Integrate $\int_0^\infty \frac{\sqrt{x}}{e^{(x-\alpha)\beta}+1}dx$ 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.
 A: Leting $u=\beta x$ and $a=\alpha\beta$,
$$\mathcal{R}(\alpha,\beta)=\int_0^\infty \frac{\sqrt{x}}{e^{(x-\alpha)\beta}+1}dx\\
=\beta^{-3/2}\int_0^\infty \frac{\sqrt{u}}{e^{(u-a)}+1}du\\
=\beta^{-3/2}\mathcal{R}(a,1)=\beta^{-3/2}\mathcal{R}(\alpha\beta,1).$$
So it suffices to evaluate the integral for the $\beta=1$ case, and the case for general $\beta>0$ will follow automatically.
For this definite integral, Wolfram Alpha gives a result 
$$\int_0^{\infty}\frac{\sqrt{u}}{e^{(u-a)}+1}du=-\frac{\sqrt{\pi}}{2}\text{Li}_{\frac32}{(-e^a)}$$
Thus,
$$\mathcal{R}(\alpha,\beta)=-\frac{\sqrt{\pi}}{2\beta^{\frac32}}\text{Li}_{\frac32}{(-e^{\beta\alpha})}.$$
I'm not sure how to invert this formula however.
A: I do not think that you could analytically invert the formula and then, the only way I can think about is numerical solution. If th value of $\mathcal{R}(\alpha,\beta)$ is given, as well as the value of $\alpha$ or $\beta$, Newton method would be quite efficient.  
If you want me to more elaborate on this topic, please give me numbers for $\mathcal{R}(\alpha,\beta)$,  $\alpha$ or $\beta$ and I should illustrate the example.
Added later to my answer
Let us suppose that $\beta$ be given. So we have to solve, for $x=\alpha \beta$, the equation $$k=-\frac{2 \beta ^{3/2}}{\sqrt{\pi }}\mathcal{R}(\alpha,\beta)=\text{Li}_{\frac{3}{2}}\left(-e^x\right)$$ Using Newton method, starting at $x=0$, the first iterate is $$x=\frac{k+\left(1-\frac{1}{\sqrt{2}}\right) \zeta
   \left(\frac{3}{2}\right)}{\left(\sqrt{2}-1\right) \zeta \left(\frac{1}{2}\right)}$$ Using Halley, we should arrive for the first iterate to $$x=\frac{\left(7+5 \sqrt{2}\right) \left(\left(\sqrt{2}-2\right) \zeta
   \left(\frac{3}{2}\right)-2 k\right) \left(\zeta \left(-\frac{1}{2}\right)
   \left(\left(4 \sqrt{2}-2\right) k+\left(5 \sqrt{2}-6\right) \zeta
   \left(\frac{3}{2}\right)\right)+4 \left(2 \sqrt{2}-3\right) \zeta
   \left(\frac{1}{2}\right)^2\right)}{8 \zeta \left(\frac{1}{2}\right)^3}$$ If you really want the true solution, forget these estimates and use Newton iterations up to convergence. If $$f(x)=\text{Li}_{\frac{3}{2}}\left(-e^x\right)-k$$ $$f'(x)=\text{Li}_{\frac{1}{2}}\left(-e^x\right)$$
A: $\newcommand{\+}{^{\dagger}}
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$\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 $\beta$.

