Definite integral: $\int_0^\infty \frac{x dx}{(1+x^2)(1+e^{\pi x})}$

Question

I need help computing the value of the following definite improper integral: $$\int_0^\infty \frac{x dx}{(1+x^2)(1+e^{\pi x})}=\text{?}$$ Here are my thoughts and attempts:

• I tried using the Laplace Transform identity for definite integrals, with no luck (since I can only compute the Laplace Transform of $\frac{1}{e^{\pi x}+1}$ in terms of the digamma function... yuck)
• I can't use the residue theorem, since the integral is from $0$ to $\infty$ and the integrand is not an even function
• I would like to expand $\frac{x}{1+x^2}=\frac{1/x}{1-(-1/x^2)}$ as a geometric series, but it wouldn't always converge since $x$ goes from $0$ to $\infty$

CONTEXT: The integral came up in Jack D'Aurizio's answer to this question.

Any ideas?

Answer 1

We use the formula (valid for $\Re (z) > 0$): $$\psi(z) = \ln z - \frac{1}{2z} - 2 \int_0^\infty \frac{x}{(x^2+1)(e^{2\pi xz}-1)} dx$$ which $\psi(z)$ is the digamma function, this follows from differentiating Binet's second formula.

Since $$\int_0^\infty \frac{x dx}{(1+x^2)(1+e^{\pi x})} = \int_0^\infty \frac{x dx}{(1+x^2)(e^{\pi x}-1)} - 2\int_0^\infty \frac{x dx}{(1+x^2)(e^{2\pi x}-1)}$$

applying the formula gives the result: $$\int_0^\infty \frac{x dx}{(1+x^2)(1+e^{\pi x})} = \frac{\ln 2 - \gamma}{2}$$

Answer 2

You were on a promising track. By the Laplace transform

$$ A_k=\int_{0}^{+\infty}\frac{x}{1+x^2}e^{-\pi k x}\,dx =(-1)^k\int_{k\pi}^{+\infty}\frac{\cos x}{x}\,dx=(-1)^k\int_{k}^{+\infty}\frac{\cos(\pi x)}{x}\,dx\tag{1}$$ hence $$ \int_{0}^{+\infty}\frac{x\,dx}{(1+x^2)(e^{\pi x}+1)}=\sum_{k\geq 1}(-1)^{k+1}A_k=-\sum_{k\geq 1}\int_{k}^{+\infty}\frac{\cos(\pi x)}{x}\,dx \\=-\frac{1}{\pi}\int_{1}^{+\infty}\frac{\lfloor x\rfloor \sin(\pi x) }{x}\,dx\tag{2}$$ and the claim follows by exploiting the Fourier sine series of $\lfloor x\rfloor-\frac{1}{2}$, Dirichlet's integral $\int_{0}^{+\infty}\frac{\sin(mx)}{x}\,dx=\frac{\pi}{2}$, its generalization $\int_{0}^{+\infty}\frac{\sin(\pi x)\sin(\pi m x)}{x}\,dx =\frac{1}{2}\log\frac{m+1}{m-1}$ granted by the complex version of Frullani's theorem, the series definition of $\gamma$.

Answer 3

Using Fourier sine transform and generalized method:

$$\color{Blue}{\int_0^{\infty } \frac{x}{\left(1+x^2\right) (1+\exp (\pi x))} \, dx}=\\\int_0^{\infty } \text{FourierSinTransform}\left[\frac{x}{1+x^2},x,s\right] \text{FourierSinTransform}\left[\frac{1}{1+\exp (\pi x)},x,s\right] \, ds=\\\int_0^{\infty } \left(\frac{e^{-s}}{2 s}-\frac{1}{2} e^{-s} \text{csch}(s)\right) \, ds=\color{blue}{\\\frac{1}{2} (-\gamma +\ln (2))}$$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{% \int_{0}^{\infty}{x\,\dd x \over \pars{1 + x^{2}}\pars{1 + \expo{\pi x}}}}} = \\[5mm] = &\ \int_{0}^{\infty}{x \over 1 + x^{2}} \pars{{1 \over \expo{\pi x} + 1} - {1 \over \expo{\pi x} - 1}}\dd x + \int_{0}^{\infty}{x \over 1 + x^{2}} {1 \over \expo{\pi x} - 1}\dd x \\[5mm] = &\ -2\int_{0}^{\infty}{x \over 1 + x^{2}} {1 \over \expo{2\pi x} - 1}\dd x + \int_{0}^{\infty}{x \over 1/4 + x^{2}} {1 \over \expo{2\pi x} - 1}\dd x \\[5mm] = &\ \int_{0}^{\infty}\pars{{x \over 1/4 + x^{2}} - {2x \over 1 + x^{2}}} {1 \over \expo{2\pi x} - 1}\dd x \\[5mm] = &\ -\,{1 \over 2}\bracks{-2\,\Im\int_{0}^{\infty}\pars{{2 \over 1 + x\ic} - {1 \over 1/2 + x\ic}} {1 \over \expo{2\pi x} - 1}\dd x} \end{align}

With the Abel-Plana Formula:

\begin{align} &\bbox[10px,#ffd]{\ds{% \int_{0}^{\infty}{x\,\dd x \over \pars{1 + x^{2}}\pars{1 + \expo{\pi x}}}}} = \\[5mm] = &\ -\,{1 \over 2}\,\lim_{N \to \infty}\bracks{% \sum_{k = 0}^{N}\pars{{2 \over k + 1} - {1 \over k + 1/2}} - \int_{0}^{N}\pars{{2 \over x + 1} - {1 \over x + 1/2}}\,\dd x} \label{1}\tag{1} \end{align}

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Note that \begin{equation} \left\{\begin{array}{rclcl} \ds{\sum_{k = 0}^{N}{2 \over k + 1}} & \ds{=} & \ds{2\sum_{k = 0}^{\infty}\pars{{1 \over k + 1} - {1 \over k + N + 2}}} & \ds{=} & \ds{2\pars{H_{N + 1} + \gamma}} \\[1mm] \ds{\sum_{k = 0}^{N}{1 \over k + 1/2}} & \ds{=} & \ds{\sum_{k = 0}^{\infty}\pars{{1 \over k + 1/2} - {1 \over k + N + 3/2}}} & \ds{=} & \ds{H_{N + 1/2} + \gamma + 2\ln\pars{2}} \end{array}\right. \label{2}\tag{2} \end{equation} where $\ds{H_{z}}$ is a Harmonic Number, $\ds{\gamma}$ is the Euler-Mascheroni Constant and \begin{equation} \int_{0}^{N}\pars{{2 \over x + 1} - {1 \over x + 1/2}}\,\dd x = 2\ln\pars{N + 1} - \ln\pars{N + {1 \over 2}} - \ln\pars{2} \label{3}\tag{3} \end{equation}

With \eqref{1}, \eqref{2} and \eqref{3}:

$$ \bbx{\int_{0}^{\infty}{x\,\dd x \over \pars{1 + x^{2}}\pars{1 + \expo{\pi x}}} = {1 \over 2}\bracks{\ln\pars{2} - \gamma}} \approx 0.0580 $$ where we used the $\ds{H_{z}}$ asymptotic behaviour as $\ds{\verts{z} \to \infty}$.