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 $$(1)$$?

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}

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.

• But once again, the integration of that last term gives me an answer dependent on $\text{Li}_2$ . Is there a way not to get this? Sep 11, 2020 at 14:47
• I don't think so. Sep 11, 2020 at 14:47
• Thank you anyway Sep 11, 2020 at 14:48
• Not that I know of, the problem originated in this other question of mine (I have written another one because maybe the problems is what I am doing and not the integration) math.stackexchange.com/questions/3822384/… Sep 11, 2020 at 15:01
• I had made a mistake in the integral and the limit should have been infinity, it changes nothing for me though as I still cannot do this. But maybe there is some "short-cut" with this limit that I don't know of Sep 11, 2020 at 15:11

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 $$\int_0^1\frac{\ln u}{1+u}du=\frac{\pi^2}{12}.$$