[Math] Improper Integral $\int_0^\infty\left(\frac{\tanh(x)}{x^3}-\frac1{x^2\cosh^2(x)}\right)dx = \frac{7\zeta(3)}{\pi^2} $

Question

$\newcommand{\sech}{\operatorname{sech}}$
$\displaystyle \int_0^{\infty}{\left(\frac{\tanh(x)}{x^3} – \frac{\sech^2(x)}{x^2} \right)\ dx }= \frac{7\zeta(3)}{\pi^2} $

What I tried

I simplified it to –

$\displaystyle \int_0^{\infty}{\frac{\sinh(2x) – 2x}{x^3 \cosh^2(x)} \ dx}$

Then I don't know how to solve. I tried Feynman's method

$\displaystyle I(a) = \int_0^{\infty}{\frac{\sinh(ax) – ax}{x^3 \cosh^2(x)} \ dx}$

But then too it didn't help much.

I thought of replacing them with trigonometric forms and then complex number real and imaginary part but wasn't helpful much.

Please try to avoid complex analysis.

Answer 1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\int_{0}^{\infty} \bracks{{\tanh\pars{x} \over x^{3}} - {1 \over x^{2}\cosh^{2}\pars{x}}}\,\dd x = {7 \over \pi^{2}}\,\zeta\pars{3}} \approx 0.8526\ \Large ?}$.

---

\begin{align} &\color{#f00}{\int_{0}^{\infty} \bracks{{\tanh\pars{x} \over x^{3}} - {1 \over x^{2}\cosh^{2}\pars{x}}}\,\dd x} \\[5mm] = &\ \int_{0}^{\infty}{\tanh\pars{x} - x \over x^{3}}\,\dd x + \int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x \\[5mm] = & -\,\half\int_{x\ =\ 0}^{x\ \to\ \infty}\bracks{\tanh\pars{x} - x} \,\dd\pars{1 \over x^{2}} + \int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x \\[5mm] = &\ \half\int_{0}^{\infty}{\mrm{sech}^{2}\pars{x} - 1 \over x^{2}}\,\dd x + \int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x = \half\int_{0}^{\infty}{\tanh^{2}\pars{x} \over x^{2}}\,\dd x \\[5mm] = &\ 32\sum_{k = 0}^{\infty}\,\sum_{n = 0}^{\infty}\,\,\ \underbrace{% \int_{0}^{\infty}{1 \over \bracks{\pars{2k + 1}\pi}^{\, 2} + 4x^{2}}\, {1 \over \bracks{\pars{2n + 1}\pi}^{\, 2} + 4x^{2}}\,\dd x} _{\ds{1 \over 8\pi^{2}\pars{2k + 1}\pars{2n + 1}\pars{k + n + 1}}} \label{1}\tag{1} \\[5mm] = &\ {4 \over \pi^{2}}\ \underbrace{\sum_{k = 0}^{\infty}{H_{k} + 2\ln\pars{2} \over \pars{2k + 1}^{2}}} _{\ds{{7 \over 4}\,\zeta\pars{3}}}\label{2}\tag{2} = \color{#f00}{{7 \over \pi^{2}}\,\zeta\pars{3}} \end{align} Note that

• In \eqref{1}, we use the identity $\ds{{\tanh\pars{x} \over x} = 8\sum_{j = 0}^{\infty}{1 \over \bracks{\pars{2j + 1}\pi}^{\, 2} + 4x^{2}}}$
• The sum over $\ds{n}$, in \eqref{1}, yields a Digamma Function term $\ds{\Psi\pars{1 + k}}$ which explains the appearance of the Harmonic Number $\ds{H_{k} = \Psi\pars{1 + k} + \gamma}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant.
• $\ds{\sum_{k = 0}^{\infty}{H_{k} \over \pars{2k + 1}^{2}} = {1 \over 4}\bracks{7\zeta\pars{3} - \pi^{2}\ln\pars{2}}}$ is a well known result.
• $\ds{\sum_{k = 0}^{\infty}{1 \over \pars{2k + 1}^{2}} = {1 \over 8}\,\pi^{2}}$.