Evaluating $\int_0^\infty \frac{\tan^{-1}(t)}{e^{2\pi t}-1}\,\mathrm{d}t$

Question

As stated in the title, I want to evaluate the integral $$\int_0^\infty \frac{\tan^{-1}(t)}{e^{2\pi t}-1}\,\mathrm{d}t$$ Because of the $e^{2\pi t}$, it seems that complex analysis techniques are required for this one, which I am not so familiar with. I have tried some substitutions and integration by parts. Besides, I tried using the property $$\int_0^\infty f(t)\,g(t) \,\mathrm{d}t =\int_0^\infty \mathscr{L}[f](s)\cdot \mathscr{L}^{-1}[g](s)\,\mathrm{d}s$$ But no luck here. I would like a solution without complex analysis techniques, if that's possible. Thank you!

Answer 1

Here is one possible way to evaluate this integral without contour integration. Consider the more general version

$$\begin{aligned}I(z)&=2\int_{0}^{\infty}\frac{\arctan(\frac{x}{z})}{e^{2\pi x}-1}dx\\ &=2\int_{0}^{\infty}\frac{1}{e^{2\pi x}-1}\left\{\int_{0}^{\infty}\frac{\sin(xt)e^{-zt}}{t}dt\right\} dx\\ &=2\int_{0}^{\infty}\frac{e^{-zt}}{t}\left\{\int_{0}^{\infty}\frac{\sin(xt)}{e^{2\pi x}-1}dx\right\} dt \\ &=2\int_{0}^{\infty}\frac{e^{-zt}}{t}\left\{\int_{0}^{\infty}\frac{e^{-2\pi x}\sin(xt)}{1-e^{-2\pi x}}dx\right\} dt\\ &=2\int_{0}^{\infty}\frac{e^{-zt}}{t}\left\{\int_{0}^{\infty}{\sin(xt)}{\sum_{k=1}^{\infty} e^{-2\pi xk}}dx\right\} dt \\ &=2\int_{0}^{\infty}\frac{e^{-zt}}{t}\left\{\sum_{k=1}^{\infty}\int_{0}^{\infty}{\sin(xt)}{ e^{-2\pi xk}}dx\right\} dt\\ &=2\int_{0}^{\infty}\frac{e^{-zt}}{t}\left\{\sum_{k=1}^{\infty}\frac{t}{(2\pi k)^{2}+t^{2}}\right\} dt \\ &=2\int_{0}^{\infty}\frac{e^{-zt}}{t}\left\{\frac{\coth\left(\frac{t}{2}\right)}{4}-\frac{1}{2t}\right\} dt \\ &=\frac{1}{2}\int_{0}^{\infty}\frac{e^{-2zy}}{y}\left\{\coth\left({y}\right)-\frac{1}{y}\right\} dy \qquad \left(\frac{t}{2} \to y \right) \end{aligned}$$

and now take the deivative of $I(z)$ with respect to z:

$$ \begin{aligned} I'(z)&=-\int_{0}^{\infty}{e^{-2zy}}\left\{\coth\left({y}\right)-\frac{1}{y}\right\} dy \\ &=\int_{0}^{\infty}{e^{-2zy}}\left\{\frac{1}{y}-\coth\left({y}\right)\right\} dy \\ &=\int_{0}^{\infty}{e^{-2zy}}\left\{\frac{1}{y}-\frac{e^{y}+e^{-y}}{e^{y}-e^{-y}}\right\} dy \\ &=\int_{0}^{\infty}{e^{-2zy}}\left\{\frac{1}{y}-\frac{1+e^{-2y}}{1-e^{-2y}}\right\} dy \\ &=\int_{0}^{\infty}{e^{-zw}}\left\{\frac{2}{w}-\frac{1+e^{-w}}{1-e^{-w}}\right\} \frac{dw}{2} \qquad \left( y \to\frac{w}{2}\right)\\ &=\int_{0}^{\infty}\left\{\frac{e^{-zw}}{w}-\frac{e^{-zw}+e^{-zw}e^{-w}}{2(1-e^{-w})}\right\} {dw}\\ &=\int_{0}^{\infty}\left\{\frac{e^{-zw}\color{red}{+e^{-w}-e^{-w}}}{w}-\frac{e^{-zw}+\color{red}{e^{-zw}-e^{-zw}}+e^{-zw}e^{-w}}{2(1-e^{-w})}\right\} {dw} \end{aligned}$$

We now rewrite this integral as the sum of three integrals:

$$ \begin{aligned} I'(z)&=\left\{\int_{0}^{\infty}\frac{e^{-w}}{w}-\frac{2e^{-zw}}{2(1-e^{-w})}dw+ \int_{0}^{\infty}\frac{e^{-zw}-e^{-zw}e^{-w}}{2(1-e^{-w})}dw+\int_{0}^{\infty}\frac{e^{-zw}-e^{-w}}{w}dw\right\} \\ &=\left\{\int_{0}^{\infty}\frac{e^{-w}}{w}-\frac{e^{-zw}}{(1-e^{-w})}dw+ \frac{1}{2}\int_{0}^{\infty}{e^{-zw}}dw+\int_{0}^{\infty}\frac{e^{-zw}-e^{-w}}{w}dw\right\} \end{aligned} $$

The first integral is a integral representation of the digamma function and the third is a integral representation of $\log(z)$, so we obtain

$$I'(z)=\psi(z)-\log(z)+\frac{1}{2z} $$

integrating back with respect to z we get

$$2\int_{0}^{\infty}\frac{\arctan(\frac{x}{z})}{e^{2\pi x}-1}dx=\log(\Gamma(z))-z\log(z)+z+\frac{1}{2}\log(z)+C \tag{1}$$

At this point, to find the constant $C$ in $(1)$ we rely on Stirlings approximation for $\log(\Gamma(z))$ namely:

$$\ln \Gamma(z)=\frac{1}{2} \ln 2 \pi+\left(z-\frac{1}{2}\right) \ln z-z \quad \text { as } z \rightarrow \infty \tag{2}$$

Plugging $(2)$ in $(1)$ and then letting $z \to \infty$ we obtain that

$$C=-\frac12 \ln(2 \pi)$$

Therefore:

$$\log(\Gamma(z))=\left(z-\frac{1}{2}\right)\log(z)-z+\frac{1}{2}\log(2\pi)+2\int_{0}^{\infty}\frac{\arctan(\frac{x}{z})}{e^{2\pi x}-1}dx \tag{3}$$

Now set $z=1$ in $(3)$ and you get the desired result.

$$ {\color{navy} {\boxed{\int_{0}^{\infty} \frac{\arctan(x)}{e^{2\pi x}-1}dx=\frac{1}{2}-\frac{1}{4}\log(2\pi)}}}$$

Answer 2

You can use the self-adjoitness of the Laplace transform. Your integral equals

$$ \int_{0}^{+\infty}\arctan(t)\sum_{n\geq 1}e^{-2\pi n t}\,dt\stackrel{\text{IBP}}{=}\int_{0}^{+\infty}\frac{1}{t^2+1}\sum_{n\geq 1}\frac{e^{-2\pi n t}}{2n}\,dt $$ or $$ \int_{0}^{+\infty}\sin(s)\sum_{n\geq 1}\frac{1}{2\pi n(2\pi n+s)}\,ds=\int_{0}^{+\infty}\frac{\sin(s)}{2\pi s}\left[\gamma+\psi\left(1+\frac{s}{2\pi}\right)\right]\,ds$$ or $$ \sum_{n\geq 1}\frac{1}{2n\pi}\left[\frac{\pi}{2}-\text{Si}(2n\pi)\right]$$ Let $f(x)$ be the sawtooth wave, i.e. the $2\pi$-periodic extension of the function which equals $\frac{\pi-x}{2}$ over $(0,2\pi)$. We have $$ f(x)=\sum_{n\geq 1}\frac{\sin(nx)}{n},\qquad \frac{1}{2}-\{x\}=\sum_{n\geq 1}\frac{\sin(2\pi n x)}{\pi n} $$ hence $$ \int_{0}^{+\infty}\frac{\arctan(t)}{e^{2\pi t}-1}\,dt = \frac{1}{2}\int_{1}^{+\infty}\frac{\frac{1}{2}-\{x\}}{x}\,dx $$ and the RHS equals $$ \frac{1}{2}\sum_{n\geq 0}\int_{0}^{1}\frac{\frac{1}{2}-x}{x+n+1}\,dx = \frac{1}{2}\sum_{n\geq 1}\left[-1+\left(n+\frac{1}{2}\right)\log\frac{n+1}{n}\right].$$ The claim now follows from summation by parts and Stirling's approximation.

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This provides an alternative proof of Binet's second $\log\Gamma$ formula. Indeed by integration by parts and the self-adjointness of the Laplace transform $$ \int_{0}^{+\infty}\frac{\arctan(t/z)}{e^{2\pi t}-1}\,dt=\int_{0}^{+\infty}\sin(sz)\sum_{n\geq 1}\frac{1}{2\pi n(2\pi n+s)}\,ds $$ equals, for any $z\in\mathbb{Z}^+/2$, $$ (-1)^{2z}\sum_{n\geq 1}\frac{1}{2\pi n}\left[\frac{\pi}{2}-\text{Si}(2\pi n z)\right] $$ where the equivalent integral $$ \int_{1}^{+\infty}\frac{\frac{1}{2}-\{zx\}}{x}\,dx=\int_{z}^{+\infty}\frac{\frac{1}{2}-\{x\}}{x}\,dx$$ can be converted into a series computable by summation by parts and Stirling's approximation. The extension of this result to $z\in\mathbb{R}^+$ follows by the Bohr-Mollerup theorem and the extension to $\text{Re}(z)>0$ follows by Schwartz' reflection principle.

Answer 3

$\displaystyle \int_{0}^{\infty}{\arctan\left(x\right) \over \mathrm{e}^{2\pi x} - 1}\,\mathrm{d}x\ =\ \bbox[10px,border:1px solid navy]{{1 \over 2} - {1 \over 4}\ln\left(2\pi\right)}$

See $\color{black}{\bf 6.1.50}$ in A & S. The result comes from the Second Binet Formula.