I need to evaluate the following $n$ real integrals: $$\int_{\frac{\pi}{2}-\frac{\pi}{n}}^{\frac{\pi}{2}+\frac{\pi}{n}}\int_0^\infty\frac{1}{\pi \sigma^2}e^{-\frac{|re^{i\theta}-i|^2}{\sigma^2}} \ dr \ d\theta$$ $$\int_{\frac{\pi}{2}+\frac{\pi}{n}}^{\frac{\pi}{2}+\frac{3\pi}{n}}\int_0^\infty\frac{1}{\pi \sigma^2}e^{-\frac{|re^{i\theta}-i|^2}{\sigma^2}} \ dr \ d\theta$$ $$\dots$$
The domain of each integration looks like a radial slice of the plane from the origin, and the integrand is a complex circularly symmetric Gaussian centered about $i$.
I am really not sure where to begin. I was hoping that the result is well-known since things like the Rayleigh distribution are commonly studied, but on an initial search turned up nothing.
Those integrals are really hard to be computed exactly, but quite easy to estimate.
For instance, if $E=\{z\in\mathbb{C}:|\operatorname{arg}(z)-\pi/2|<\pi/n\}$ and $B_\rho$ is the ball centered in $i$ having radius $\rho=\sin\frac{\pi}{n}$, then $B\subset E$ and:
$$\int_{B}\frac{1}{\pi\sigma^2}\exp\left(-\frac{1}{\sigma^2}\|z-i\|^2\right)d\mu=1-e^{-\rho^2/\sigma^2}.$$
For any $R\in[\rho,+\infty)$, let
$$\delta(R)=\frac{\nu(E\cap B_R)}{2\pi R}$$ where $\nu$ is the $1$-dimensional Lebesgue measure. $\delta(R)$ just gives the percentage of points of $B_R$ that lie inside $E$: it is a function that quickly decreases from $1$ to $\frac{1}{n}$. We have:
$$I_1=\int_{E}\frac{1}{\pi\sigma^2}\exp\left(-\frac{1}{\sigma^2}\|z-i\|^2\right)d\mu=1-e^{-\rho^2/\sigma^2}+\int_{1}^{+\infty}\frac{R\,\delta(R)}{\pi\sigma^2}\exp\left(-\frac{R^2}{\sigma^2}\right) dR$$ so: $$ I_1 \geq \frac{1}{n}+\frac{n-1}{n}\left(1-e^{-\rho^2/\sigma^2}\right), $$ for instance. Tighter bounds can be derived from more careful approximations of $\delta(R)$, and the same approach works also for the second integral. By integrating along horizontal lines, we have that the exact value of the first integral is given by: $$ I_1=\frac{1}{\sqrt{\pi\sigma}}\int_{0}^{+\infty}\operatorname{Erf}\left(\frac{l\tan\frac{\pi}{n}}{\sigma}\right)\exp\left(-\frac{(l-1)^2}{\sigma}\right)\,dl.$$