calculus - Improper Integral $int_0^inftyleft(frac{tanh(x)}{x^3}-frac1{x^2cosh^2(x)}right)dx = frac{7zeta(3)}{pi^2} $

$\newcommand{\sech}{\operatorname{sech}}$
$\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

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$\ds{\int_{0}^{\infty}
\bracks{{\tanh\pars{x} \over x^{3}} - {1 \over x^{2}\cosh^{2}\pars{x}}}\,\dd x
:\ ?}$.

\begin{align}
&\color{#f00}{\int_{0}^{\infty}
\bracks{{\tanh\pars{x} \over x^{3}} - {1 \over x^{2}\cosh^{2}\pars{x}}}\,\dd x} =
\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.



• $\ds{\sum_{k = 0}^{\infty}{1 \over \pars{2k + 1}^{2}} =
  {1 \over 8}\,\pi^{2}}$.