I came across the following integral in polar form- $$ \int\limits_0^{2\pi}\int\limits_{0}^{\infty}\exp\bigg(-\frac{1}{2(1-\alpha^2)}(r^2-2\alpha r^2\cos\theta\sin\theta)\bigg)rdrd\theta$$.
Is there any way to solve this integral or any substitution that we can make to solve the above integral? Thanks in advance for any help.
The inner integral is immediate because, factorizing the exponent you get an integral in the form
$$\int_0^{\infty}x e^{-Ax^2}dx=\left.-\frac{1}{2A}e^{-Ax^2}\right]_{0}^{\infty}$$
where
$$A=\frac{1-2\alpha \cos\theta\sin\theta}{2(1-\alpha^2)}$$