HYPERBOLIC INTEGRAL WITH GLAISHER CONSTANT

Today we will prove the following result seen in this Twitter post:



\int_{0}^{\infty} \frac{x \sin x \ln x}{\cosh x-\cos x} d x=\frac{\pi^{2}}{6}\left(1-12 \ln A+\frac{\ln 2}{2}+\ln \pi\right)


First we need a Lemma:

Lemma 1:

\frac{\sin(ax)}{\cos(ax)+\cosh(bx)}&=-2\sum_{k=1}^\infty (-1)^{k} e^{-k b x}\sin\left(k a x \right)\\(1)

Proof:


\begin{align*} \frac{1}{\cos(ax)+\cosh(bx)}&=\frac{2}{e^{iax}+e^{-iax}+e^{bx}+e^{-bx}}\\ &=\frac{2}{e^{bx}\left(1+e^{-2bx}+e^{-bx+iax}+e^{-bx-iax}\right)}\\ &=\frac{2e^{-bx}}{\left(1+e^{-bx+iax}\right)\left(1+e^{-bx-iax}\right)}\\ &=\frac{2e^{-bx}}{2ie^{-bx}\sin(cx)}\left(\frac{1}{1+e^{-bx-iax}}-\frac{1}{1+e^{-bx+iax}} \right)\\ &=\frac{2e^{-bx}}{2ie^{-bx}\sin(ax)}\left(\sum_{k=0}^\infty (-1)^k e^{-kbx-ikax}-\sum_{k=0}^\infty (-1)^k e^{-kbx+ikax} \right)\\ &=\frac{2}{2i\sin(ax)}\sum_{k=0}^\infty (-1)^{k+1} e^{-k b x}\left(e^{i k a x}-e^{-i k a x} \right)\\ &=\frac{2}{\sin(ax)}\sum_{k=0}^\infty (-1)^{k+1} e^{-k b x}\sin\left(k a x \right) \\ &=-\frac{2}{\sin(ax)}\sum_{k=1}^\infty (-1)^{k} e^{-k b x}\sin\left(k a x \right) \qquad \blacksquare\\ \end{align*}

Corollary:


If we let ax= ax-i\pi in (1) we obtain


\frac{\sin(ax)}{\cosh(bx)-\cos(ax)}&=2\sum_{k=1}^\infty e^{-k b x}\sin\left(k a x \right)\\
(2)

Now, consider the following integral


\int_0^\infty \frac{x^{s-1}\sin(x)}{\cosh(x)-\cos(x)}\,dx &=2^{1-s/2}\sin \left(\frac{s \pi}{4} \right)\Gamma(s)\zeta(s)(3)


Proof:


\begin{align*} \int_0^\infty \frac{x^{s-1}\sin(x)}{\cosh(x)-\cos(x)}\,dx&=2\sum_{k=1}^\infty \int_0^\infty e^{-k x}\sin\left(k x \right) x^{s-1}\,dx &(\text{by}\,(2))\\ &=2\sum_{k=1}^\infty \Gamma(s) \frac{\sin \left(s \arctan \left(\frac{k}{k}\right)\right)}{\left(k^{2}+k^{2}\right)^{\frac{s}{2}}} &(*) \\ &=2\Gamma(s)\sin \left(\frac{s \pi}{4} \right)\sum_{k=1}^\infty \frac{1}{\left(2k^{2}\right)^{\frac{s}{2}}} \\ &=2^{1-s/2}\sin \left(\frac{s \pi}{4} \right)\Gamma(s)\zeta(s) \end{align*}


Where in (*) we used the following result proved here


\begin{align*} &\int_{0}^{\infty} e^{-a t} t^{x-1} \sin (b t) d t=\Gamma(x) \frac{\sin \left(x \arctan \left(\frac{b}{a}\right)\right)}{\left(a^{2}+b^{2}\right)^{\frac{x}{2}}} \\ & \text{for}\qquad \operatorname{Re}(x)>0, b>0, a>0 \notag \end{align*}

Now if we differentiate (3) with respect to s and let s \to 2 we get


\begin{aligned} &\int_0^\infty \frac{x\sin(x) \ln(x)}{\cosh(x)-\cos(x)}\,dx\\ &=2^{-s / 2-1} \Gamma(s)\left(4 \sin \left(\frac{\pi s}{4}\right) \zeta^{\prime}(s)+\zeta(s)\left(\pi \cos \left(\frac{\pi s}{4}\right)-\log (4) \sin \left(\frac{\pi s}{4}\right)+4 \sin \left(\frac{\pi s}{4}\right) \psi(s)\right)\right)\Big|_{s=2}\\ &=\frac14\left(4 \zeta^{\prime}(2)+\frac{\pi^2}{6}\left(-2\ln(2) +4 \psi(2)\right)\right)\\ &= \zeta^{\prime}(2)+\frac{\pi^2}{6}\left(-\frac{\ln(2)}{2} + \psi(2)\right)\right)\\ &= \frac{\pi^2}{6}\left(\ln(2)+\ln(\pi)+\gamma-12\ln(A) \right)+\frac{\pi^2}{6}\left(-\frac{\ln(2)}{2}-\gamma + 1 \right)\right)\\ &=\frac{\pi^{2}}{6}\left(1-12 \ln (A)+\frac{\ln (2)}{2}+\ln (\pi)\right) \qquad \blacksquare \end{aligned}


where we used that (see here)


\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln(2)+\ln(\pi)+\gamma-12\ln(A) \right)

and


\psi(2)=1-\gamma



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