Research article

Definite integral of the logarithm hyperbolic secant function in terms of the Hurwitz zeta function

  • We evaluate definite integrals of the form given by 0R(a,x)log(cos(α)sech(x)+1)dx. The function R(a,x) is a rational function with general complex number parameters. Definite integrals of this form yield closed forms for famous integrals in the books of Bierens de Haan [4] and Gradshteyn and Ryzhik [5].

    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 evaluate definite integrals of the form given by 0R(a,x)log(cos(α)sech(x)+1)dx. The function R(a,x) is a rational function with general complex number parameters. Definite integrals of this form yield closed forms for famous integrals in the books of Bierens de Haan [4] and Gradshteyn and Ryzhik [5].


    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!=12πiCewywk+1dw. (1.2)

    We use the method in [6]. Here the contour is in the upper left quadrant with (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!=12πiCawwk1cos(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=12πi0Cawwk1cos(wx)log(cos(α)sech(x)+1)dwdx=12πiC0awwk1cos(wx)log(cos(α)sech(x)+1)dxdw=12πiCπawwk2cosh(πw2)csch(πw)dw12πiCπawwk2csch(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 |(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 yπ/2 for the first equation and y by y+π/2 for second then add these two equations to get

    (yπ2)k+(y+π2)kk!=12πiC2wk1ewycosh(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!=12πip=0C4πwk1cosh(πw2)ew(log(a)+π(2p+1))dw=12πiCp=04πwk1cosh(πw2)ew(log(a)+π(2p+1))dw (3.2)

    where (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(k1,2log(a)+π4π)+ζ(k1,2log(a)+3π4π))=12πiCπawwk2cosh(πw2)csch(πw)dw (3.3)

    Then following the procedure of (3.1) and (3.2) we replace y by y+α and yα to get the second equation for the contour integral given by

    (yα)k+(α+y)kk!=12πiC2wk1ewycosh(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!=12πip=0C2wk1cosh(αw)ew(log(a)+π(2p+1))dw=12πiCp=02wk1cosh(αw)ew(log(a)+π(2p+1))dw (3.5)

    Then we simplify to get

    2k+1πk+2(k+1)!(ζ(k1,α+log(a)+π2π)+ζ(k1,α+log(a)+π2π))=12πiCπ(aw)wk2csch(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(k1,α+log(a)+π2π)+ζ(k1,α+log(a)+π2π))k+1)2(2k+1πk+2(ζ(k1,2log(a)+π4π)+ζ(k1,2log(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 (z)>1,q0,1,.. 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,πα2π)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(απ)+πlog(απ)πlogΓ(α+π2π)+πlogΓ(πα2π)))=csc(α)(αlog(2π)+πlog(Γ(α+π2π)Γ(πα2π))) (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=12π(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=2k1πkcsc2(α)sec(πk2)(k(ζ(1k,πα2π)+ζ(1k,α+π2π)))2k1πkcsc2(α)sec(πk2)(2πcot(α)(ζ(k,πα2π)ζ(k,α+π2π))) (5.6)

    from (7.102) in [3]. Next we use L'Hopital's rule and take the limit as α0 to get

    0xk(cosh(x)+1)2dx=1321k((2k8)ζ(k2)(2k2)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=1324kkΓ(k)ζ(k2)23kΓ(k)ζ(k2)1322kkΓ(k)ζ(k)+23kΓ(k)ζ(k)1321kklog(256)ζ(k2)Γ(k)+1321kklog(4)ζ(k)Γ(k)+1324kkζ(k2)Γ(k)ψ(0)(k+1)23kζ(k2)Γ(k)ψ(0)(k+1)1322kkζ(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 k1 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(π212aπΓ(aπ+12)Γ(aα+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 (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))=2π2 (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(γψ(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))=π2coth1(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 (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.



    [1] F. Oberhettinger, Tables of Fourier Transforms and Fourier Transforms of Distributions, 1st ed.; Springer-Verlag, Berlin Heidelberg, 1990.
    [2] M. Abramowitz, I. A. Stegun (Eds), Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing, New York, Dover, 1982.
    [3] D. H. Bailey, J. M. Borwein, N. J. Calkin, R. Girgensohn, D. R. Luke, V. H. Moll, Experimental Mathematics in Action, Wellesley, MA: A K Peters, 2007.
    [4] D. Bierens de Haan, Nouvelles Tables d'intégrales définies, Amsterdam, 1867
    [5] I. S. Gradshteyn, I. M. Ryzhik, Tables of Integrals, Series and Products, 6 Ed, Academic Press, USA, 2000.
    [6] R. Reynolds, A. Stauffer, A Method for Evaluating Definite Integrals in Terms of Special Functions with Examples, International Mathematical Forum, 15 (2020), 235-244. doi: 10.12988/imf.2020.91272
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