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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 $(1)$?

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3 Answers 3

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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 $$ \begin{align} \int_0^\infty\frac pT\frac1{e^{p/T}+1}\,\mathrm{d}p &=T\int_0^\infty\frac{x}{e^x+1}\,\mathrm{d}x\\ &=T\int_0^\infty x\sum_{k=1}^\infty(-1)^{k-1}e^{-kx}\,\mathrm{d}x\\ &=T\sum_{k=1}^\infty(-1)^{k-1}\frac1{k^2}\int_0^\infty xe^{-x}\,\mathrm{d}x\\[3pt] &=T\eta(2)\Gamma(2)\\[6pt] &=T\frac{\pi^2}{12} \end{align} $$

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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.

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  • $\begingroup$ 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? $\endgroup$ Sep 11, 2020 at 14:47
  • $\begingroup$ I don't think so. $\endgroup$ Sep 11, 2020 at 14:47
  • $\begingroup$ Thank you anyway $\endgroup$ Sep 11, 2020 at 14:48
  • $\begingroup$ 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/… $\endgroup$ Sep 11, 2020 at 15:01
  • $\begingroup$ 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 $\endgroup$ Sep 11, 2020 at 15:11
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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 $$ \int_0^1\frac{\ln u}{1+u}du=\frac{\pi^2}{12}. $$

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