How to solve or workaround the integration? From statistical mechanics I've found $ \left( c,\alpha \in \mathbb{R}^+ \right) $:
$ lnZ = c\int_0^\infty dx\ x^2ln\left( 1 + e^{-\alpha x^2}\right) $
what I performed a integration by parts to get:
$ lnZ = \dfrac{2\alpha}{3}c\int_0^\infty dx\dfrac{x^4}{1 + e^{\alpha x^2}} $.
Every exercise in quantum gases (fermions/bosons) lead me to this integration in classical limit. I try to solve it via Taylor's polynomium around $x=0$ because it's a integration limit value. The first term which will not vanish (get 0 value in $x=0$) is in $\frac{d^4f}{dx^4}\left( x\right) = \frac{4!}{24!}x^4\ $ what take me a divergence in the integration.
Other way I've tried is apply Taylor's to $ ln\left( 1+e^{-\alpha x^2}\right) $ at $x=0 $ what is the same to span $ ln\left(1+x\right) $ around $ x=1 $:
$ \int ln\left(1+x\right) dx = ln2 +\sum_{k\in\mathbb{N}}\left( -1 \right)^{k+1}\dfrac{\left( x-1 \right)^{k}}{k\left( k+1 \right)} $.
Using test of $\ \ lim_{k\rightarrow\infty}\left| \frac{A_k}{A_{k+1}} \right| < 1 \ \ $ I conclude this converges on $ x\in\left( 0,2\right) $.
My questions:


*

*The integration was solved and the result is valid just for $ x\in\left( 0,2\right) \Rightarrow e^{-\alpha x^2}\in\left( e^{-4\alpha}, 1\right) $. Is this conclusion right?

*I feel Like is something going wrong in my solution. Else way to solve this integration or workaround it?

 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\ln\pars{Z} = {2\alpha \over 3}\,c
\int_{0}^{\infty}\dd x\,{x^{4} \over 1 + \expo{\alpha x^{2}}}:\ 
{\LARGE ?}}$.

\begin{align}
\ln\pars{Z} & =
{2\alpha \over 3}\,c\int_{0}^{\infty}\dd x\,{x^{4} \over 1 + \expo{\alpha x^{2}}}
\,\,\,\stackrel{\alpha x^{2}\ \mapsto\ x}{=}\,\,\,
{c \over 3\alpha^{3/2}}\int_{0}^{\infty}
{x^{3/2} \over \expo{x} + 1}\,\dd x
\\[5mm] & =
{c \over 3\alpha^{3/2}}\int_{0}^{\infty}
x^{3/2}\pars{{1 \over \expo{x} + 1} - {1 \over \expo{x} - 1}}
\,\dd x +
{c \over 3\alpha^{3/2}}\int_{0}^{\infty}
{x^{3/2} \over \expo{x} - 1}\,\dd x
\\[5mm] & =
-\,{2c \over 3\alpha^{3/2}}\int_{0}^{\infty}
{x^{3/2} \over \expo{2x} - 1}
\,\dd x +
{c \over 3\alpha^{3/2}}\int_{0}^{\infty}
{x^{3/2} \over \expo{x} - 1}\,\dd x
\\[5mm] & =
-\,{2c \over 3\alpha^{3/2}}\bracks{2^{-5/2}\int_{0}^{\infty}
{x^{3/2} \over \expo{x} - 1}
\,\dd x} +
{c \over 3\alpha^{3/2}}\int_{0}^{\infty}
{x^{3/2} \over \expo{x} - 1}\,\dd x
\\[5mm] & =
{\pars{1 - 2^{-3/2}}c \over 3\alpha^{3/2}}\
\underbrace{\int_{0}^{\infty}
{x^{3/2} \over \expo{x} - 1}\,\dd x}
_{\ds{\underbrace{\Gamma\pars{5 \over 2}}_{\ds{3\root{\pi}/4}}\zeta\pars{5 \over 2}}}\ =
\bbx{{1 - 2^{-3/2} \over 4}\root{\pi}\zeta\pars{5 \over 2}
\,{c \over \alpha^{3/2}}}
\end{align}

See Mellin-type Integrals.

