Integral $\int_{0}^{\infty }\!{\frac {t\ln \left( {t}^{2}+1 \right) }{{{\rm e}^ {4\,\pi\,t}}+1}}\,{\rm d}t$

Question

I have the general formulas about:

$$\begin{align} \int_{0}^{\infty}\!{\frac {t\ln \left( {t}^{2}+{z}^{2} \right) }{{ {\rm e}^{\pi\,t}}+1}}\,{\rm d}t &= -4\,\ln \left( \Gamma (z/2 ) \right) -4\,{\it ln}\,G (z/2) +2\,\ln \left( \Gamma \left( z \right) \right)+{\frac{1}{6}}-2\,\ln \left( A \right) \\ &\hphantom{{}={}} + {\frac { \left( -6\,{z}^{2}+4 \right) \ln \left( 2 \right) }{12}}-{\frac {{z}^{2}\ln \left( z \right) }{2}}+{\frac {3\,{z}^{2}}{4}}+2\,{\it ln}\,G \left( z \right)\end{align}$$

And $$\begin{align} \int_{0}^{\infty }\!{\frac {t\ln \left( {t}^{2}+{z}^{2} \right) }{{ {\rm e}^{2\,\pi\,t}}+1}} \, {\rm d}t &= -{\frac {{z}^{2}\ln \left( z \right) }{2}}-\ln \left( 2 \right) {z}^ {2}-{\it ln}\,G \left( z \right) +{\frac {{\it ln}\,G \left( 2\,z \right) }{2}} \\ &\hphantom{{}={}}+{\frac {3\,{z}^{2}}{4}}-\ln \left( \Gamma \left( z \right) \right) +{\frac{1}{24}}-{\frac {\ln \left( A \right) }{2}}+ {\frac {\ln \left( \Gamma \left( 2\,z \right) \right) }{2}}+{\frac { \ln \left( 2 \right) }{24}}\end{align}$$

Where G(z) is the Barnes G-function and A is the Glaisher-Kinkelin constant.

I am interested about the general integral: $$\int_{0}^{\infty }\!{\frac {t\ln \left( {t}^{2}+{z}^{2} \right) }{{ {\rm e}^{4\,\pi\,t}}+1}}\,{\rm d}t$$

I find for example $$\int_{0}^{\infty }\!{\frac {t\ln \left( {t}^{2}+1 \right) }{{{\rm e}^ {4\,\pi\,t}}+1}}\,{\rm d}t = -{\frac {\ln \left( A \right) }{8}}+{\frac{73}{96}}-{\frac {5\,\ln \left( 2 \right) }{4}}+{\frac {\ln \left( 3 \right) }{8}}$$

But be careful, Wolfram give a false closed form.

Please, can someone prove it ?

Answer 1

Edit: I did not really make much progress. Random Variable’s answer is very good! I leave this here as the working is interesting but I don’t think it can be salvaged.

I spent a while on this, and made progress. Perhaps someone can further the work!

$\newcommand{\d}{\,\mathrm{d}}\newcommand{\L}{\operatorname{Log}}\newcommand{\res}{\operatorname{Res}}$I will be adapting the work from this paper by Blagouchine, of which I have lately been enamoured. This integral is similar in spirit to the exercises he sets in the paper!

Let $L_\beta$, $\beta\in\Bbb R^+$ be the rectangular contour which runs from vertices $\{0,\beta,\beta+2i/n,2i/n\}$ for an $n\in\Bbb N$. Let $\Lambda$ denote the $\L\Gamma$ function. Recall from Binet's formulas that we have, for $\Re w\gt0$, $\Lambda(w)=w\cdot\L w+\mathcal{O}(w)$. I use an $a\ge0$.

$$\begin{align}\oint_{L_\beta}\frac{w}{e^{n\pi w}+1}\Lambda\left(\frac{nw}{2 i}+a\right)\d w&=\int_{0}^\beta\cdots\d w-\int_{2i/n}^{\beta+2i/n}\cdots\d w+\\&\int_{2i/n}^{0}\cdots\d w+\int_{\beta}^{\beta+2i/n}\cdots\d w\end{align}$$

As $\beta\to\infty$, consider the final integral. We have something like:

$$\left|\int\cdots\right|\lt2\pi\cdot\sup_{0\le\phi\le2/n}\left|\frac{\beta+\phi i}{e^{\pi n\beta+\phi i}+1}\cdot(\beta\L \beta+\mathcal{O}(\beta))\right|\to0$$

Since $e^\beta$ dominates $\beta,\,\L\beta$. Using the substitution $w\mapsto w+2i/n$ in the imaginary integral involving $\beta$, and writing $J$ for the integral from $0\to2i/n$ we find:

$$\begin{align}\lim_{\beta\to\infty}\oint_{L_\beta}\frac{w}{e^{n\pi w}+1}\Lambda\left(\frac{nw}{2i}+a\right)\d w&=\int_{0}^\infty\frac{w}{e^{n\pi w}+1}\Lambda\left(\frac{nw}{2i}+a\right)\d w\\&-\int_{0}^\infty\frac{w+2i/n}{e^{n\pi w}e^{2\pi i}+1}\Lambda\left(\frac{nw+2i}{2i}+a\right)\d w-J\\&=\int_{0}^\infty\frac{w}{e^{n\pi w}+1}\left[\Lambda\left(\frac{nw}{2i}+a\right)-\Lambda\left(\frac{nw}{2i}+a+1\right)\right]\d w\\&-\frac{2i}{n}\int_{0}^\infty\frac{1}{e^{n\pi w}+1}\Lambda\left(\frac{nw}{2i}+a+1\right)\d w\end{align}$$

Now I use the fact that $\Lambda(s+1)=\L s+\Lambda(s)$ to obtain:

$$\begin{align}\oint_{L_\infty}\frac{w}{e^{n\pi w}+1}\Lambda\left(\frac{nw}{2i}+a\right)\d w&=-\int_{0}^\infty\frac{w}{e^{n\pi w}+1}\L\left(\frac{nw}{2i}+a\right)\d w\\&-\frac{2i}{n}\int_{0}^\infty\frac{1}{e^{n\pi w}+1}\Lambda\left(\frac{nw}{2i}+a+1\right)\d w-J\end{align}$$

The integrand has poles when $e^{\pi nw}=-1$, that is when $n\pi w=\pi(2k+1)i$, or: $w=\frac{2k+1}{n}i$. In the strip under consideration, we have $0\le\frac{2k+1}{n}\le2/n$ true only for $k=0$. The only pole is then at $\zeta=i/n$. Employ the residue theorem, and take real parts:

$$\begin{align}\Re\{J\}-2\pi\cdot\Im\left\{\res_{\zeta=i/n}\left[\frac{\zeta}{e^{n\pi \zeta}+1}\Lambda\left(\frac{n}{2i}\cdot\zeta+a\right)\right]\right\}&=-\int_{0}^\infty\frac{w}{e^{n\pi w}+1}\Re\left\{\L\left(\frac{nw}{2i}+a\right)\right\}\d w\\&+\frac{2}{n}\int_{0}^\infty\frac{1}{e^{n\pi w}+1}\Im\left\{\Lambda\left(\frac{nw}{2i}+a+1\right)\right\}\d w\\&=-\int_{0}^\infty\frac{w}{e^{n\pi w}+1}\frac{1}{2}\L\left(\frac{n^2w^2}{4}+a^2\right)\d w\\&+\frac{2}{n}\int_{0}^\infty\frac{1}{e^{n\pi w}+1}\Im\left\{\Lambda\left(\frac{nw}{2i}+a+1\right)\right\}\d w\\&=-\frac{1}{2}\int_{0}^\infty\frac{w}{e^{n\pi w}+1}\L\left(w^2+4a^2/n^2\right)\d w\\&-\frac{\L(n^2/4)}{2}\int_{0}^\infty\frac{w}{e^{\pi n w}+1}\d w\\&+\frac{2}{n}\int_{0}^\infty\frac{1}{e^{n\pi w}+1}\Im\left\{\Lambda\left(\frac{nw}{2i}+a+1\right)\right\}\d w\end{align}$$

Re-arranging, we find that the integral you're interested in is:

$$\begin{align}\int_{0}^\infty\frac{w}{e^{n\pi w}+1}\L\left(w^2+4a^2/n^2\right)\d w&=4\pi\cdot\Im\left\{\res_{\zeta=i/n}\left[\frac{\zeta}{e^{n\pi \zeta}+1}\Lambda\left(\frac{n}{2i}\cdot\zeta+a\right)\right]\right\}\\&+\frac{4}{n}\int_{0}^\infty\frac{1}{e^{n\pi w}+1}\Im\left\{\Lambda\left(\frac{nw}{2i}+a+1\right)\right\}\d w\\&-2\L(n/2)\int_{0}^\infty\frac{w}{e^{\pi n w}+1}\d w-2\Re\{J\}\end{align}$$

I number the four expressions on the right hand side of the last equation as $1,2,3,4$ in the order they're written. Let us first deal with $(1)$:

Near $\zeta$ the denominator has the expansion: $$e^{n\pi w}+1=(w-\zeta)n\pi\cdot(-1)+\mathcal{O}((w-\zeta)^2)$$

Then: $$\begin{align}\res_{\zeta=i/n}\left[\frac{\zeta}{e^{n\pi \zeta}+1}\Lambda\left(\frac{n}{2i}\cdot\zeta+a\right)\right]&=\lim_{w\to\zeta}\frac{w-\zeta}{e^{n\pi w}+1}\left(w\cdot\Lambda\left(\frac{n}{2i}\cdot w+a\right)\right)\\&=-n\pi\cdot\frac{i}{n}\Lambda\left(\frac{1}{2}+a\right)\\&=-\pi i\Lambda\left(\frac{1}{2}+a\right)\end{align}$$

So $(1)$ resolves to $4\pi\cdot\Im\left\{-\pi i\Lambda\left(\frac{1}{2}+a\right)\right\}=-4\pi^2\cdot\Lambda\left(\frac{1}{2}+a\right)$.

Now let's tackle $(3)$ with Torelli's integral-sum exchange theorem and geometric series.

$$\int_0^\infty\frac{x}{e^{\pi nx}+1}\d x=\sum_{k=0}^\infty(-1)^k\int_0^\infty xe^{-\pi n(k+1)x}\d x$$Using the substitution $\pi n(k+1)x\mapsto x$, this is:$$\sum_{k=0}^\infty(-1)^k\frac{1}{(\pi n(k+1))^2}\int_0^\infty xe^{-x}\d x=\frac{1}{\pi^2n^2}\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}$$We can cite this answer to find the latter series as, with $x=0$, as $\pi^2/12$, giving us: $$-2\L(n/2)\int_0^\infty\frac{w}{e^{n\pi w}+1}\d w=-\frac{\L(n/2)}{6n^2}$$

It remains to tackle $2$ and $4$. For $2$:

$$\Im\{\Lambda(s+1)\}=\Im\left\{-\gamma s+\sum_{k=1}^\infty\frac{s}{k}-\L\left(1+\frac{s}{k}\right)\right\}$$Which is, for $s=\sigma+i\tau$, $\sigma\gt 0$: $$-\gamma\cdot\tau+\sum_{k=1}^\infty\frac{\tau}{k}-\arctan\left(\frac{\tau}{k+\sigma}\right)$$

Now let's perform a lazy integral-sum exchange again, and letting $s=\frac{nw}{2i}+a$ for brevity, the integral in $(2)$ becomes:

$$-\gamma\int_0^\infty\frac{\tau}{e^{n\pi w}+1}\d w+\sum_{k=1}^\infty\int_0^\infty\frac{1}{e^{n\pi w}+1}\left(\frac{\tau}{k}-\arctan\left(\frac{\tau}{k+\sigma}\right)\right)\d w$$We must realise that $\sigma=a,\tau=-\frac{nw}{2}$. Using the same approach as I did to evaluate $(3)$, we find the first term to be: $$\frac{\gamma}{24n}$$Using geometric series yet again in the summation, we find: $$\sum_{k=1}^\infty\sum_{m=1}^\infty(-1)^{m-1}\left[\frac{-n}{2k}\int_0^\infty we^{-mn\pi w}\d w\right]+\sum_{k=1}^\infty\int_0^\infty\frac{\arctan\left(\frac{nw}{2k+2a}\right)}{e^{n\pi w}+1}\d w$$The first summand, using the same mapping as before, resolves to: $$\sum_{m=1}^\infty(-1)^{m-1}\left[-\frac{n}{2k}\cdot\frac{1}{(mn\pi)^2}\int_0^\infty we^{-w}\d w\right]=-\frac{1}{2nk\pi^2}\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m^2}=-\frac{1}{24nk}$$In the arctangent integral, first make the exponential purely in terms of $w$: $$\int_0^\infty\cdots\d w=\frac{1}{n\pi}\int_0^\infty\frac{\arctan\left(\frac{w}{2\pi(k+a)}\right)}{e^w+1}\d w$$I will call this integral $\mathcal{A}(k,n,a)$ as I am for the moment stuck on it!

So far we have found:

$$\begin{align}\int_{0}^\infty\frac{w}{e^{n\pi w}+1}\L\left(w^2+4a^2/n^2\right)\d w&=-4\pi^2\Lambda\left(\frac{1}{2}+a\right)+\frac{\gamma}{24n}\\&+\frac{4}{n}\sum_{k=1}^\infty\left[\mathcal{A}(k,n,a)-\frac{1}{24nk}\right]-\frac{\L(n)-\L(2)}{6n^2}\\&+2\int_0^{2i/n}\Re\left\{\frac{w}{e^{n\pi w}+1}\Lambda\left(\frac{n}{2i}\cdot w+a\right)\right\}\d w\end{align}$$

Approaching $(4)$, I make the substitution $w=ix$.

$$\begin{align}2\int_0^{2i/n}\Re\left\{\frac{w\cdot\Lambda\left(\frac{n}{2i}\cdot w+a\right)}{e^{n\pi w}+1}\right\}\d w&=-2\int_0^{2/n}\frac{x\cdot\Lambda\left(\frac{n}{2}x+a\right)}{\Re\{e^{n\pi ix}+1\}}\d x\\&=-2\int_0^{2/n}\frac{x\cdot\Lambda\left(\frac{n}{2}x+a\right)}{\cos(n\pi x)+1}\d x\\&=-\frac{8}{n^2}\int_0^1\frac{x\cdot\Lambda\left(x+a\right)}{\cos(2\pi x)+1}\d x\end{align}$$