How do I integrate $\int^{\infty}_0 \frac{p}{T} \frac{1}{e^{p/T}+1} dp$ by parts? (if it is by parts) I've been trying to integrate this equation:
$$\int^{\infty}_0 \frac{p}{T} \frac{1}{e^{p/T}+1} dp \tag{1} $$
and attempted to do it by parts following : $u=p^3$ and $dv=\frac{1}{e^{p/T}+1}$
where therefore $du= 3p^2$ and $v= -T\ln(1+e^{p/T})+p$
But when I try integrating the $\int v du$ by parts I'm unable to integrate $dv$.
On Wolframalpha it simply says that the integral gives :  $$\int -T\ln(1+e^{p/T})+p dp =\frac{x^2}{2} - T^2 \text{Li}_2(-e^{p/T}) \tag{2}$$ but this doesn't look like the way in which I should write it.
How do integrate $(1)$?
 A: I don't see an easy way to evaluate the integral using integration by parts, at least not until it has been reduced to a sum of Gamma integrals.
The usual way I have seen this integral evaluated is to expand
$$
\frac1{e^x+1}=\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}
$$
as follows
$$
\begin{align}
\int_0^\infty\frac pT\frac1{e^{p/T}+1}\,\mathrm{d}p
&=T\int_0^\infty\frac{x}{e^x+1}\,\mathrm{d}x\\
&=T\int_0^\infty x\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}\,\mathrm{d}x\\
&=T\sum_{k=1}^\infty(-1)^{k-1}\frac1{k^2}\int_0^\infty xe^{-x}\,\mathrm{d}x\\[3pt]
&=T\eta(2)\Gamma(2)\\[6pt]
&=T\frac{\pi^2}{12}
\end{align}
$$
A: Forgetting about the limits, an approach I would use would be to first do a $w$-substitution $w=e^{p/T}+1$. Where
$$I=\int \frac{p}{T}\frac{1}{e^{p/T}+1}\,dp,$$
I believe this gives
$$I=T\int \ln(w-1)\cdot \frac{1}{w(w-1)}\,dw.$$
Do partial fractions on $\frac{1}{w(w-1)}$ to get:
$$I=T\int \left(\frac{\ln(w-1)}{w-1}-\frac{\ln(w-1)}{w}\right)\,dw.$$
The first terms should yield easily to $u=w-1$ but the second probably needs some special functions.
I see no use for parts here.
A: Letting $u=e^{-p/T}$, one has
$$ \int^{\infty}_0 \frac{p}{T} \frac{1}{e^{p/T}+1} dp =-T\int_0^1\frac{\ln u}{1+u}du=\frac{\pi^2}{12}T $$
where
$$ \int_0^1\frac{\ln u}{1+u}du=\frac{\pi^2}{12}. $$
