Here some context , after computing some integral in the form $\int_{0}^{\infty} \frac{\ln(a^2+x^2)}{\cosh(x)+\cos(b)}dx \,\,\, , \int_{0}^{\infty} \frac{x\ln(a^2+x^2)}{\sinh(x)}dx \,\,\, , \int_{0}^{\infty} \frac{x\ln(a^2+x^2)}{e^x+1}dx \,\,\, ,\int_{0}^{\infty} \frac{x\ln(a^2+x^2)}{e^x-1}dx$ , i asked myself about this one : $\int_{0}^{\infty} \frac{\ln(a^2+x^2)}{e^x+1}dx$ and especially $\int_{0}^{\infty} \frac{\ln(x)}{e^x+1}dx$ to begin, but i'm extremely strugguling on this one.
Here my attemp , and i just have to introduce 2 Lemma to help the proof :
(1): $\forall z \in \mathbb{R}^*_+ \,\,\,, \,\, \ln(z)= \int_{0}^{\infty} \frac{e^{-t}-e^{-zt}}{t}dt$
(2): $\forall z \in \mathbb{R}_+ \,\,\,, \,\, \int_{0}^{\infty} \frac{e^{-zt}}{e^t+1}dt=H_{z/2}-H_{z}+\ln(2)$ , where $H_z=\int_{0}^{1}\frac{1-t^z}{1-t}dt=\int_{0}^{\infty}\frac{1-e^{-zt}}{e^t-1}dt$, (the harmonic number)
(3): $\forall z \in \mathbb{R}_+ \,\,\,, \,\, \tan^{-1}(z)= \int_{0}^{\infty} \frac{\sin(zt)e^{-t}}{t}dt$
So ,
$I=\int_{0}^{\infty} \frac{\ln(x)}{e^x+1}dx=\int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-s}-e^{-xs}}{e^x+1}dxds =\int_{0}^{\infty} \ln(2)\frac{e^{-s}-1}{s}-\frac{H_{s/2}-H_s}{s} \, ds$ , Using Lemma (1) and (2)
Now , since $1=\frac{2}{\pi} \int_{0}^{\infty} \frac{sin(u)}{u}du$ , we can re-write :
$I=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\frac{sin(u)}{u} \left[\ln(2)\frac{e^{-s}-1}{s}-\frac{H_{s/2}-H_s}{s}\right] \, dsdu$ $\,\,\,\,\,\,u \mapsto us$
$=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\frac{sin(us)}{u} \left[\ln(2)\frac{e^{-s}-1}{s}-\frac{H_{s/2}-H_s}{s}\right] \, dsdu$
And using Lemma (3) and $H_z=\int_{0}^{\infty}\frac{1-e^{-zv}}{e^v-1}dv$ :
$I=\frac{2}{\pi}\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{u} \left[\ln(2)\sin(us)\frac{e^{-s}-1}{s}-\int_{0}^{\infty}\sin(us)\frac{e^{-sv}-e^{-sv/2}}{s(e^v-1)}dv\right] \, dsdu$
$=\frac{2}{\pi}\int_{0}^{\infty}\frac{1}{u} \left[-\ln(2)\tan^{-1}(1/u)-\int_{0}^{\infty}\frac{\tan^{-1}(2v/u)-\tan^{-1}(v/u)}{e^v-1}dv\right] \, du$
And from here im totally blocked , and i think its maybe a good way since :
$-\frac{1}{2} \ln^2(2)=\left( -\frac{\pi}{2} \ln(2)\right )\left( \frac{1}{\pi} \ln(2)\right)=\left(\int_{0}^{\infty} \frac{\tan^{-1}(1/p)-\tan^{-1}(2/p)}{p}dp\right )\left(\int_{0}^{\infty} \frac{1}{e^{\pi w}+1}dw \right )$
$=\int_{0}^{\infty}\int_{0}^{\infty} \frac{\tan^{-1}(1/p)-\tan^{-1}(2/p)}{p}\frac{1}{e^{\pi w}+1} dpdw$ , $\,\,\,\, p \mapsto p/w $
$=\int_{0}^{\infty}\int_{0}^{\infty} \frac{\tan^{-1}(w/p)-\tan^{-1}(2w/p)}{p}\frac{1}{e^{\pi w}+1} dpdw$ , $\,\,\,\, w \mapsto w/\pi \,\,\, $ and $\,\,\,\, p \mapsto p/\pi$
$=\int_{0}^{\infty}\int_{0}^{\infty} \frac{\tan^{-1}(w/p)-\tan^{-1}(2w/p)}{p}\frac{1}{e^{w}+1} dpdw\,\,\,\,\, $ which strongly resembles the integral above
