’s rule to the right-hand side as k → Z ∞ x sinh( x ) sinh(2 x )(cos( t ) + cosh(2 x )) dx = 12 π sec (cid:18) t (cid:19) ζ (cid:18) − , π − t π (cid:19) + 12 π sec (cid:18) t (cid:19) ζ (cid:18) − , t + π π (cid:19) − π sec (cid:18) t (cid:19) ζ (cid:18) − , − t π (cid:19) − π sec (cid:18) t (cid:19) ζ (cid:18) − , (cid:18) tπ + 3 (cid:19)(cid:19) + 14 csc (cid:18) t (cid:19) log (cid:18) tan (cid:18) t + π (cid:19)(cid:19) from entry (1) in Table (64:4:2) in [11], where − π < Re ( t ) < π .
11 Definite Integral in terms of the log-gamma log(Γ( x )) and Harmonic number H k functions Using equation (19) taking the first partial derivative with respect to k andapplying L’Hopitals’ rule as k → ∞ log( x ) sinh( x ) sinh(2 x )(cos( t ) + cosh(2 x )) dx = 116 csc (cid:18) t (cid:19) (cid:18) H − t + π π − H − t +3 π π + ψ (0) (cid:18) t + π π (cid:19) − ψ (0) (cid:18) (cid:18) tπ + 3 (cid:19)(cid:19)(cid:19) + 2 π sec (cid:18) t (cid:19) log π Γ (cid:0) − t π (cid:1) Γ (cid:0) (cid:0) tπ + 3 (cid:1)(cid:1) Γ (cid:0) π − t π (cid:1) Γ (cid:0) t + π π (cid:1) !! (21)from equations (64:4:1), (64:9:2), and (64:10:2) in [11]. Using equation (21) and setting t = π/ Z ∞ log( x ) sinh( x ) tanh(2 x )sech(2 x ) dx = 18 − (1)+ √ π log π Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) !! Using equation (21) and setting t = π/ Z ∞ log( x ) sinh( x ) sinh(2 x )(2 cosh(2 x ) + 1) dx = 1288 √ π log(2) + 6 log(64)+ 9 √ π log( π ) + 6 √ π log Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) !! Using equation (21) and setting t = π/ Z ∞ log( x ) sinh( x ) sinh(2 x )(2 cosh(2 x ) + 1) dx = 1288 √ π log(2) + 6 log(64)+ 9 √ π log( π ) + 6 √ π log Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) !! Using equation (21) and setting t = 2 π/ Z ∞ log( x ) sinh( x ) sinh(2 x )(2 cosh(2 x ) − dx = 116 − (cid:16) √ (cid:17) + π log π Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) !! Using equation (21) and setting t = 0 and applying L’Hopital’s rule as t → Z ∞ log( x ) tanh ( x )sech( x ) dx = 2 Cπ + 14 π log π Γ (cid:0) (cid:1) Γ (cid:0) (cid:1) !
12 Derivation of hyperbolic and algebraic forms
Using equation (12) setting k = − t = π/ α by e iβ simplifyingwe get Z ∞ x tanh( ax )sech( ax ) cosh( mx ) β + x dx = e − iπm a πa (cid:18) − iπm Φ (cid:18) e − imπa , , aβ π + 34 (cid:19) + e iπma (cid:18) iπm Φ (cid:18) e − imπa , , aβ + π π (cid:19) + a Φ (cid:18) e − imπa , , aβ + π π (cid:19)(cid:19) − a Φ (cid:18) e − imπa , , aβ π + 34 (cid:19) + e iπma (cid:18) a Φ (cid:18) e imπa , , aβ + π π (cid:19) − iπm Φ (cid:18) e imπa , , aβ + π π (cid:19)(cid:19) + ie iπma (cid:18) πm Φ (cid:18) e imπa , , aβ π + 34 (cid:19) + ia Φ (cid:18) e imπa , , aβ π + 34 (cid:19)(cid:19)(cid:19) (27)Next we take the first partial derivative with respect to m and simplifyingto get 11 ∞ x tanh( ax )sech( ax ) cosh( mx ) β + x dx = e − iπm a πa (cid:18) − iπm Φ (cid:18) e − imπa , , aβ π + 34 (cid:19) + e iπma (cid:18) iπm Φ (cid:18) e − imπa , , aβ + π π (cid:19) + a Φ (cid:18) e − imπa , , aβ + π π (cid:19)(cid:19) − a Φ (cid:18) e − imπa , , aβ π + 34 (cid:19) + e iπma (cid:18) a Φ (cid:18) e imπa , , aβ + π π (cid:19) − iπm Φ (cid:18) e imπa , , aβ + π π (cid:19)(cid:19) + ie iπma (cid:18) πm Φ (cid:18) e imπa , , aβ π + 34 (cid:19) + ia Φ (cid:18) e imπa , , aβ π + 34 (cid:19)(cid:19)(cid:19) (28)from equation (9.550) in [1]. Next we set m = 0 simplifying in terms of theTrigamma function ψ (1) ( z ) to get (29) Z ∞ x tanh( ax )sech( ax ) β + x dx = ψ (1) (cid:16) aβ + π π (cid:17) − ψ (1) (cid:16) aβ π + (cid:17) π from equation (64:4:1) in [11]. Using equation (12) and setting k = − , t = π/ α by e iβ simpli-fying we get (30) Z ∞ (cid:18) x + iβ ) + 1( x − iβ ) (cid:19) tanh( ax )sech( ax ) sinh( mx ) dx = e − iπm a π (cid:18) πm Φ (cid:18) e − imπa , , aβ π + 34 (cid:19) + e iπma (cid:18) ia Φ (cid:18) e − imπa , , aβ + π π (cid:19) − πm Φ (cid:18) e − imπa , , aβ + π π (cid:19)(cid:19) − ia Φ (cid:18) e − imπa , , aβ π + 34 (cid:19) − e iπma (cid:18) πm Φ (cid:18) e imπa , , aβ + π π (cid:19) + ia Φ (cid:18) e imπa , , aβ + π π (cid:19)(cid:19) + e iπma (cid:18) πm Φ (cid:18) e imπa , , aβ π + 34 (cid:19) + ia Φ (cid:18) e imπa , , aβ π + 34 (cid:19)(cid:19)(cid:19) m and setting m = 0simplifying to get Z ∞ x ( x − β )( β + x ) tanh( ax )sech( ax )( β + x ) dx = πψ (1) (cid:16) aβ + π π (cid:17) − πψ (1) (cid:16) aβ π + (cid:17) + aβ (cid:16) ζ (cid:16) , aβ π + (cid:17) − ζ (cid:16) , aβ + π π (cid:17)(cid:17) π (31)from equations (64:12:1) (64:13:3) and (64:4:1) in [11].
13 Discussion
In this article we derived the integrals of hyperbolic and logarithmic functionsin terms of the Lerch function. Then we used these integral formula to de-rive known and new results. We were able to produce a formal derivation forequation (27) Table 27 in Bierens de Haan [9] and equation (3.514.4) in [1] notpreviously published. The results presented were numerically verified for bothreal and imaginary values of the parameters in the integrals using Mathematicaby Wolfram. In this work we used Mathematica software to numerically evalu-ate both the definite integral and associated Special function for complex valuesof the parameters k , α , a , m and t . We considered various ranges of theseparameters for real, integer, negative and positive values. We compared theevaluation of the definite integral to the evaluated Special function and ensuredagreement.
14 Conclusion
In this paper, we have derived a method for expressing definite integrals in termsof Special functions using contour integration. The contour we used was specificto solving integral representations in terms of the Hurwitz zeta function. Weexpect that other contours and integrals can be derived using this method.
15 Acknowledgments
This paper is fully supported by the Natural Sciences and Engineering ResearchCouncil (NSERC) Grant No. 504070.
References [1] Gradshteyn I.S & Ryzhik I.M,
Tables of Integrals, Series and Products, 6Ed , Academic Press (2000), USA.132] Wang, Z.X., Guo, D.R.
Special Functions , World Scientific (1989).[3] Krantz, S.G.
Handbook of Complex Variables , Springer Science, New York(1999).[4] Prodanov, D.,
Regularized Integral Representations of the ReciprocalGamma Function.
Fractal Fract , 3, 1.[5] Reynolds, R.; Stauffer, A.,
Definite Integral of Arctangent and Polyloga-rithmic Functions Expressed as a Series.
Mathematics , 7, 1099.[6] Reynolds, R.; Stauffer, A.
A Definite Integral Involving the LogarithmicFunction in Terms of the Lerch Function.
Mathematics , 7, 1148.[7] Reynolds, R.; Stauffer, A.
Derivation of Logarithmic and Logarithmic Hy-perbolic Tangent Integrals Expressed in Terms of Special Functions.
Math-ematics , 8, 687.[8] Reynolds, R.; Stauffer, A.
A Method for Evaluating Definite Integrals in-Terms of Special Functions with Examples
International MathematicalForum, Vol. 15, , no. 5, 235- 244[9] Bierens de Haan, D.,
Nouvelles Tables d’int´egrales d´efinies , Amsterdam,1867[10] Choi, Junesang. and Srivastava, H.M.,
A family of log-gamma integralsand associated results , Journal of Mathematical Analysis and Applica-tions, 303 (2005) 436-449, Science Direct, Elsevier.[11] Jan Myland, Keith B Oldham, Jerome Spanier,
An Atlas of Functions:With Equator, the Atlas Function Calculator , Springer; 2nd ed. 2009 edi-tion (Dec 2 2008)[12] Whittaker, E. T. and Watson, G. N.
A Course in Modern Analysis , 4thed. Cambridge, England: Cambridge University Press.[13] Yu. A. Brychkov, O.I. Marichev, N.V. Savischenko,
Handbook of MellinTransforms , CRC Press, Taylor & Francis Group, Boca Raton, FL., (2019)[14] Prudnikov, A.P., Brychkov, Yu. A., Marichev, O.I.
Integrals and Series,More Special Functions , USSR Academy of Sciences, Vol. 1, Moscow(1990).[15] Abramowitz, M. and Stegun, I.A.(Eds),