How to calculate integral $\int_{0}^{\infty}x\ \operatorname{sech^3}(x)dx$

Question

For this integral, I used the identity $$ \operatorname{sech}^3(x) = \frac{8e^{-3x}}{(1 + e^{-2x})^3} = 4\sum_{n=1}^{\infty}(-1)^{n-1}n(n+1)e^{-(2n+1)x} $$

After replacing this expression in the given integral, using Laplace transforms, I obtained a divergent series. I know the answer is G (Catalan constant) - $\frac{1}{2}.$

Answer 1

Recall the following integral representation of the Catalan constant \begin{eqnarray*} \frac{1}{2} \int_{0}^{\infty} \frac{x}{\cosh(x)} dx =G. \end{eqnarray*} Using $\operatorname{sech}^2(x)+\tanh^2(x)=1$ gives \begin{eqnarray*} \int_{0}^{\infty} x \operatorname{sech}^3(x) dx + \int_{0}^{\infty} x \operatorname{sech}(x) \tanh^2(x) dx = \int_{0}^{\infty} x \operatorname{sech}(x) dx. \end{eqnarray*} Integration by parts on the middle integral \begin{eqnarray*} &\int_{0}^{\infty} \color{blue}{x \tanh(x)} \operatorname{sech} (x) \tanh(x) dx = \\ & -\color{blue}{x \tanh(x)} \operatorname{sech}(x) ]_{0}^{\infty}+ \int_{0}^{\infty} (\color{blue}{x \operatorname{sech}^2(x)+ \tanh(x)})\operatorname{sech}(x) dx. \end{eqnarray*} Now \begin{eqnarray*} \int_{0}^{\infty} \tanh(x)\operatorname{sech}(x) dx=1. \end{eqnarray*} Rearrange and the result follows.

Answer 2

Integrate by parts

\begin{align} &\int_{0}^{\infty}x\operatorname{sech^3} xdx\\ =&\int_0^{\infty}\frac{x \>\text{sech}\>x}{2\tanh x}d(\tanh^2 x) =- \frac12\int_0^{\infty} (\tanh x \>\text{sech}\>x -x \>\text{sech}\>x)dx\\ =&-\frac12 + \frac12\int_0^{\infty} x \>\text{sech}\>x \overset{t=e^{-x}}{dx} = -\frac12 + \int_0^{1} \frac{\ln t}{1+t^2}dt = -\frac12 + G \end{align}