Integrate $\int_0^\infty \frac{\sqrt{x}}{e^{(x-\alpha)\beta}+1}dx$

Question

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.

Answer 1

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.

Answer 2

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 $$f(x)=\text{Li}_{\frac{3}{2}}\left(-e^x\right)-k$$ $$f'(x)=\text{Li}_{\frac{1}{2}}\left(-e^x\right)$$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large a}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\cal R}\pars{\alpha,\beta} \equiv \int_{0}^{\infty}{% \root{x} \over \expo{\pars{x - \alpha}\beta} + 1}\,\dd x\,,\quad \beta > 0}$

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 $\beta$.