Integral of exponential function with trigonometric identities

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

• hint Do a change of variable from Polar to Cartesian coordinates, then do a translation in the $x$ direction, then (optionally) change back to Polar coordinates. – Willie Wong Sep 1 '11 at 1:56
• I have changed the above problem to the definite integral using cdf, here is the link. math.stackexchange.com/questions/61103/…. Help needed... – shaikh Sep 1 '11 at 3:02

Integration with respect to $\theta$ can be carried out explicitly in terms of modified Bessel function of the first kind.
$$\int_0^{2 \pi} \exp\left( \frac{ \alpha r}{2 \sigma^2} \cos \theta \right) \, \mathrm{d} \theta = 2 \pi I_0 \left( \frac{ \alpha r}{2 \sigma^2} \right)$$
$$\mathcal{I}_d = 2 \pi \int_0^d r \cdot \mathrm{e}^{-\frac{r^2}{4 \sigma^2}} \cdot I_0 \left( \frac{ \alpha r}{2 \sigma^2} \right) \mathrm{d} r = 4 \pi \sigma^2 \exp\left( \frac{\alpha^2}{4 \sigma^2} \right) \left(1 - Q_1\left( \frac{\alpha}{\sqrt{2} \sigma}, \frac{d}{\sqrt{2} \sigma} \right)\right)$$ where $Q_1(a,b)$ is the Marcum Q-function.
Notice that $$\lim_{d \to \infty} \mathcal{I}_d = 4 \pi \sigma^2 \exp\left( \frac{\alpha^2}{4 \sigma^2} \right)$$