Compute $I=\int_0^{+\infty}\frac{\arctan(t)}{e^{\pi t}-1}dt$

Question

I would like to compute $\displaystyle I=\int_0^{+\infty}\frac{\arctan(t)}{e^{\pi t}-1}dt$

Let $D=(0,+\infty)$, I have $\frac{1}{e^{-\pi t}-1}=\frac{e^{-\pi t}}{1-e^{-\pi t}}$

So $$\frac{\arctan(t)}{e^{\pi t}-1}=\sum_{k=1}^{+\infty}\arctan(t)e^{-k \pi t}$$

Now, I can use integration term by term theorem, And finally I have,

$$ I=\sum_{k=1}^{+\infty}\int_0^{+\infty}\arctan(t)e^{-k \pi t}dt $$

By integration by parts I get, $$ \int_0^{X}\arctan(t)e^{k \pi t}dt\rightarrow \frac{1}{k\pi}\int_0^{+\infty}\frac{e^{-k \pi t}}{1+t^2}dt $$

Therefore, $$ I=\sum_{k=1}^{+\infty}\frac{1}{k\pi}\int_0^{+\infty}\frac{e^{-k \pi t}}{1+t^2}dt $$

Now, I am stuck.

I would like to find a closed form,

Thank you in advance for your time.

Answer 1

Let $ \displaystyle I(z) = \int_{0}^{\infty}\frac{\arctan \frac{x}{z}}{e^{\pi x}-1} \ dx$.

Then

$$\begin{align} I(z) &= \int_{0}^{\infty} \int_{0}^{\infty}\frac{1}{e^{\pi x}-1} \frac{\sin (xt)}{t}e^{-zt} \ dt \ dx \tag{1} \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \frac{\sin (tx)}{e^{\pi x}-1} \ dx \ dt \\&= \int_{0}^{\infty} \frac{e^{-zt}}{t} \int_{0}^{\infty} \sin (tx) \sum_{n=1}^{\infty} e^{-\pi nx}\ dx \ dt \\ &=\int_{0}^{\infty} \frac{e^{-zt}}{t} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin(tx) \ e^{-\pi n x} \ dx \ dt \\ &= \int_{0}^{\infty} \frac{e^{-zt}}{t} \sum_{n=1}^{\infty} \frac{t}{t^{2} + \pi^{2}n^{2}} \\&= \frac{1}{2}\int_{0}^{\infty} \frac{e^{-zt}}{t} \Big(\coth t -\frac{1}{t}\Big) \ dt \tag{2} \end{align}$$

Now differentiate inside of the integral with respect to $z$.

$$ I'(z) = - \frac{1}{2} \int_{0}^{\infty}e^{-zt} \Big( \coth t - \frac{1}{t}\Big) \ dt $$

And then integrate by parts.

$$ \begin{align} I'(z) &= -\frac{1}{2} e^{-zt} \Big(\log(\sinh t) - \log t \Big) \Big|^{\infty}_{0} - \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big( \log (\sinh t) - \log(t) \Big) \ dt \\ &=- \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big( \log (\sinh t) - \log(t) \Big) \ dt \\ &= - \frac{z}{2} \int_{0}^{\infty} e^{-zt} \Big(t - \log (2) + \log(1-e^{-2t}) - \log(t) \Big) \ dt \\ &= -\frac{z}{2} \Big( \frac{1}{z^{2}} - \frac{\log(2)}{z} + \frac{\log(z) + \gamma}{z} \Big) + \frac{z}{2} \int_{0}^{\infty} e^{-zt} \sum_{n=1}^{\infty} \frac{e^{-2tn}}{n} \ dt \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} +\frac{z}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\infty} e^{-(2n+z)t} \ dt \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{z/2}{n(n+z/2)} \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} - \frac{\gamma}{2} + \frac{1}{2}\psi\left( \frac{z}{2}+1 \right) + \frac{\gamma}{2} \tag{3} \\ &= - \frac{1}{2z} + \frac{\log 2}{2} - \frac{\log z}{2} + \frac{1}{2} \psi \Big( \frac{z}{2}+1 \Big) \end{align}$$

Then integrating back,

$$I(z) = - \frac{\log z}{2} + \frac{z \log 2}{2} - \frac{z \log z}{2} + \frac{z}{2} + \log \Gamma \left( \frac{z}{2} +1 \right) + C$$

where using Stirling's formula the constant of integration is found to be $ \displaystyle -\frac{\log(\pi)}{2}$.

Therefore,

$$ \begin{align} \int_{0}^{\infty} \frac{\arctan x}{e^{\pi x}-1} \ dx &= I(1) \\ &= \frac{\log 2}{2} + \frac{1}{2} + \log \Gamma \left(\frac{3}{2} \right)- \frac{\log \pi}{2} \\ &= \frac{\log 2}{2} + \frac{1}{2} + \log \left( \frac{\sqrt{\pi}}{2} \right) -\frac{\log \pi}{2} \\ &=\frac{1}{2} - \frac{\log 2}{2} \end{align}$$

$ $

$(1)$ $\int_{0}^{\infty} \frac{e^{-x} \sin(x)}{x} dx$ Evaluate Integral

$(2)$ Series expansion of $\coth x$ using the Fourier transform

$(3)$ http://en.wikipedia.org/wiki/Digamma_function#Series_formula

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I \equiv \int_{0}^{\infty}{\arctan\pars{t} \over \expo{\pi t} - 1}\,\dd t}$

\begin{align} I &=\int_{0}^{\infty}{1 \over \expo{\pi t} - 1} \int_{0}^{t}{\dd x \over x^{2} + 1}\,\dd t =\int_{0}^{\infty}{\dd x \over x^{2} + 1} \int_{x}^{\infty}{\expo{-\pi t} \over 1 - \expo{-\pi t}}\,\dd t \\[3mm]&=-\,{1 \over \pi}\int_{0}^{\infty} {\ln\pars{1 - \expo{-\pi x}} \over x^{2} + 1}\,\dd x =-\,{1 \over \pi}\,\Im\int_{0}^{\infty} {\ln\pars{1 + \expo{-\pi\bracks{x - \ic}}} \over x - \ic}\,\dd x \end{align}

Let's consider $$ {\cal J}\pars{\mu} \equiv -\,{1 \over \pi}\,\int_{0}^{\infty}\Im\pars{% {\ln\pars{1 + \expo{-\mu\bracks{x - \ic}}}\over x - \ic}}\,\dd x\quad \mbox{such that}\quad I = {\cal J}\pars{\pi}\tag{2} $$

\begin{align} {\cal J}'\pars{\mu}&= {1 \over \pi}\,\Im\int_{0}^{\infty} {\expo{-\mu\bracks{x - \ic}} \over 1 + \expo{-\mu\bracks{x - \ic}}}\dd x =\left.-\,{1 \over \pi\mu}\,\Im\ln\pars{1 + \expo{-\mu\bracks{x - \ic}}} \right\vert_{x\ = 0}^{x\ \to\ \infty} \\[3mm]&={1 \over \pi\mu}\,\Im\ln\pars{1 + \expo{\ic \mu}} ={1 \over \pi\mu}\,\Im\ln\pars{2\expo{\ic\mu/2}\cos\pars{\mu \over 2}} ={1 \over 2\pi}\tag{3} \end{align}

Notice that $\ds{{\cal J}\pars{0} = -\,\half\,\ln\pars{2}}$ \begin{align} I&={\cal J}\pars{\pi} =-\,\half\,\ln\pars{2} + \bracks{{\cal J}\pars{\pi} - {\cal J}\pars{0}} =-\,\half\,\ln\pars{2} + \int_{0}^{\pi}\overbrace{{1 \over 2\pi}}^{\ds{\mbox{From}\ \pars{3}}}\,\dd t \\[3mm]&=-\,\half\,\ln\pars{2} + \half \end{align}

$$\color{#00f}{\large% I \equiv \int_{0}^{\infty}{\arctan\pars{t} \over \expo{\pi t} - 1}\,\dd t =\half\bracks{1 - \ln\pars{2}}} $$