I am dealing with the following integral $$\int_{0}^{\infty}\frac{dx}{x^2}\left(\frac{1}{\cosh^2x}-\frac{\tanh{x}}{x}\right).$$ My attempt to calculate this integral: calculate residues of $$f(x)=\frac{x-\sinh x\cosh x}{x^3\cosh^2 x},$$ then use Cauchy theorem about residues (integration over the contour over Im axis). I know that the answer is $-7\zeta(3)/\pi^2$, but I don't understand how to check it.

The function $f(x)$ has the second order pole at $x_0=i\pi/2+i\pi n$ (also the third order poles at $x=0$ but it's not important). To calculate residue, I use $$\mathrm{res}\,f(x)=\lim\limits_{x\rightarrow x_0}\frac{1}{2}\left[f(x)(x-x_0)^2\right]'.$$

Can anyone help with this integral?


Nice question. We may start with the Weierstrass product for the hyperbolic cosine: $$ \cosh(x) = \prod_{n\geq 0}\left(1+\frac{4x^2}{\pi^2(2n+1)^2}\right) \tag{1}$$ and apply $\frac{d}{dx}\log(\cdot)$ and $\frac{d^2}{dx^2}\log(\cdot)$ to both sides, getting: $$ \frac{\tanh x}{x}=\frac{8}{\pi^2}\sum_{n\geq 0}\frac{1}{\pi^2+(2n+1)^2 x^2} $$

$$I= \int_{0}^{+\infty}\frac{dx}{x^2}\left(\frac{1}{\cosh^2(x)}-\frac{\tanh(x)}{x}\right) = -\sum_{n\geq 0}\int_{0}^{+\infty}\frac{64\,dx}{\left(4x^2+\pi^2(2n+1)^2\right)^2}\tag{2} $$ Computing the inner integrals we get $$ I = -\sum_{n\geq 0}\frac{8}{\pi^2(2n+1)^3}=-\frac{8}{\pi^2}\left[\zeta(3)-\frac{1}{8}\zeta(3)\right]=\color{red}{\frac{-7\zeta(3)}{\pi^2}}.\tag{3} $$

  • $\begingroup$ It is great, thank You! $\endgroup$ – Artem Alexandrov Dec 6 '18 at 17:21
  • $\begingroup$ @ArtemAlexandrov: you're welcome. The consequences of Weierstrass products will never cease to amaze us :) $\endgroup$ – Jack D'Aurizio Dec 6 '18 at 18:25

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