TITLE: Integrate $\int_0^\infty \frac{\sqrt{x}}{e^{(x-\alpha)\beta}+1}dx$
QUESTION [5 upvotes]: I need to solve for the parameter $\alpha$ after I calculate the integral.$$
\mathcal{R}(\alpha,\beta)=\int_0^\infty \frac{\sqrt{x}}{e^{(x-\alpha)\beta}+1}dx, \ \ \beta >0
$$ The result of this integral is
$$
\mathcal{R}(\alpha,\beta)=-\frac{\sqrt{\pi}}{2\beta^{\frac32}}\text{Li}_{\frac32}{(-e^{\beta\alpha})}
$$ Thanks to David H solution below.  However, I need to invert this result now and solve for $\alpha$.
Thanks for the help.

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{{\cal R}\pars{\alpha,\beta} \equiv \int_{0}^{\infty}{%
     \root{x} \over \expo{\pars{x - \alpha}\beta} + 1}\,\dd x\,,\quad \beta > 0}$

I found that the OP result is true  whenever $\ds{\large\alpha < 0\mbox{!!!}}$:

\begin{align}
\color{#00f}{\large{\cal R}\pars{\alpha,\beta}}&=\int_{0}^{\infty}\root{x}\,{%
\expo{-\pars{x - \alpha}\beta} \over 1 + \expo{-\pars{x - \alpha}\beta}}\,\dd x
=\beta^{-1}\int_{x = 0}^{x \to \infty}\root{x}\,
\dd\ln\pars{{1 + \expo{-\pars{x - \alpha}\beta}}}
\\[3mm]&=\beta^{-1}\int_{0}^{\infty}\ln\pars{{1 + \expo{-\pars{x - \alpha}\beta}}}\,
\half\,x^{-1/2}\,\dd x
\\[3mm]&=\half\,\beta^{-1}\int_{0}^{\infty}x^{-1/2}\sum_{\ell = 1}^{\infty}
{\pars{-1}^{\ell + 1} \over \ell}\,\expo{-\ell\pars{x - \alpha}\beta}\,\dd x
\\[3mm]&=\half\,\beta^{-1}\sum_{\ell = 1}^{\infty}{\pars{-1}^{\ell + 1} \over \ell}\,
\expo{\ell\alpha\beta}\int_{0}^{\infty}x^{-1/2}
\expo{-\ell\beta x}\,\dd x
\\[3mm]&=\half\,\beta^{-1}\sum_{\ell = 1}^{\infty}{\pars{-1}^{\ell + 1} \over \ell}\,
\expo{\ell\alpha\beta}\,{1 \over \pars{\ell\beta}^{1/2}}\
\overbrace{\int_{0}^{\infty}x^{-1/2}\expo{-x}\,\dd x}
^{\ds{\Gamma\pars{\half} = \root{\pi}}}
\\[3mm]&=-\,{\root{\pi} \over 2\beta^{3/2}}\sum_{\ell = 1}^{\infty}
{\pars{-\expo{\alpha\beta}}^{\ell} \over \ell^{3/2}}
=\color{#00f}{\large -\,{\root{\pi} \over 2\beta^{3/2}}\,
{\rm Li}_{3/2}\pars{-\expo{\alpha\beta}}}
\end{align}

Since this integral is linked to a Fermi-Dirac statistics, $\ds{\alpha < 0}$ corresponds to a non degenerated Fermi gas which should be true for 'small' values of $\beta$.