Integral $\int_{0}^{\infty}\frac{1}{x^2}(\frac{1}{\cosh^2x}-\frac{\tanh{x}}{x})dx$

Question

I am dealing with the following integral from superconductivity theory $$\int_{0}^{\infty}\frac{1}{x^2}\left(\frac{1}{\cosh^2x}-\frac{\tanh{x}}{x}\right)dx$$ 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?

Answer 1

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 $$ I = -\sum_{n\geq 0}\frac{8}{\pi^2(2n+1)^3}=-\frac{8}{\pi^2}\left[\zeta(3)-\frac{1}{8}\zeta(3)\right]=\color{red}{\frac{-7\zeta(3)}{\pi^2}}.\tag{3} $$

Answer 2

Differentiate $\int_0^\infty \frac{\sin(x y)}{\sinh\frac{\pi y}2 } dy = \tanh x$ twice with respect to $x$ to get $$\int_0^\infty \frac{y^2\sin(x y)}{\sinh\frac{\pi y}2 } dy =- \frac{d^2 (\tanh x)}{dx^2}= 2\tanh x \ \text{sech}^2 \ x$$ which is utilized to integrate \begin{align} &\int_{0}^{\infty}\frac{1}{x^2}\left(\frac{1}{\cosh^2x}-\frac{\tanh{x}}{x}\right)dx\\ = &\ \frac12\int_{0}^{\infty}\left(\tanh{x}-x \ \text{sech}^2 \ x\right)d\left(\frac1{x^2}\right)\\ \overset{ibp}=& -\int_{0}^{\infty}\frac{\tanh{x}\ \text{sech}^2 \ x}{x}dx = -\int_{0}^{\infty}\frac1x \bigg(\int_0^\infty \frac{y^2 \sin(x y)}{2\sinh\frac{\pi y}2 } dy\bigg) dx\\ =& -\frac12 \int_{0}^{\infty} \frac{y^2}{\sinh\frac{\pi y}2 }\int_0^\infty \frac{\sin(x y)}{x} dx\ dy = -\frac\pi4\int_{0}^{\infty} \frac{y^2}{\sinh\frac{\pi y}2 }\overset{\ln t={-{\pi y}/2}}{dy}\\ =&- \frac4{\pi^2} \int_0^1 \frac{\ln^2t}{1-t^2}dt = - \frac4{\pi^2}\cdot \frac74\zeta(3)=-\frac7{\pi^2}\zeta(3) \end{align}