Integral of exponential function with trigonometric identities

Question

I need help in solving the following definite integral. I could not find any example like this $$\int_{0}^{2\pi}\int_{0}^{d}\exp\!\Big(\frac{-r^2 +2\alpha\; r\cos\theta}{4\;\sigma^2}\Big)r\; dr\; d\theta$$

Answer 1

Integration with respect to $\theta$ can be carried out explicitly in terms of modified Bessel function of the first kind.

$$ \int_0^{2 \pi} \exp\left( \frac{ \alpha r}{2 \sigma^2} \cos \theta \right) \, \mathrm{d} \theta = 2 \pi I_0 \left( \frac{ \alpha r}{2 \sigma^2} \right) $$

The remaining integral

$$ \mathcal{I}_d = 2 \pi \int_0^d r \cdot \mathrm{e}^{-\frac{r^2}{4 \sigma^2}} \cdot I_0 \left( \frac{ \alpha r}{2 \sigma^2} \right) \mathrm{d} r = 4 \pi \sigma^2 \exp\left( \frac{\alpha^2}{4 \sigma^2} \right) \left(1 - Q_1\left( \frac{\alpha}{\sqrt{2} \sigma}, \frac{d}{\sqrt{2} \sigma} \right)\right) $$ where $Q_1(a,b)$ is the Marcum Q-function.

Notice that $$ \lim_{d \to \infty} \mathcal{I}_d = 4 \pi \sigma^2 \exp\left( \frac{\alpha^2}{4 \sigma^2} \right) $$