Can we find the asymptotic behavior of this $f(x) =\int_{0}^{\infty}\frac{u^2}{1+\frac{e^{u^2}}{x}}du$? I encountered this function in Statistical Mechanics.  $$f(x) =\int_{0}^{\infty}\frac{u^2}{1+\frac{e^{u^2}}{x}}du$$ 
For $x=0$, we define its value to be zero.
I wanted to see it's asymptotic behavior in the limit x tending to $\infty$. Can we express the asymptotic behavior of this function, in the limit x tending to $\infty$, in terms of other known mathematical functions (if possible; in elementary functions)? For some physical reasons (which are irrelevant here), $x$ belongs to $[0,\infty)$.

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\begin{align}
\mrm{f}\pars{x} & \equiv
\int_{0}^{\infty}{u^{2} \over 1 + \expo{u^{2}}/x}\,\dd u 
\,\,\,\stackrel{u^{2}\ \mapsto\ u}{=}\,\,\,
{1 \over 2}\int_{0}^{\infty}{u^{1/2} \over \expo{u}/x + 1}\,\dd u
\\[5mm] & =
-\,{1 \over 2}\,\Gamma\pars{3 \over 2}\mrm{Li}_{3/2}\pars{-x}
=
-\,{1 \over 4}\,\root{\pi}\,\mrm{Li}_{3/2}\pars{-x}\label{1}\tag{1}
\end{align}

See the Polylogarithm Integral Representation.

By using the Polylogarithm Asymptotic Expansion, \eqref{1}
becomes ( $\ds{B_{n}}$ is a Bernoulli Number )
\begin{align}
\mrm{f}\pars{x} & =
-\,{1 \over 4}\,\root{\pi}\sum_{k = 0}^{\infty}\pars{-1}^{k}\pars{1 - 2^{1 - 2k}}\pars{2\pi}^{2k}\,
{B_{2k} \over \pars{2k}!}\,{\ln^{3/2 - 2k}\pars{x} \over \Gamma\pars{5/2 - 2k}}
\\[5mm] & =
\bbx{{1 \over 3}\,\ln^{3/2}\pars{x} + {\pi^{2} \over 24}\,\ln^{-1/2}\pars{x} +
{7\pi^{4} \over 1920}\,\ln^{-5/2}\pars{x} +
\mrm{O}\pars{\ln^{-9/2}\pars{x}}}
\\ & \mbox{as}\ x\ \to\ \infty
\end{align}
