While I was performing experiments and change of variables in integrals involving the Apéry's constant with the help of Wolfram Alpha online calculator, I've written with the help of this online calculator an integral for $\zeta(3)$ in terms of the Gudermannian function $\operatorname{gd}(x)$. See the definition of this special function for example in this Wikipedia.
See it
int ye^y/(1+e^(2y))gd(y)dy, from y=-infinite to 0
Question. I would like to know a simple and rigurous way to prove $$\int_{-\infty}^0\frac{ye^y\operatorname{gd}(y)}{1+e^{2y}}\mathrm dy=\frac78\zeta(3)$$ Many thanks.
You already asked ten similar questions. Expand $\frac{1}{\cosh y}$ in powers of $e^y$, integrate term by term, expand $\frac{1}{1+e^{2y}}$ in powers of $e^y$, multiply the two series, and use that $\int_0^\infty y e^{-ny}\mathrm dy = \int_0^\infty (x/n) e^{-n(x/n)}\mathrm d(x/n)=n^{-2} \int_0^\infty x e^{-x}\mathrm dx$ $ = n^{-2} (-xe^{-x}|_0^\infty +\int_0^\infty e^{-x}\mathrm dx) = n^{-2}$