# Integral with rational functions of powers and exponentials

Any ideas how to solve: \begin{equation} \int_0^\infty x^{n+\frac{1}{2}} (e^{a x }-1)^{-\frac{1}{2}} e^{i x t} dx \end{equation} where $a$ and $t$ are real, positive constants; $n$ is a positive integer.

I think the problem comes from having rational functions in both powers and exponential functions.

I tried to get ride of the rational power, but it didn't really help \begin{equation} \frac{\partial}{\partial q } \int_0^\infty x^{n} (e^{a x }-1)^{-\frac{1}{2}} e^{i x t+ q x^{1/2}} dx \end{equation}

Having an hint how to solve this for $t=0$ would already be useful. Thanks!

• Out of curiousity, in what context did this integral appear? – Semiclassical Sep 12 '14 at 19:09
• It a function of my own that I need for my research. But it is well defined, and should have a finite integrale value. I just don't know how to find the analytical form. – Aurelia Sep 12 '14 at 19:20
• Ok. I ask because my physics brain sees the $n=0$ case as the Fourier transform of the density of states for a boson gas (with zero chemical potential). Probably just a coincidence. – Semiclassical Sep 12 '14 at 19:23
• This is where is comes from indeed. I normally use the DOS square, i.e. $\frac{x}{e^{ax}-1}$, in which case the integral is easy to solve – Aurelia Sep 12 '14 at 19:32
• Oof, I'd quite missed that square-root. That indeed makes life interesting. I'll see what I can do... – Semiclassical Sep 12 '14 at 19:36

It's actually enough to resolve the $n=0$ case, since $t$-derivatives of $$F(t)=\int_0^\infty \frac{e^{i x t}}{\sqrt{e^{2x}-1}} x^{1/2}\,dx$$ will bring down more powers of $x$; I've chosen units for so that $a=1$ for convenience. Even with these simplifications, I don't know how to compute a closed-form and so will have to settle for an appropriate series expansion. We rewrite the fraction in the integrand and expand in powers of exponenetials $$\frac{e^{i x t}}{\sqrt{e^{x}-1}}=\frac{e^{i x t-x/2}}{\sqrt{1-e^{-x}}}=\sum_{k=0}^\infty \binom{2k}{k}\left(\frac{e^{-x}}{4}\right)^k e^{i x t-x/2}=\sum_{k=0}^\infty \binom{2k}{k}4^{-k} e^{i x t-(k+1/2)x}.$$

Integrating term-by-term then gives

$$F(t)=\sum_{k=0}^\infty \binom{2k}{k}4^{-k}\cdot \frac{1}{2}\pi^{1/2} [(k+\frac{1}{2})-i x]^{-3/2}=\sum_{k=0}^\infty \binom{2k}{k}\frac{ 2^{-2k-1}\pi^{1/2}}{(k+\frac{1}{2}-i t)^{3/2}}$$

which is a rather formidable result. It's possible that this can be resummed as a hypergeometric series of some sort; I'll see what I can find.

$x^{n+\frac{1}{2}} (1-e^{-a x })^{-\frac{1}{2}} e^{x(-\frac{a}{2})} dx=x^{n+\frac{1}{2}}(\sum_p\frac{(2p-1)!!}{2p!!}e^{-a(p+\frac{1}{2})x})$.

So $\int x^{n+\frac{1}{2}} (1-e^{-a x })^{-\frac{1}{2}} e^{x(-\frac{a}{2})} dx=\Gamma(n+1.5).\sum_p \frac{(2p-1)!!}{2p!!}\frac{1}{a(p+\frac{1}{2})}^{n+1.5}$

But I guess you already had this, and this is not really practical...