Prove that $\int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2+2xy \cos \theta)} \,dx dy = \frac{\theta}{2\sin\theta}$

Question

Prove that the following integral: $$\int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2+2xy \cos \theta)} \,dx dy = \frac{\theta}{2\sin\theta}$$

The hints written on the book are beta function and to suppose the slanted $xy$ plane. However, I cannot figure out a thing.

Answer 1

Assume $\theta\in\left[0,\frac{\pi}{2}\right]$. Then, by setting $z=x+y\cos\theta$ we have: $$ I = \int_{0}^{+\infty}\int_{y\cos\theta}^{+\infty} e^{-(z^2+y^2\sin^2\theta)}\, dz \,dy = \frac{1}{\sin\theta}\int_{0}^{+\infty}\int_{w\cot\theta}^{+\infty}e^{-(z^2+w^2)}\,dz\,dw.$$ Switching now to polar coordinates we get $I=\frac{\theta}{2\sin\theta}$ as stated.