Integral of a gaussian over a slice of the plane 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.
 A: 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:
$$ I_1=\frac{1}{\sqrt{\pi\sigma}}\int_{0}^{+\infty}\operatorname{Erf}\left(\frac{l\tan\frac{\pi}{n}}{\sigma}\right)\exp\left(-\frac{(l-1)^2}{\sigma}\right)\,dl.$$
