Double integral using polar coordinates with e to the power The integral I have is $$\int_0^{+\infty}\int_0^{+\infty}\exp(-x^2 - 2xy\cos(a) - y^2)\,\mathrm{d}x\mathrm{d}y.$$ I plan to use the change to polar coordinates, but as I do it, I get stuck with $\cos(a)$. What do I do with it? 
 A: Due to the periodicity of the cosine we can restrict our attention to $a \in [0, \pi]$. Note that the integral diverges for $a = \pi$, so we want to compute
$$ f \colon [0,\pi) \to (0,\infty) \, , \, f(a) = \int \limits_0^\infty \int \limits_0^\infty \mathrm{e}^{-[x^2 + y^2 + 2 \cos(a) x y ]} \, \mathrm{d} x \, \mathrm{d} y \, . $$
Introducing polar coordinates we find
$$ f(a) = \int \limits_0^{\pi/2} \int \limits_0^\infty r \mathrm{e}^{- r^2 [1 + 2 \cos(a) \sin(\phi) \cos(\phi)]} \, \mathrm{d} r \, \mathrm{d} \phi  = \frac{1}{2} \int \limits_0^{\pi/2} \frac{\mathrm{d} \phi}{1 + 2 \cos(a) \sin(\phi) \cos(\phi)} \, , \, a \in [0,\pi) .$$
Now the substitution $t = \tan(\phi)$ is helpful. It leads to
$$ f(a) = \frac{1}{2} \int \limits_0^\infty \frac{\mathrm{d} t}{1 + t^2 + 2 \cos(a) t} \, , \, a \in [0,\pi) .$$
In particular,
$$ f(0) = \frac{1}{2} \int \limits_0^\infty \frac{\mathrm{d} t}{(1+t)^2} = \frac{1}{2} \, . $$
For $a \in (0,\pi)$ we can complete the square in the denominator and let $t + \cos(a) = \sin(a) u$:
\begin{align}
f(a) &= \frac{1}{2} \int \limits_0^\infty \frac{\mathrm{d} t}{[t + \cos(a)]^2 + \sin^2 (a)} = \frac{1}{2 \sin(a)} \int \limits_{\cot(a)}^\infty \frac{\mathrm{d} u}{1 + u^2} \\
&= \frac{1}{2 \sin(a)} \operatorname{arccot}[\cot(a)] = \frac{a}{2 \sin(a)} \, .
\end{align}
By the way, the value of the integral over the entire plane is a simple consequence of this result:
$$ \int \limits_{-\infty}^\infty \int \limits_{-\infty}^\infty \mathrm{e}^{-[x^2 + y^2 + 2 \cos(a) x y ]} \, \mathrm{d} x \, \mathrm{d} y = 2 [f(a) + f(\pi - a)] = \frac{\pi}{\sin(a)} \, , \, a \in (0,\pi) . $$
