How to calculate integral $\int_{0}^{\infty}x\ \operatorname{sech^3}(x)dx$ 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) - $\frac{1}{2}.$
 A: 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.
A: 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}
