# Evaluate $\int_0^\infty (u\log(a^2+u^2))/(e^u-1)\, du$

Does the following definite integral have a known "closed form" value?

$$\int_0^\infty\frac{u\log(a^2+u^2)}{e^u-1}~du,$$

or can anyone see a way to integrate it?

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Wolfram Alpha found a value in terms of many special functions for $a=1$ and nothing for $a=2$. I would strongly suspect there is not a closed form. – Potato Aug 11 '12 at 22:08
Mathematica (the Alpha engine) often does not find integrals which are known to have closed form, so I wouldn't be too sure! – poirot Aug 11 '12 at 22:11
Matlab also couldnt find but of course it doesnt prove that there isnt. – Seyhmus Güngören Aug 11 '12 at 22:28

I was only able to find the following formula: for $a = 2\pi\alpha$,

\begin{align*} \int_{0}^{\infty} \frac{u \log(4\pi^2\alpha^2+u^2)}{e^u - 1} \; du &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\ &\quad + \pi^2 \alpha \left(2\alpha \log\alpha - \alpha + 2 - 4 \log\Gamma(\alpha+1)\right) \\ &\quad + 4\pi^2 \int_{0}^{\alpha} \log\Gamma(u+1)\;du. \end{align*}

Indeed, let

$$I(a) = \int_{0}^{\infty} \frac{u \log(a^2+u^2)}{e^u - 1} \; du.$$

Then we have

\begin{align*}I(0) & = 2 \int_{0}^{\infty} \frac{u \log u}{e^u - 1} \; du = 2 \left. \frac{d}{ds}\zeta(s)\Gamma(s) \right|_{s=2} \\ & = 2(\zeta'(2) + \zeta(2)\psi_0(2)) = 2\left(\zeta'(2) + \frac{\pi^2}{6}(1-\gamma)\right). \end{align*}

Also,

\begin{align*} I'(a) &= \int_{0}^{\infty} \frac{2au}{u^2+a^2} \frac{du}{e^u-1} \\ &= 2a \int_{0}^{\infty} \sum_{n=1}^{\infty} e^{-nu} \left(\int_{0}^{\infty} \sin(ux)e^{-ax}\;dx\right)\;du \\ &= 2a \int_{0}^{\infty} \sum_{n=1}^{\infty} \left(\int_{0}^{\infty} \sin(xu)e^{-nu}\;du\right)e^{-ax}\;dx \\ &= 2a \int_{0}^{\infty} \left(\sum_{n=1}^{\infty} \frac{x}{x^2+n^2}\right)e^{-ax}\;dx \\ &= a \int_{0}^{\infty} \left(\pi\coth(\pi x) - \frac{1}{x}\right)e^{-ax}\;dx \end{align*}

Proceeding,

\begin{align*} I'(a) &= a \left[ \left(\log\sinh(\pi x)-\log x \right)e^{-ax} \right]_{0}^{\infty} + a^2 \int_{0}^{\infty} \left(\log\sinh(\pi x)-\log x \right)e^{-ax} \; dx \\ &= -a\log\pi + a^2 \int_{0}^{\infty} \left(\pi x + \log\left(1 - e^{-2\pi x}\right)-\log2 + \log a -\log ax \right)e^{-ax} \; dx \\ &= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right)+a^2 \int_{0}^{\infty} e^{-ax}\log\left(1 - e^{-2\pi x}\right) \; dx \\ &= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a^2 \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n} e^{-(a+2n\pi)x} \; dx \\ &= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a \sum_{n=1}^{\infty} \frac{\frac{a}{2\pi}}{n\left(n + \frac{a}{2\pi}\right)} \\ &= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a \left(\gamma + \psi_0 \left( \frac{a}{2\pi} +1 \right)\right) \\ &= \pi + a\left(\log\left(\frac{a}{2\pi}\right) - \psi_0 \left( \frac{a}{2\pi} +1 \right)\right). \end{align*}

Thus integrating,

\begin{align*} I(a) &= I(0) + \int_{0}^{a} I'(t) \; dt \\ &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) + \pi a + \int_{0}^{a} t \left(\log\left(\frac{t}{2\pi}\right) - \psi_0 \left( \frac{t}{2\pi} +1 \right)\right) \; dt \\ &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) + 2 \pi^2 \alpha + 4\pi^2 \int_{0}^{\alpha} u \left(\log u - \psi_0 (u+1)\right) \; du \\ &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\ &\quad + \pi^2 \alpha \left(2 \alpha \log \alpha - \alpha + 2 \right) - 4\pi^2 \int_{0}^{\alpha} u \psi_0 (u+1) \; du \\ &= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\ &\quad + \pi^2 \alpha \left(2 \alpha \log \alpha - \alpha + 2 \right) - 4\pi^2 \left[ u \log\Gamma(u+1) \right]_{0}^{\alpha} \\ &\quad + 4\pi^2 \int_{0}^{\alpha} \log\Gamma(u+1) \; du, \end{align*}

which gives the desired result. You may verify this formula numerically with the following Mathematica code

a = 8;
NIntegrate[(u Log[4 Pi^2 a^2 + u^2])/(Exp[u] - 1), {u, 0, Infinity}]
2 Zeta'[2] + Pi^2/3 (1 - EulerGamma) +
Pi^2 a (2 a Log[a] - a + 2 - 4 LogGamma[1 + a]) +
4 Pi^2 NIntegrate[LogGamma[1 + u], {u, 0, a}]
Clear[a];


In fact, we can give a neater form as follows:

\begin{align*} \int_{0}^{\infty} \frac{u \log(a^2+u^2)}{e^{2\pi u} - 1} \; du &= \zeta'(-1) + \frac{a}{2}\left(1 - \frac{a}{2}\right) + \frac{a^2}{2}\log a - a \log\Gamma(a+1)\\ &\quad + \int_{0}^{a} \log\Gamma(t+1) \; dt. \end{align*}

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Note that the integral of the log-gamma function is expressible in terms of the Barnes function. – J. M. Aug 12 '12 at 3:08
Using that identity, we have $$\frac{\pi^2}{3}((6a+1)\log(2\pi)+6 a^2\log(a)+12\left(\zeta'(-1)-\log G(1+a)\right)-9a^2)$$ and $\zeta'(-1)$ is of course related to the Glaisher-Kinkelin constant. – J. M. Aug 12 '12 at 4:50

The result involves the Glaisher constant denoted here by $A$, so I doubt that there exist a simple proof for this formula

For the beginning $$\int\limits_{0}^{+\infty}\frac{u\log(a^2+u^2)}{e^{u}-1}du= \int\limits_{0}^{+\infty}\frac{u\log(a^2)}{e^{u}-1}du+ \int\limits_{0}^{+\infty}\frac{u\log(1+a^{-2}u^2)}{e^{u}-1}du=$$ $$\log(a^2)\int\limits_{0}^{+\infty}\frac{u}{e^{u}-1}du+ \int\limits_{0}^{+\infty}\frac{at\log(1+t^2)}{e^{at}-1}adt= \log(a^2)\int\limits_{0}^{+\infty}\frac{u}{e^{u}-1}du+ a^2\int\limits_{0}^{+\infty}\frac{t\log(1+t^2)}{e^{at}-1}dt$$ The first integral in the last expression is well know, and its computation you can find here. But as for the second... here what Mathematica 8 gives $$-\frac{1}{12 a^2}(3 a^2-6 a^2 \log (a)+6 a^2 \log (\pi )+a^2 \log (64)+24 \pi a \text{log\Gamma }\left(\frac{a}{2 \pi }\right)+12 \pi a+4 \pi ^2 \log (a)-48 \pi ^2 \psi ^{(-2)}\left(\frac{a}{2 \pi }\right)-48 \pi ^2 \log (A)+4 \pi ^2-4 \pi ^2 \log (\pi )-\pi ^2 \log (16))$$ So the result is $$\frac{1}{12} (-3 a^2+6 a^2 \log (a)-6 a^2 \log (\pi )-a^2 \log (64)-24 \pi a \text{log\Gamma }\left(\frac{a}{2 \pi }\right)-12 \pi a+48 \pi ^2 \psi ^{(-2)}\left(\frac{a}{2 \pi }\right)+48 \pi ^2 \log (A)-4 \pi ^2+4 \pi ^2 \log (\pi )+\pi ^2 \log (16))$$

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