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An Alternate proof for a case of a Malmsten integral

Abdulhafeez Ayinde Abdulsalam
Abstract

In this paper, a direct proof is presented for a case of a Malmsten integral. The method used in solving the integral is a direct one that the author has not come accross in any old or recent publication. Integration by parts, Laplace transform, an integral representation for the hyperbolic secant function, and the digamma representation for an alternating series are employed to derive the result.

AMS Subject Classification: 33-XX, 26A42.

Keywords: Laplace transform, logarithmic integrals, integration by parts, gamma function, digamma function, rediscovery, direct proof, Euler, Carl Malmsten, Vardi, Iaroslav Blagouchine, colleagues.

1 Introduction

The integral evaluated in this paper is a case of the logarithmic integrals recently treated in Mathematical literature. Iaroslav V. Blagouchine in 2014 argued that the problem to be presented in this article was actually more older than reported (see [1]). He further clarified with sufficient and very convincing evidences that the so-called Vardi’s integral (see [2]) was actually a particular case of the considered family of integrals, first evaluated by Carl Malmsten and colleagues in 1842 (see [3]). Blagouchine in his fascinating article on the rediscovery of Malmsten’s integrals presented several generalizations of the integral by contour integration method (see [1]). In this work, a direct proof is established while abstaining from methods used by Malmsten, Vardi, or Blagouchine. We hereby begin with the following proposition.

{proposition*}
  1. (a)

    For a∈ℝa\in\mathbb{R}

    ∫0∞ln⁡(x2+a2)cosh⁡(π​x)=2​ln⁡(2​Γ​(|a|2+34)Γ​(|a|2+14)).\int_{0}^{\infty}\frac{\ln\left(x^{2}+a^{2}\right)}{\cosh\left(\pi x\right)}=2\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{\left|a\right|}{2}+\frac{3}{4}\right)}{\Gamma\left(\frac{\left|a\right|}{2}+\frac{1}{4}\right)}\right).
  2. (b)
    ∫0∞ln⁡x​sech⁡(x)​d​x=π​ln⁡(2​π​Γ​(34)Γ​(14))=π​ln⁡(2​π32Γ​(14)2).\int_{0}^{\infty}\ln{x}\operatorname{sech}\left(x\right)\mathrm{d}x=\pi\ln\left(\frac{\sqrt{2\pi}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)=\pi\ln\left(\frac{2\pi^{\frac{3}{2}}}{\Gamma\left(\frac{1}{4}\right)^{2}}\right).
  3. (c)

    For a∈ℝ+a\in\mathbb{R}^{+}, ℜ⁡(b)>0\Re(b)>0

    ∫0∞ln⁡(a​x)​sech⁡(b​x)​dx=πb​ln⁡(2​a​π32b​Γ​(14)2).\int_{0}^{\infty}\ln\left(ax\right)\operatorname{sech}\left(bx\right)\,\mathrm{d}x=\frac{\pi}{b}\ln\left(\frac{2\sqrt{a}\pi^{\frac{3}{2}}}{\sqrt{b}\Gamma\left(\frac{1}{4}\right)^{2}}\right).
Proof.

Let Δ​(a)=∫0∞ln⁡(x2+a2)cosh⁡(π​x)​dx\displaystyle\Delta\left(a\right)=\int_{0}^{\infty}\frac{\ln\left(x^{2}+a^{2}\right)}{\cosh\left(\pi x\right)}\,\mathrm{d}x.

Then

Δ​(a)\displaystyle\Delta\left(a\right) =∫0∞ln⁡(|a|−i​x)cosh⁡(π​x)​dx+∫0∞ln⁡(|a|+i​x)cosh⁡(π​x)​dx\displaystyle=\int_{0}^{\infty}\frac{\ln\left(\left|a\right|-ix\right)}{\cosh\left(\pi x\right)}\,\mathrm{d}x+\int_{0}^{\infty}\frac{\ln\left(\left|a\right|+ix\right)}{\cosh\left(\pi x\right)}\,\mathrm{d}x

where i=−1i=\sqrt{-1}.

Δ​(a)\displaystyle\Delta\left(a\right) =2​∫0∞ln⁡(|a|−i​x)e−2​π​x+1​e−π​x​dx+2​∫0∞ln⁡(|a|+i​x)e−2​π​x+1​e−π​x​dx\displaystyle=2\int_{0}^{\infty}\frac{\ln\left(\left|a\right|-ix\right)}{e^{-2\pi x}+1}e^{-\pi x}\,\mathrm{d}x+2\int_{0}^{\infty}\frac{\ln\left(\left|a\right|+ix\right)}{e^{-2\pi x}+1}e^{-\pi x}\,\mathrm{d}x
=−2π​∫0∞ln⁡(|a|−i​x)​d​(arctan⁡(e−π​x))\displaystyle=-\frac{2}{\pi}\int_{0}^{\infty}\ln\left(\left|a\right|-ix\right)\,\,\mathrm{d}\left(\arctan\left(e^{-\pi x}\right)\right)
−2π​∫0∞ln⁡(|a|+i​x)​d​(arctan⁡(e−π​x))\displaystyle\qquad\qquad-\frac{2}{\pi}\int_{0}^{\infty}\ln\left(\left|a\right|+ix\right)\,\,\mathrm{d}\left(\arctan\left(e^{-\pi x}\right)\right)
=−2​iπ​∫0∞arctan⁡(e−π​x)|a|−i​x​dx+2​iπ​∫0∞arctan⁡(e−π​x)|a|+i​x​dx+ln⁡a.\displaystyle=\frac{-2i}{\pi}\int_{0}^{\infty}\frac{\arctan\left(e^{-\pi x}\right)}{\left|a\right|-ix}\,\mathrm{d}x+\frac{2i}{\pi}\int_{0}^{\infty}\frac{\arctan\left(e^{-\pi x}\right)}{\left|a\right|+ix}\,\mathrm{d}x+\ln{a}.

It follows by Laplace transform that

Δ​(a)−ln⁡a\displaystyle\Delta\left(a\right)-\ln{a} =−2​iπ​∫0∞arctan⁡(e−π​x)​∫0∞e−t​(|a|−i​x)​dt​dx\displaystyle=\frac{-2i}{\pi}\int_{0}^{\infty}\arctan\left(e^{-\pi x}\right)\int_{0}^{\infty}e^{-t\left(\left|a\right|-ix\right)}\,\,\mathrm{d}t\,\mathrm{d}x
+2​iπ​∫0∞arctan⁡(e−π​x)​∫0∞e−t​(|a|+i​x)​dt​dx\displaystyle\qquad\qquad+\frac{2i}{\pi}\int_{0}^{\infty}\arctan\left(e^{-\pi x}\right)\int_{0}^{\infty}e^{-t\left(\left|a\right|+ix\right)}\,\,\mathrm{d}t\,\mathrm{d}x
=−2​iπ​∫0∞e−|a|​t​∫0∞ei​t​x​arctan⁡(e−π​x)​dx​dt\displaystyle=\frac{-2i}{\pi}\int_{0}^{\infty}e^{-\left|a\right|t}\int_{0}^{\infty}e^{itx}\arctan\left(e^{-\pi x}\right)\mathrm{d}x\,\,\mathrm{d}t
+2​iπ​∫0∞e−|a|​t​∫0∞e−i​t​x​arctan⁡(e−π​x)​dx​dt.\displaystyle\qquad\qquad+\frac{2i}{\pi}\int_{0}^{\infty}e^{-\left|a\right|t}\int_{0}^{\infty}e^{-itx}\arctan\left(e^{-\pi x}\right)\mathrm{d}x\,\,\mathrm{d}t.
Δ​(a)−ln⁡a=−2​iπ​∫0∞e−|a|​t​∫0∞(ei​t​x−e−i​t​x)​arctan⁡(e−π​x)​dx​dt.\displaystyle\Delta\left(a\right)-\ln{a}=\frac{-2i}{\pi}\int_{0}^{\infty}e^{-\left|a\right|t}\int_{0}^{\infty}\left(e^{itx}-e^{-itx}\right)\arctan\left(e^{-\pi x}\right)\mathrm{d}x\,\,\mathrm{d}t.

By Euler’s formula,

ei​t​x−e−i​t​x=2​i​sin⁡(t​x).e^{itx}-e^{-itx}=2i\sin{\left(tx\right)}.

Therefore

Δ​(a)−ln⁡a\displaystyle\Delta\left(a\right)-\ln{a} =4π​∫0∞e−|a|​t​∫0∞sin⁡(t​x)​arctan⁡(e−π​x)​dx​dt\displaystyle=\frac{4}{\pi}\int_{0}^{\infty}e^{-\left|a\right|t}\int_{0}^{\infty}\sin{\left(tx\right)}\arctan\left(e^{-\pi x}\right)\mathrm{d}x\,\,\mathrm{d}t
=4π​∫0∞e−|a|​t​∫0∞d​(−cos⁡(t​x)t)​arctan⁡(e−π​x)​dt\displaystyle=\frac{4}{\pi}\int_{0}^{\infty}e^{-\left|a\right|t}\int_{0}^{\infty}\mathrm{d}\left(\frac{-\cos{\left(tx\right)}}{t}\right)\arctan\left(e^{-\pi x}\right)\,\,\mathrm{d}t
=4π​∫0∞e−|a|​t​(π4​t−π2​t​∫0∞cos⁡(t​x)cosh⁡(π​x)​dx)​dt\displaystyle=\frac{4}{\pi}\int_{0}^{\infty}e^{-\left|a\right|t}\left(\frac{\pi}{4t}-\frac{\pi}{2t}\int_{0}^{\infty}\frac{\cos{\left(tx\right)}}{\cosh\left(\pi x\right)}\,\mathrm{d}x\right)\,\,\mathrm{d}t
=4π​∫0∞e−|a|​t​(π4​t−12​t​∫0∞cos⁡(t​xπ)cosh⁡(x)​dx)​dt\displaystyle=\frac{4}{\pi}\int_{0}^{\infty}e^{-\left|a\right|t}\left(\frac{\pi}{4t}-\frac{1}{2t}\int_{0}^{\infty}\frac{\cos{\left(\frac{tx}{\pi}\right)}}{\cosh\left(x\right)}\,\mathrm{d}x\right)\,\,\mathrm{d}t
=4π​∫0∞e−|a|​t​(π4​t−π4​t​sech⁡(t2))​dt\displaystyle=\frac{4}{\pi}\int_{0}^{\infty}e^{-\left|a\right|t}\left(\frac{\pi}{4t}-\frac{\pi}{4t}\operatorname{sech}\left(\frac{t}{2}\right)\right)\,\,\mathrm{d}t
=∫0∞e−|a|​t​(1t−1t​sech⁡(t2))​dt\displaystyle=\int_{0}^{\infty}e^{-\left|a\right|t}\left(\frac{1}{t}-\frac{1}{t}\operatorname{sech}\left(\frac{t}{2}\right)\right)\,\,\mathrm{d}t
=∫0∞(e−|a|​tt−2​e−(|a|+12)​tt​(1+e−t))​dt=t→2​t∫0∞(e−2​|a|​tt−2​e−(2​|a|+1)​tt​(1+e−2​t))​dt\displaystyle=\int_{0}^{\infty}\left(\frac{e^{-\left|a\right|t}}{t}-\frac{2e^{-\left(\left|a\right|+\frac{1}{2}\right)t}}{t\left(1+e^{-t}\right)}\right)\,\,\mathrm{d}t{\stackrel{{\scriptstyle\,\,t\rightarrow 2t}}{{=}}}\int_{0}^{\infty}\left(\frac{e^{-2\left|a\right|t}}{t}-\frac{2e^{-\left(2\left|a\right|+1\right)t}}{t\left(1+e^{-2t}\right)}\right)\,\,\mathrm{d}t
=z→e−t−∫01(z2​|a|−2​z2​|a|+11+z2)​d​zz​ln⁡z=−∫01z2​|a|ln⁡z​(1z−21+z2)​dz\displaystyle{\stackrel{{\scriptstyle z\rightarrow e^{-t}}}{{=}}}-\int_{0}^{1}\left(z^{2\left|a\right|}-\frac{2z^{2\left|a\right|+1}}{1+z^{2}}\right)\,\,\frac{\mathrm{d}z}{z\ln{z}}=-\int_{0}^{1}\frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{1}{z}-\frac{2}{1+z^{2}}\right)\,\,\mathrm{d}z
=−∫01z2​|a|ln⁡z​(1+z2−2​zz​(1+z2))​dz=−∫01z2​|a|ln⁡z​((1−z)2z​(1+z2))​dz\displaystyle=-\int_{0}^{1}\frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{1+z^{2}-2z}{z\left(1+z^{2}\right)}\right)\,\,\mathrm{d}z=-\int_{0}^{1}\frac{z^{2\left|a\right|}}{\ln{z}}\left(\frac{\left(1-z\right)^{2}}{z\left(1+z^{2}\right)}\right)\,\,\mathrm{d}z
=∫01z2​|a|−1​(1−z)1+z2​∫01zp​dp​dz=∫01∫01z2​|a|+p−1​(1−z)1+z2​dp​dz\displaystyle=\int_{0}^{1}\frac{z^{2\left|a\right|-1}\left(1-z\right)}{1+z^{2}}\int_{0}^{1}z^{p}\mathrm{d}p\,\,\mathrm{d}z=\int_{0}^{1}\int_{0}^{1}\frac{z^{2\left|a\right|+p-1}\left(1-z\right)}{1+z^{2}}\mathrm{d}p\,\,\mathrm{d}z
=∫01∫01∑k=0∞(−1)k​z2​|a|+p+2​k−1​(1−z)​d​z​d​p\displaystyle=\int_{0}^{1}\int_{0}^{1}\sum_{k=0}^{\infty}\left(-1\right)^{k}z^{2\left|a\right|+p+2k-1}\left(1-z\right)\mathrm{d}z\,\,\mathrm{d}p
=∫01∑k=0∞(−1)k​∫01z2​|a|+p+2​k−1​(1−z)​dz​dp\displaystyle=\int_{0}^{1}\sum_{k=0}^{\infty}\left(-1\right)^{k}\int_{0}^{1}z^{2\left|a\right|+p+2k-1}\left(1-z\right)\mathrm{d}z\,\,\mathrm{d}p
=∫01∑k=0∞(−1)k​(12​|a|+p+2​k−12​|a|+p+2​k+1)​d​p\displaystyle=\int_{0}^{1}\sum_{k=0}^{\infty}\left(-1\right)^{k}\left(\frac{1}{2\left|a\right|+p+2k}-\frac{1}{2\left|a\right|+p+2k+1}\right)\,\mathrm{d}p
=12​∫01∑k=0∞(−1)k​(1k+2​|a|+p2−1k+2​|a|+p+12)​d​p\displaystyle=\frac{1}{2}\int_{0}^{1}\sum_{k=0}^{\infty}\left(-1\right)^{k}\left(\frac{1}{k+\frac{2\left|a\right|+p}{2}}-\frac{1}{k+\frac{2\left|a\right|+p+1}{2}}\right)\,\mathrm{d}p
=−14∫01(ψ0(2​|a|+p4)−ψ0(2​|a|+p4+12)\displaystyle=-\frac{1}{4}\int_{0}^{1}\left(\psi_{0}\left(\frac{2\left|a\right|+p}{4}\right)-\psi_{0}\left(\frac{2\left|a\right|+p}{4}+\frac{1}{2}\right)\right.
−ψ0(2​|a|+p+14)+ψ0(2​|a|+p+14+12))dp\displaystyle\qquad\qquad\qquad\left.-\psi_{0}\left(\frac{2\left|a\right|+p+1}{4}\right)+\psi_{0}\left(\frac{2\left|a\right|+p+1}{4}+\frac{1}{2}\right)\right)\mathrm{d}p
=−ln⁡(Γ​(2​|a|+p4)​Γ​(2​|a|+p+14+12)Γ​(2​|a|+p4+12)​Γ​(2​|a|+p+14))|01\displaystyle=-\ln\left(\frac{\Gamma\left(\frac{2\left|a\right|+p}{4}\right)\Gamma\left(\frac{2\left|a\right|+p+1}{4}+\frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right|+p}{4}+\frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right|+p+1}{4}\right)}\right)\biggr{|}_{0}^{1}
=−ln⁡(Γ​(2​|a|+14)​Γ​(2​|a|+24+12)Γ​(2​|a|+14+12)​Γ​(2​|a|+24))+ln⁡(Γ​(2​|a|4)​Γ​(2​|a|+14+12)Γ​(2​|a|4+12)​Γ​(2​|a|+14))\displaystyle=-\ln\left(\frac{\Gamma\left(\frac{2\left|a\right|+1}{4}\right)\Gamma\left(\frac{2\left|a\right|+2}{4}+\frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right|+1}{4}+\frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right|+2}{4}\right)}\right)+\ln\left(\frac{\Gamma\left(\frac{2\left|a\right|}{4}\right)\Gamma\left(\frac{2\left|a\right|+1}{4}+\frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right|}{4}+\frac{1}{2}\right)\Gamma\left(\frac{2\left|a\right|+1}{4}\right)}\right)
=−ln⁡(Γ​(2​|a|+14)2​Γ​(2​|a|+24+12)Γ​(2​|a|+14+12)2​Γ​(2​|a|+24))+ln⁡(Γ​(2​|a|4)Γ​(2​|a|4+12))\displaystyle=-\ln\left(\frac{\Gamma\left(\frac{2\left|a\right|+1}{4}\right)^{2}\Gamma\left(\frac{2\left|a\right|+2}{4}+\frac{1}{2}\right)}{\Gamma\left(\frac{2\left|a\right|+1}{4}+\frac{1}{2}\right)^{2}\Gamma\left(\frac{2\left|a\right|+2}{4}\right)}\right)+\ln\left(\frac{\Gamma\left(\frac{2\left|a\right|}{4}\right)}{\Gamma\left(\frac{2\left|a\right|}{4}+\frac{1}{2}\right)}\right)
=−ln⁡(Γ​(2​|a|+14)2​Γ​(|a|2+1)Γ​(2​|a|+14+12)2​Γ​(2​|a|+24))+ln⁡(Γ​(|a|2)Γ​(2​|a|+24))\displaystyle=-\ln\left(\frac{\Gamma\left(\frac{2\left|a\right|+1}{4}\right)^{2}\Gamma\left(\frac{\left|a\right|}{2}+1\right)}{\Gamma\left(\frac{2\left|a\right|+1}{4}+\frac{1}{2}\right)^{2}\Gamma\left(\frac{2\left|a\right|+2}{4}\right)}\right)+\ln\left(\frac{\Gamma\left(\frac{\left|a\right|}{2}\right)}{\Gamma\left(\frac{2\left|a\right|+2}{4}\right)}\right)
=−ln⁡(|a|​Γ​(2​|a|+14)22​Γ​(2​|a|+14+12)2)=2​ln⁡(2​Γ​(2​|a|+14+12)|a|​Γ​(2​|a|+14)).\displaystyle=-\ln\left(\frac{\left|a\right|\Gamma\left(\frac{2\left|a\right|+1}{4}\right)^{2}}{2\Gamma\left(\frac{2\left|a\right|+1}{4}+\frac{1}{2}\right)^{2}}\right)=2\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{2\left|a\right|+1}{4}+\frac{1}{2}\right)}{\sqrt{\left|a\right|}\Gamma\left(\frac{2\left|a\right|+1}{4}\right)}\right).

Hence

Δ​(a)\displaystyle\Delta\left(a\right) =2​ln⁡(2​Γ​(2​|a|+14+12)|a|​Γ​(2​|a|+14))+ln⁡|a|=2​ln⁡(2​Γ​(|a|2+34)Γ​(|a|2+14)),\displaystyle=2\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{2\left|a\right|+1}{4}+\frac{1}{2}\right)}{\sqrt{\left|a\right|}\Gamma\left(\frac{2\left|a\right|+1}{4}\right)}\right)+\ln{\left|a\right|}=2\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{\left|a\right|}{2}+\frac{3}{4}\right)}{\Gamma\left(\frac{\left|a\right|}{2}+\frac{1}{4}\right)}\right), (1)

thereby proving (a) in section 1. To prove (b) in section 1, take the limit of (1) as a→0a\to 0.

Thus

∫0∞ln⁡(x)cosh⁡(π​x)​dx=ln⁡(2​Γ​(34)Γ​(14)),\int_{0}^{\infty}\frac{\ln\left(x\right)}{\cosh\left(\pi x\right)}\,\mathrm{d}x=\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right),

which implies

∫0∞ln⁡x​sech⁡(π​x)​d​x=ln⁡(2​Γ​(34)Γ​(14))\int_{0}^{\infty}\ln{x}\operatorname{sech}\left(\pi x\right)\mathrm{d}x=\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)
1π​∫0∞ln⁡(xπ)​sech⁡(x)​dx=ln⁡(2​Γ​(34)Γ​(14))\frac{1}{\pi}\int_{0}^{\infty}\ln\left(\frac{x}{\pi}\right)\operatorname{sech}\left(x\right)\mathrm{d}x=\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)
∫0∞ln⁡(xπ)​sech⁡(x)​dx=π​ln⁡(2​Γ​(34)Γ​(14))\int_{0}^{\infty}\ln\left(\frac{x}{\pi}\right)\operatorname{sech}\left(x\right)\mathrm{d}x=\pi\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)
∫0∞ln⁡x​sech⁡(x)​d​x−ln⁡π​∫0∞sech⁡(x)​dx=π​ln⁡(2​Γ​(34)Γ​(14))\int_{0}^{\infty}\ln{x}\operatorname{sech}\left(x\right)\mathrm{d}x-\ln{\pi}\int_{0}^{\infty}\operatorname{sech}\left(x\right)\mathrm{d}x=\pi\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)
∫0∞ln⁡x​sech⁡(x)​d​x−π2​ln⁡π=π​ln⁡(2​Γ​(34)Γ​(14))\int_{0}^{\infty}\ln{x}\operatorname{sech}\left(x\right)\mathrm{d}x-\frac{\pi}{2}\ln{\pi}=\pi\ln\left(\frac{\sqrt{2}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)
∫0∞ln⁡x​sech⁡(x)​d​x=π​ln⁡(2​π​Γ​(34)Γ​(14))=π​ln⁡(2​π32Γ​(14)2).\int_{0}^{\infty}\ln{x}\operatorname{sech}\left(x\right)\mathrm{d}x=\pi\ln\left(\frac{\sqrt{2\pi}\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right)=\pi\ln\left(\frac{2\pi^{\frac{3}{2}}}{\Gamma\left(\frac{1}{4}\right)^{2}}\right).

To prove (c) in section 1,

∫0∞ln⁡(a​x)​sech⁡(b​x)​dx\displaystyle\int_{0}^{\infty}\ln\left(ax\right)\operatorname{sech}\left(bx\right)\,\mathrm{d}x =∫0∞ln⁡(a​x)​sech⁡(b​x)​dx\displaystyle=\int_{0}^{\infty}\ln\left(ax\right)\operatorname{sech}\left(bx\right)\,\mathrm{d}x
=1b​∫0∞ln⁡(a​ub)​sech⁡(u)​dx\displaystyle=\frac{1}{b}\int_{0}^{\infty}\ln\left(\frac{au}{b}\right)\operatorname{sech}\left(u\right)\,\mathrm{d}x
=1b​∫0∞ln⁡(u)​sech⁡(u)​dx−ln⁡(ba)b​∫0∞sech⁡(u)​dx\displaystyle=\frac{1}{b}\int_{0}^{\infty}\ln\left(u\right)\operatorname{sech}\left(u\right)\,\mathrm{d}x-\frac{\ln{\left(\frac{b}{a}\right)}}{b}\int_{0}^{\infty}\operatorname{sech}\left(u\right)\,\mathrm{d}x
=πb​ln⁡(2​π32Γ​(14)2)−ln⁡(ba)b⋅π2\displaystyle=\frac{\pi}{b}\ln\left(\frac{2\pi^{\frac{3}{2}}}{\Gamma\left(\frac{1}{4}\right)^{2}}\right)-\frac{\ln{\left(\frac{b}{a}\right)}}{b}\cdot\frac{\pi}{2}
=πb​ln⁡(2​a​π32b​Γ​(14)2).\displaystyle=\frac{\pi}{b}\ln\left(\frac{2\sqrt{a}\pi^{\frac{3}{2}}}{\sqrt{b}\Gamma\left(\frac{1}{4}\right)^{2}}\right).

∎

References

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