$$
I =
\int_{0}^{\infty} \frac{\tanh\left(x\right)\operatorname{sech}\left(x\right)}{x}\,{\rm d}x
$$
I started with the substitution $\cosh x=t \implies [0,\infty] \to[1,\infty]$
$$I=\int_1^\infty \frac{1}{t^2 \ln(t+\sqrt{t^2-1})}\,dt$$
then the substitution $t+\sqrt{t^2-1}=z\implies [1,\infty] \to[1,\infty]$
$$I=\int_1^\infty \frac{2(z^2-1)}{(z^2+1)^2\ln(z)}\,dz$$
I am stuck on this integral and I have tried some more substitutions but they seem to make the integral harder to evaluate, I am interested to see how to solve this.
Best Answer
Let $J(a)=\int_0^\infty \frac{z^{a+1}}{(1+z^2)^2\ln z}dz$. Then $$J’(a) = \int_0^\infty \frac{t^{a+1}}{(1+t^2)^2} dt\overset{ibp} =-\frac a2 \int_0^\infty \frac{t^{a+1}}{1+t^2}dt =\frac{\pi a}{4}\csc\frac{\pi a}2 $$ and
$$I=\int_1^\infty \frac{2(z^2-1)}{(z^2+1)^2\ln z}\,dz =\int_0^\infty \frac{z^2-1}{(z^2+1)^2\ln z}\,dz$$
$$=\int_{-1}^{1}J’(a)da \overset{t=\tan\frac {\pi a}4} =\frac 4\pi \int_0^{1}\frac {\tan^{-1}t}{t}dt =\frac{4}{\pi}G \\ $$