I'm having difficulty with a question. It says
By putting $x=r\cos(\theta), y=r\sin(\theta)$, prove that $$\int_0^{\infty}\int_0^{\infty}e^{-(x^2 + 2xy\cos(\alpha)+y^2)}\mathrm dx\mathrm dy=\frac{\alpha}{2\sin(\alpha)}$$
Making the substitution we get \begin{align*}&\int_0^{\pi/2}\int_0^{\infty}re^{-r^2 (\sin(2\theta) \cos(\alpha) + 1)}dr\ d\theta \\ \\ =&\int_0^{\pi/2} \frac{1}{2(\sin(2\theta) \cos(\alpha) + 1)} d\theta\\ \\ =&\frac{\alpha}{2\sin(\alpha)} \int_0^{\pi/2} \frac{\sin(\alpha)}{2 \alpha \cos(\alpha) \sin(\theta) \cos(\theta) + \alpha} d\theta\\ \\ =&\frac{\alpha}{2\sin(\alpha)} \int_0^{\pi/2}\frac{\tan(\alpha) \sec^2(\theta)}{2 \alpha \tan(\theta) + \alpha \sec(\alpha) \sec^2(\theta)} d\theta \end{align*} setting $u=\tan(\theta),\ du=\sec^2(\theta)\ d\theta$ $$=\frac{\alpha}{2\sin(\alpha)}\int_0^{\infty}\frac{\tan(\alpha)\ du}{\alpha \sec(\alpha) + 2\alpha u+ \alpha \sec(\alpha) u^2}$$
Now \begin{align*}u&=\frac{\cos(\alpha)\{-2\alpha\pm\sqrt{4\alpha ^2 - 4\alpha ^2\sec^2(\alpha)}\}}{2\alpha}\\ \\ &=-\cos(\alpha) \mp i\sin(\alpha)\\ &=-e^{\pm i\alpha} \end{align*}
So substituting in
\begin{align*} =&\frac{\alpha}{2\sin(\alpha)}\int_0^{\infty}\frac{\tan(\alpha)\ du}{(u-e^{i\alpha})(u-e^{- i\alpha})}\\ \\ =&\frac{\alpha}{2\sin(\alpha)}\left[ \frac{\tan(\alpha)\{\log(1-ue^{i\alpha}) - \log(-u+e^{i\alpha})\}}{-1+e^{2i\alpha}} \right]^{\infty}_{u\, =\, 0} \end{align*}
which I then can't evaluate (and certainly isn't $1$). Any help?