Recently I got stuck on a 2 dimensional integral in polar coordinate,
the expression is the following:
$I(x)=\lim_{\xi\rightarrow0^+}\int_0^\infty dr\int_{-\pi/2}^{\pi/2}dt\frac{2\xi ^{2-2 x}r^{2x+1} \cos (t)^{2 x}}{\left(e^{r^2}-1\right) \left(e^{\xi ^2+r^2+2 \xi r \sin (t)}-1\right)}$
Basically, it is integral in the right-half plane. The result should be finite for $0<x<1$.
I was trying to integral over the angular part first and then do the radial integral using a contour in the complex plane, but the calculation becomes hard for general $x$. I still could not calculate or even guess the correct form of the result. I tried combinations of $\Gamma$ function, $\zeta$ function and $\sin(\pi(x-1))$ and so on…
While I could not solve the integral for general $x$, I obtained some result for specific $x$ using numerical method:
$I(1/2)=I(3/4)=\pi^2$, $I(1/4)=2\pi^2$. Furthermore, it seems that $I(1/6)=2\sqrt{3}I(2/3)$.
Despite the unfruitful attempts, I strongly believe that there should be a unified and elegant result, base on the calculation for specific $x$ values.
Any help will be greatly appreciated!
Edit:
Seems the result can be easily obtained by integrating over radial direction first and then doing the angular part.
The result is indeed simple:
$I=\frac{2\pi^2}{4^{x}\sin(\pi x)}$
Best Answer
Since I figured out the answer myself after posting the question, let me provide my tentative solution to this problem.
This integral can be evaluated in polar coordinate: $I= \lim_{\kappa\rightarrow0}\int_0^\infty dr\int_{0}^{\pi} d\theta f(r,\theta)$ with
\begin{equation} f(r,\theta)=\frac{\kappa^{2-2x}}{\exp(r^2)-1}\frac{2r(r\sin(\theta))^{2x}}{\exp(r^2-2r\kappa\cos(\theta)+\kappa^2)-1} \end{equation}
First perform the integral over $r$. The function $f(r,\theta)$ has a branch cut from $r=0$ to $r=+ \infty$. It also has poles at $r_{\pm,n}=\kappa\cos(\theta)\pm\sqrt{-\kappa^2\sin(\theta)^2+2\pi ni}$ with $n$ being an integer. The residues at $r=r_{\pm.n}$ are: \begin{equation} \text{Res}f(r_{\pm,n}) = \pm\frac{2\kappa^{2-2x}\sin(\theta)^{2x}}{\exp((r_\pm)^2)-1}\frac{(r_\pm)^{2x+1}}{r_+-r_-} \end{equation}
In the limit of $\kappa\rightarrow0$, the residues vanishes if $n\neq0$. When $n=0$, the residues remain finite: \begin{equation} \text{Res}f(r_{\pm,0}) = \pm\frac{2\kappa^{2-2x}\sin(\theta)^{2x}}{\exp(\kappa^2\exp(\pm2i\theta))-1}\frac{(\kappa\exp(\pm i\theta))^{2x+1}}{2i\kappa\sin(\theta)}\approx-\pm i(\exp(\pm i\theta)\sin(\theta))^{2x-1} \end{equation} Because of the branch cut, the integral can be related to the sum of all residues in the complex plane by: \begin{equation} I(1-\exp(2i\pi x))=2\pi i\sum_n\int_0^{\pi}d\theta\text{Res} f(r_{\pm,n},\theta) \end{equation} Thus, the integral can be calculated as: \begin{equation} I=\frac{2\pi i\int_0^{\pi} d\theta(- i(\exp(i\theta)\sin(\theta))^{2x-1}+ i(\exp(- i\theta)\sin(\theta))^{2x-1})}{1-\exp(2i\pi x)}=\frac{2\pi^2}{4^{x}\sin(\pi x)} \end{equation}