How to prove a generalized integral identity

Question

$$ \int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})}dt=-\frac{1}{4}+\frac{\gamma}{2} $$ where $\gamma$ = Euler Gamma $$ \int_{0}^{\infty }\frac{t}{( e^{2\pi t}-1)(1+t^{2}) ^{2}}dt=\frac{\pi^2}{24} -\frac{3}{8} $$ $$ \int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)( 1+t^{2})^{3}}dt=\frac{\pi^2}{96} +\frac{\zeta(3)}{8} -\frac{7}{32} $$

Answer 1

Case $n\ge2$

We can write (for $n\ge 2$) \begin{align} \int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})^n}dt=-\frac12\int_0^{\infty}\frac{tdt}{(1+t^2)^n}+\frac{1}{2}\int_0^{\infty}\frac{t\coth\pi t\,dt}{(1+t^2)^n}, \end{align}

• The first integral can be calculated in terms of elementary functions: $$\int_0^{\infty}\frac{tdt}{(1+t^2)^n}=\frac12\left[\frac{(1+t^2)^{1-n}}{1-n}\right]_0^{\infty}=\frac{1}{2n-2}.$$

• In the second integral, we have an even function, so we can write it as $\displaystyle\frac12\int_{-\infty}^{\infty}$. Next it becomes easy to evaluate it by residues by pulling the integration contour to $i\infty$. In this way it will be given by the contribution of poles at $t_k=ik$, $k\in\mathbb{Z}_{>0}$, where in addition all the poles except $t_1=i$ are simple, and $t_1$ is of order $n+1$. Explicitly, $$\int_{-\infty}^{\infty}\frac{t\coth\pi t\,dt}{(1+t^2)^n}=-\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^n}+2\pi i \cdot\mathrm{res}_{t=i}\frac{t\coth\pi t}{(1+t^2)^n}.$$ It is rather clear that the first sum can be expressed in terms of $\zeta$-function evaluated at integer values and the second term can be obtained by calculating a sufficient number of derivatives.

• For example, \begin{align} &\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^2}=\frac{5}{8},\\ &\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^3}=\frac{7-8\zeta(3)}{16},\\ &\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^4}=\frac{45-32\zeta(3)}{128},\\ &\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^5}=\frac{77-40\zeta(3)-32\zeta(5)}{256},\\ &\\ &\ldots\;\ldots\;\ldots \end{align} and also \begin{align} 2\pi i \cdot\mathrm{res}_{t=i}\frac{t\coth\pi t}{(1+t^2)^2}&=\frac{4\pi^2+3}{24},\\ 2\pi i \cdot\mathrm{res}_{t=i}\frac{t\coth\pi t}{(1+t^2)^3}&=\frac{2\pi^2+3}{48}, \end{align} and so on.

Combining all these results, one can calculate the initial integral for any integer $n\ge 2$: $$\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})^n}dt=\frac14\left[2\pi i \cdot\mathrm{res}_{t=i}\frac{t\coth\pi t}{(1+t^2)^n}-\sum_{k=2}^{\infty}\frac{2k}{(1-k^2)^n}-\frac{1}{n-1}\right].$$

The case $n=1$ should be treated separately, I will do it later (or maybe someone will come out with his own proof before).

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Added: Case $n=1$

(I don't really like this derivation, there should be a simpler one; however, I am quite satisfied with my previous derivation for $n\ge 2$)

Here the problem is that we cannot separate the integral into two parts as before. We will remedy this by considering a slightly more general integral $$I(\alpha)=\int_{0}^{\infty }\frac{t}{(e^{2\pi t}-1)(1+t^{2})(1+\alpha^2 t^2)}dt,\tag{1}$$ and recovering the result we are interested in in the limit $\alpha\rightarrow 0$. Repeating our previous manipulations with (1), we find \begin{align} I(\alpha)=-\frac12\int_0^{\infty}\frac{tdt}{(1+t^2)(1+\alpha^2t^2)}+\frac{1}{4}\int_{-\infty}^{\infty}\frac{t\coth\pi t\,dt}{(1+t^2)(1+\alpha^2 t^2)}, \end{align}

Evaluating these integrals (the first directly, the second by residues), one obtains (for $\alpha>0$) $$I(\alpha)=\frac{1}{2}\frac{\ln\alpha}{1-\alpha^2}-\frac14 \sum_{k=2}^{\infty}\frac{2k}{(1-k^2)(1-\alpha^2 k^2)}+\frac{1+3\alpha^2}{8(1-\alpha^2)^2}-\frac{\pi\coth\frac{\pi}{a}}{4(1-\alpha^2)},$$ where the last two terms correspond to the contributions of the poles at $t=i$ and $t=i/\alpha$. Hence the integral we are interested in is given by $$I(0)=\frac{1}{2}\lim_{\alpha\rightarrow 0}\left[\ln\alpha-\sum_{k=2}^{\infty}\frac{k}{(1-k^2)(1-\alpha^2 k^2)}\right]+\frac{1}{8}-\frac{\pi}{4}.\tag{2}$$ The sum inside the limit can be calculated in closed form for any $\alpha$ (recall that e.g. $\sum_{k=2}^N\frac{1}{k+a}=\psi(1+N+a)-\psi(2+a)$): \begin{align}\sum_{k=2}^{\infty}\frac{k}{(1-k^2)(1-\alpha^2 k^2)}&=-\frac{4\gamma-3+2\psi(2+\alpha^{-1})+2\psi(2-\alpha^{-1})}{4(1-\alpha^2)}=\\ &=-\frac{4\gamma-3+2\psi(2+\alpha^{-1})+2\psi(-1+\alpha^{-1})+2\pi\cot\frac{\pi}{\alpha}}{4(1-\alpha^2)} \end{align} Therefore, to compute the limit in (2), it suffices to know the asymptotics of digamma function for large argument. It is given by $$\psi(\Lambda)=\ln\Lambda+O\left(\Lambda^{-1}\right),$$ and therefore we get that, as $\alpha\rightarrow 0$, $$\sum_{k=2}^{\infty}\frac{k}{(1-k^2)(1-\alpha^2 k^2)}=\ln\alpha-\frac{4\gamma-3+2\pi}{4}+O\left(\alpha\right).$$ Being combined with (2), this gives the result: $I(0)=\frac{\gamma}{2}-\frac14$. $\blacksquare$

Answer 2

By $$ \int_{0}^{\infty }\frac{\arctan \frac{t}{z}}{e^{2\pi t}-1}dt=\frac{z-z\ln z}{2}-\frac{1}{4}\ln \frac{2\pi }{z}+\frac{1}{2}\ln \Gamma (z) [{Re}z>0] $$ we have $$ \int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left( z^{2}+t^{2}\right) }dt=\frac{\ln z}{4}-\frac{1}{4z}-\frac{1}{2}\psi \left( z\right), $$ $$ \int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left( z^{2}+t^{2}\right) ^{2}}dt=\frac{1}{4z}\psi ^{\prime }\left( z\right) -\frac{% 1}{4z^{2}}-\frac{1}{8z^{3}} , $$ $$ \int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left( z^{2}+t^{2}\right) ^{3}}dt=\frac{1}{16z^{3}}\psi ^{\prime }\left( z\right) - \frac{1}{16z^{2}}\psi ^{\prime \prime }\left( z\right) -\frac{1}{8z^{4}}- \frac{3}{32z^{5}} $$ and the recurrence formula $$ I_{p}(z)=\int_{0}^{\infty }\frac{t}{\left( e^{2\pi t}-1\right) \left( z^{2}+t^{2}\right)^p }dt=\frac{-1}{2pz}I_{p-1}^{\prime }(z). $$ For $\psi ^{(k)}(z)$ we have the following identities \begin{equation} \begin{array}{c} \psi ^{(k)}(n+z)=\psi ^{(k)}(z)+k!(-1)^{k}\sum\limits_{l=1}^{n-1}\frac{1}{% \left( {l+z}\right) ^{k+1}}, \\ \psi ^{(k)}(z-n)=\psi ^{(k)}(z)+k!\sum\limits_{l=1}^{n}\frac{1}{\left( {l-n-z% }\right) ^{k+1}}.% \end{array} \end{equation} and \begin{equation} \begin{array}{c} \psi ^{(k)}(0)=\left\{ \begin{array}{c} -\gamma ,k=0 \\ k{{!(-1)^{k+1}}\zeta (k+1)},k>0% \end{array}% \right. , \\ \psi ^{(k)}(\frac{1}{2})=\left\{ \begin{array}{c} -\gamma -2\ln 2,k=0 \\ k{{!(-1)^{k+1}}}\left( 2^{k+1}-1\right) {\zeta (k+1)},k>0% \end{array} \right. , \\ \psi ^{(k)}(\frac{1}{4})=\left\{ \begin{array}{c} -\gamma -2\ln 2,k=0 \\ -\frac{k{!}}{2\pi }\left( \left( k+2+4^{k+2}\right) {\zeta }% (k+2)-2\sum\limits_{l=0}^{k/2-1}4^{k-2l}{{{\zeta }(k-2l)\zeta }}(2l+2)\right) \\ -k{!}2^{k}(2^{k+1}-1){\zeta }(k+1),k\text{ is even}% \end{array} \right. , \\ \psi ^{(k)}(\frac{3}{4})=\left\{ \begin{array}{c} -\gamma -2\ln 2,k=0 \\ \frac{k{!}}{2\pi }\left( \left( k+2+4^{k+2}\right) {\zeta }% (k+2)-2\sum\limits_{l=0}^{k/2-1}4^{k-2l}{{{\zeta }(k-2l)\zeta }}(2l+2)\right) \\ -k{!}2^{k}(2^{k+1}-1){\zeta }(k+1),k\text{ is even}% \end{array} \right. \end{array} \end{equation}