Citation: Robert Reynolds, Allan Stauffer. Definite integral of the logarithm hyperbolic secant function in terms of the Hurwitz zeta function[J]. AIMS Mathematics, 2021, 6(2): 1324-1331. doi: 10.3934/math.2021082
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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 [5]. In 1867 David Bierens de Haan derived hyperbolic integrals of the form
∫∞0((log(a)−ix)k+(log(a)+ix)k)log(cos(α)sech(x)+1)dx | (1.1) |
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 [6]. The generalized Cauchy's integral formula is given by
ykk!=12πi∫Cewywk+1dw. | (1.2) |
We use the method in [6]. Here the contour is in the upper left quadrant with ℜ(w)<0 and going round the origin with zero radius. Using a generalization of Cauchy's integral formula we first replace y by ix+log(a) for the first equation and then y by −ix+log(a) to get the second equation. Then we add these two equations, followed by multiplying both sides by 12log(cos(α)sech(x)+1) to get
((log(a)−ix)k+(log(a)+ix)k)log(cos(α)sech(x)+1)2k!=12πi∫Caww−k−1cos(wx)log(cos(α)sech(x)+1)dw | (2.1) |
where the logarithmic function is defined in Eq (4.1.2) in [2]. We then take the definite integral over x∈[0,∞) of both sides to get
∫∞0((log(a)−ix)k+(log(a)+ix)k)log(cos(α)sech(x)+1)2k!dx=12πi∫∞0∫Caww−k−1cos(wx)log(cos(α)sech(x)+1)dwdx=12πi∫C∫∞0aww−k−1cos(wx)log(cos(α)sech(x)+1)dxdw=12πi∫Cπaww−k−2cosh(πw2)csch(πw)dw−12πi∫Cπaww−k−2csch(πw)cosh(αw)dw | (2.2) |
from Eq (1.7.7.120) in [1] and the integral is valid for α, a, and k complex and |ℜ(α)|<π.
In this section we will again use the generalized Cauchy's integral formula to derive equivalent contour integrals. First we replace y by y−π/2 for the first equation and y by y+π/2 for second then add these two equations to get
(y−π2)k+(y+π2)kk!=12πi∫C2w−k−1ewycosh(πw2)dw | (3.1) |
Next we replace y by log(a)+π(2p+1) then we take the infinite sum over p∈[0,∞) to get
∞∑p=02π((log(a)+π(2p+1)−π2)k+(log(a)+π(2p+1)+π2)k)k!=12πi∞∑p=0∫C4πw−k−1cosh(πw2)ew(log(a)+π(2p+1))dw=12πi∫C∞∑p=04πw−k−1cosh(πw2)ew(log(a)+π(2p+1))dw | (3.2) |
where ℜ(w)<0 according to (1.232.3) in [5]. Then we simplify the left-hand side to get the Hurwitz zeta function
−2k+1πk+2(k+1)!(ζ(−k−1,2log(a)+π4π)+ζ(−k−1,2log(a)+3π4π))=12πi∫Cπaww−k−2cosh(πw2)csch(πw)dw | (3.3) |
Then following the procedure of (3.1) and (3.2) we replace y by y+α and y−α to get the second equation for the contour integral given by
(y−α)k+(α+y)kk!=12πi∫C2w−k−1ewycosh(αw)dw | (3.4) |
next we replace y by log(a)+π(2p+1) and take the infinite sum over p∈[0,∞) to get
∞∑p=0(−α+log(a)+π(2p+1))k+(α+log(a)+π(2p+1))kk!=12πi∞∑p=0∫C2w−k−1cosh(αw)ew(log(a)+π(2p+1))dw=12πi∫C∞∑p=02w−k−1cosh(αw)ew(log(a)+π(2p+1))dw | (3.5) |
Then we simplify to get
2k+1πk+2(k+1)!(ζ(−k−1,−α+log(a)+π2π)+ζ(−k−1,α+log(a)+π2π))=12πi∫Cπ(−aw)w−k−2csch(πw)cosh(αw)dw | (3.6) |
Since the right-hand sides of Eqs (2.2), (3.3) and (3.5) are equivalent we can equate the left-hand sides to get
∫∞0((log(a)−ix)k+(log(a)+ix)k)log(cos(α)sech(x)+1)dx=2(2k+1πk+2(ζ(−k−1,−α+log(a)+π2π)+ζ(−k−1,α+log(a)+π2π))k+1)−2(2k+1πk+2(ζ(−k−1,2log(a)+π4π)+ζ(−k−1,2log(a)+3π4π))k+1) | (4.1) |
from (9.521) in [5] where ζ(z,q) is the Hurwitz zeta function. Note the left-hand side of Eq (4.1) converges for all finite k. The integral in Eq (4.1) can be used as an alternative method to evaluating the Hurwitz zeta function. The Hurwitz zeta function has a series representation given by
ζ(z,q)=∞∑n=01(q+n)z | (4.2) |
where ℜ(z)>1,q≠0,−1,.. and is continued analytically by (9.541.1) in [5] where z=1 is the only singular point.
In this section we have evaluated integrals and extended the range of the parameters over which the integrals are valid. The aim of this section is to derive a few integrals in [5] in terms of the Lerch function. We also present errata for one of the integrals and faster converging closed form solutions.
Using Eq (4.1) and taking the first partial derivative with respect to α and setting a=1 and simplifying the left-hand side we get
∫∞0xkcos(α)+cosh(x)dx=2kπk+1csc(α)sec(πk2)(ζ(−k,π−α2π)−ζ(−k,α+π2π)) | (5.1) |
from Eq (7.102) in [3].
Using Eq (5.1) and taking the first partial derivative with respect to k and setting k=0 and simplifying the left-hand side we get
∫∞0log(x)cos(α)+cosh(x)dx=csc(α)(αlog(2π)−πlog(−α−π)+πlog(α−π)−πlogΓ(−α+π2π)+πlogΓ(−π−α2π)))=csc(α)(αlog(2π)+πlog(Γ(α+π2π)Γ(π−α2π))) | (5.2) |
from (7.105) in [3].
Using Eq (5.2) and setting α=π/2 and simplifying we get
∫∞0log(x)sech(x)dx=πlog(√2πΓ(34)Γ(14)) | (5.3) |
Using Eq (5.2) and taking the first derivative with respect to α and setting α=π/2 and simplifying we get
∫∞0log(x)sech2(x)dx=log(π4)−γ | (5.4) |
where γ is Euler's constant.
Using Eq (5.1) and taking the first partial derivative with respect to k then setting k=−1/2 and α=π/2 and simplifying we get
∫∞0log(x)sech(x)√xdx=12√π(−2ζ′(12,14)+2ζ′(12,34)+(ζ(12,34)−ζ(12,14))(π+log(14π2))) | (5.5) |
The expression in [4] is correct but converges much slower than Eq (5.5).
Using Eq (5.1) and taking the first partial derivative with respect to α we get
∫∞0xk(cos(α)+cosh(x))2dx=−2k−1πkcsc2(α)sec(πk2)(k(ζ(1−k,π−α2π)+ζ(1−k,α+π2π)))−2k−1πkcsc2(α)sec(πk2)(2πcot(α)(ζ(−k,π−α2π)−ζ(−k,α+π2π))) | (5.6) |
from (7.102) in [3]. Next we use L'Hopital's rule and take the limit as α→0 to get
∫∞0xk(cosh(x)+1)2dx=−1321−k((2k−8)ζ(k−2)−(2k−2)ζ(k))Γ(k+1) | (5.7) |
Then we take the first partial derivative with respect to k to get
∫∞0xklog(x)(cosh(x)+1)2dx=1324−kkΓ(k)ζ′(k−2)−23kΓ(k)ζ′(k−2)−1322−kkΓ(k)ζ′(k)+23kΓ(k)ζ′(k)−1321−kklog(256)ζ(k−2)Γ(k)+1321−kklog(4)ζ(k)Γ(k)+1324−kkζ(k−2)Γ(k)ψ(0)(k+1)−23kζ(k−2)Γ(k)ψ(0)(k+1)−1322−kkζ(k)Γ(k)ψ(0)(k+1)+23kζ(k)Γ(k)ψ(0)(k+1) | (5.8) |
Finally we set k=0 to get
∫∞0log(x)(cosh(x)+1)2dx=13(14ζ′(−2)−γ+log(π2)) | (5.9) |
The integral listed in [4] appears with an error in the integrand.
Using Eq (4.1) we first take the limit as k→−1 by applying L'Hopital's rule and using (7.105) in [3] and simplifying the right-hand side we get
∫∞0log(cos(α)sech(x)+1)a2+x2dx=πalog(√π212−aπΓ(aπ+12)Γ(a−α+π2π)Γ(a+α+π2π)) | (5.10) |
Next we take the first partial derivative with respect to α and set α=0 to get
∫∞0sech(x)a2+x2dx=−ψ(0)(2a+π4π)−ψ(0)(2a+3π4π)2a | (5.11) |
from Eq (8.360.1) in [5] where ℜ(a)>0, next we replace x with bx to get
∫∞0sech(bx)a2+b2x2dx=−ψ(0)(2a+π4π)−ψ(0)(2a+3π4π)2ab | (5.12) |
Next we set a=b=π to get
∫∞0sech(πx)x2+1dx=12(ψ(0)(54)−ψ(0)(34))=2−π2 | (5.13) |
from Eq (8.363.8) in [5].
Using Eq (5.12) and setting a=b=π/2 we get
∫∞0sech(πx2)x2+1dx=12(−γ−ψ(0)(12))=log(2) | (5.14) |
from Eq (8.363.8) in [5].
Using Eq (5.12) and setting a=b=π/4 we get
∫∞0sech(πx4)x2+1dx=12(ψ(0)(78)−ψ(0)(38))=π−2coth−1(√2)√2 | (5.15) |
from Eq (8.363.8) in [5].
Using Eq (5.10) and taking the second partial derivative with respect to α we get
∫∞0sech2(x)a2+x2dx=ψ(1)(aπ+12)πa | (5.16) |
from Eq (8.363.8) in [5] where ℜ(a)>0.
In this paper we were able to present errata and express our closed form solutions in terms of special functions and fundamental constants such π, Euler's constant and log(2). The use of the trigamma function is quite often necessary in statistical problems involving beta or gamma distributions. This work provides both an accurate and extended range for the solutions of the integrals derived.
We have presented a novel method for deriving some interesting definite integrals by Bierens de Haan using contour integration. The results presented were numerically verified for both real and imaginary and complex values of the parameters in the integrals using Mathematica by Wolfram.
This research is supported by Natural Sciences and Engineering Research Council of Canada NSERC Canada under Grant 504070.
The authors declare there are no conflicts of interest.
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