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I am trying to find the integral of the following expression:- $$\int_{-\infty}^{\infty} \frac{x^2}{e^{k}+e^{x^2}}dx$$ I think this can be solved using contour integration. The poles will be the points with $z^2 = k+i(2n+1)\pi$ according to me. But this results in infinite poles if you choose a contour covering the entire real axis. Does anybody have any insights to solve this integral?

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    $\begingroup$ It is not a problem of infinitely poles. You can't use contour integration because the function doesn't $\to 0$ for $|x|$ large on an half-plane. $\endgroup$
    – reuns
    Sep 24, 2019 at 13:12
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    $\begingroup$ You never really choose the contour to be infinite. You always choose a finite contour (say a semi circle of radius $R$), then apply the residue theorem and take the limit of $R$ going to infinity in the end. And for any finite value of the radius $R$ there will only be a finite number of poles inside the contour. Thus an infinite number of poles in the upper half plane is not an issue as it only appears in the limit. $\endgroup$
    – Winther
    Sep 24, 2019 at 18:17
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    $\begingroup$ @reuns That doesn't necessarily mean that it can't be solved using contour integration. It may require a nastier contour than the standard large semicircle, but, for example, $e^{-x^2}$ can be done using a parallelogram, combined with some modification to give the integrand poles in useful places. $\endgroup$
    – Chappers
    Sep 25, 2019 at 3:13
  • $\begingroup$ @Chappers I don't see what you mean, we need to close the contour with some curve where it $\to 0$, here there is not $\endgroup$
    – reuns
    Sep 25, 2019 at 11:08
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    $\begingroup$ @reuns See, e.g. math.stackexchange.com/q/1266856/221811 $\endgroup$
    – Chappers
    Sep 25, 2019 at 13:37

3 Answers 3

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Let's assume, for the moment, that $k\leq 0$. Then we can rewrite our integral in the following way:

$$\int_{-\infty}^\infty \frac{x^2}{e^k + e^{x^2}}dx = e^{-k}\int_{-\infty}^\infty \frac{x^2e^ke^{-x^2}}{e^ke^{-x^2}+1}dx = e^{-k}\sum_{n=1}^\infty (-1)^{n+1}e^{nk}\int_{-\infty}^\infty x^2e^{-nx^2}dx$$

Then, by Feynman's trick

$$= e^{-k}\sum_{n=1}^\infty (-1)^{n+1} e^{nk}\Biggr(-\frac{d}{dn}\int_{-\infty}^\infty e^{-nx^2}dx\Biggr) = e^{-k}\sum_{n=1}^\infty (-1)^{n+1} e^{nk} \left(-\frac{d}{dn}\sqrt{\frac{\pi}{n}}\right)$$

$$ = -\frac{\sqrt{\pi}}{2}e^{-k} \sum_{n=1}^\infty \frac{(-e^k)^n}{n^{\frac{3}{2}}} \equiv -\frac{\sqrt{\pi}}{2}e^{-k} \text{Li}_{\frac{3}{2}}(-e^k)$$

While these specific steps don't apply when $k>0$, we can use analytic continuation to extend the polylogarithm (and taking the antilimit).

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    $\begingroup$ Nice use of Feynman's trick. $\to +1$ $\endgroup$ Sep 24, 2019 at 9:40
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    $\begingroup$ Alternatively I think you could use the identity $$\frac{\Gamma(s/2)}{n^s\pi^{s/2}}=\int_0^\infty x^{s/2-1}e^{-n^2\pi x}dx,$$ instead of Feynman's trick since your integral is symmetric about $x=0$. You'd need to apply a change of variable and substitute a value for $s$ first. $\endgroup$
    – pshmath0
    Sep 24, 2019 at 10:15
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Using a CAS, I did not find any antiderivative. However, for the integral $$\int_{-\infty}^{\infty} \frac{x^2}{K+e^{x^2}}\,dx=-\frac{\sqrt{\pi } }{2 K} \text{Li}_{\frac{3}{2}}(-K)$$ where appears the polylogarithm function.

$$\int_{-\infty}^{\infty} \frac{x^2}{e^{k}+e^{x^2}}\,dx=-\frac{\sqrt{\pi }}{2} e^{-k} \,\text{Li}_{\frac{3}{2}}\left(-e^k\right)$$

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    $\begingroup$ Hmm, interesting. Originally, I came across the above integral while manipulating a function related to fermi-Dirac statistics. I think if we do go along the road of expanding the above function in terms of the infinite poles, we may end up in the polylogarithmic function only. $\endgroup$ Sep 24, 2019 at 9:25
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I would like to add a different approach, directly connecting the integral to the field of polylogarithms (and I don't know why it wasn't mentioned within the two excellent answers already given).

Clearly, the integrand is an even function of $x$ and thus we only need to be concerned about the interval $[0,\infty]$. Enforcing the substitution $x^2\mapsto x$ reveals the integral representation of the Polylogarithm Function as \begin{align*} \int_{-\infty}^\infty \frac{x^2}{e^{x^2}+e^k}\mathrm dx&=2\int_0^\infty \frac{x^2}{e^{x^2}+e^k}\mathrm dx\\ &=\int_0^\infty\frac{x^\frac12}{e^x+e^k}\mathrm dx\\ &=-e^{-k}\int_0^\infty\frac{x^{\frac32-1}}{e^x/(-e^k)-1}\mathrm dx\\ &=-e^{-k}\left[\Gamma\left(\frac32\right)\operatorname{Li}_\frac32\left(-e^k\right)\right] \end{align*}

$$\therefore~\int_{-\infty}^\infty \frac{x^2}{e^{x^2}+e^k}\mathrm dx~=~-\frac{\sqrt\pi }2e^{-k}\operatorname{Li}_\frac32\left(-e^k\right)$$

The last step follows by basic properties of the Gamma Function. Anyway, note that the conditions - under which this representation holds - are met and at the same time the final expression matches the ones given by the other answers. Integral representation, as the one used here, are quite helpful when dealing with integrals of the "Mellin type" as they are sometimes called (due to their relation to the Mellin Transform).

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    $\begingroup$ Nice and "simple" solution, indeed ! $\to +1$ $\endgroup$ Sep 25, 2019 at 6:53
  • $\begingroup$ @ClaudeLeibovici Thank you kindly! $\endgroup$
    – mrtaurho
    Sep 26, 2019 at 21:15

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