Computing $\iint_{\mathbb R^2}\exp\left(u\frac{xy}{\sqrt{x^2+y^2}}+v\frac{x^2-y^2}{2\sqrt{x^2+y^2}}-\frac12(x^2+y^2)\right)dxdy$ I'm trying to work on another solution to this question by computing the moment generating function:
$$M(u,v)=\iint_{\mathbb{R}^2}\frac1{2\pi}\exp\left(u\frac{xy}{\sqrt{x^2+y^2}}+v\frac{x^2-y^2}{2\sqrt{x^2+y^2}}\right)\cdot\exp\left(-\frac{1}{2}(x^2+y^2)\right)dxdy$$ which should integrate to : $$M(u,v)=\exp\left(\frac18(u^2+v^2)\right)$$
Polar coordinates look like they could help but I'm helpless (maybe completing the squares in terms of $r$ ?):
$$M(u,v)=\int_0^{+\infty}\int_{0}^{2\pi}\frac1{2\pi}\exp\left(\frac{r}2(u \sin2\theta + v\cos 2\theta)\right) r\cdot\mathrm{e}^{-\frac{1}{2}r^2}\:\mathrm{d}r\,\mathrm{d}\theta$$
 A: After changing to polar coordinates, make the change of variables $\theta=\phi/2$, to obtain that $$\begin{align*}M(u,v)&=\int_0^{\infty}\int_{0}^{2\pi}\frac1{2\pi}\exp\left(\frac{r}2(u \sin2\theta + v\cos 2\theta)\right) r\cdot\mathrm{e}^{-\frac{1}{2}r^2}\,dr\,d\theta\\&=\int_0^{\infty}\int_{0}^{4\pi}\frac1{4\pi}\exp\left(\frac{r}2(u \sin\phi + v\cos \phi)\right) r\cdot\mathrm{e}^{-\frac{1}{2}r^2}\,dr\,d\phi\\&=\frac{1}{2\pi}\int_0^{\infty}\int_{0}^{2\pi}\exp\left(\frac{r}2(u \sin\phi + v\cos \phi)\right) r\cdot\mathrm{e}^{-\frac{1}{2}r^2}\,dr\,d\phi\end{align*}.$$ Changing back to cartesian coordinates, we obtain that $$\begin{align*}M(u,v)&=\frac{1}{2\pi}\int_{\mathbb R^2}\exp\left(\frac{xu+yv}{2}\right)\exp\left(-\frac{x^2+y^2}{2}\right)\,dx\,dy\\
&=\frac{1}{2\pi}\int_{\mathbb R}\exp\left(\frac{ux}{2}-\frac{x^2}{2}\right)\,dx\cdot\int_{\mathbb R}\exp\left(\frac{vy}{2}-\frac{y^2}{2}\right)\,dy\\
&=\frac{1}{2\pi}e^{u^2/8}\int_{\mathbb R}\exp\left(-(u/\sqrt{8}-x/\sqrt{2})^2\right)\,dx\cdot e^{v^2/8}\int_{\mathbb R}\exp\left(-(v/\sqrt{8}-y/\sqrt{2})^2\right)\,dy\\&=e^{(u^2+v^2)/8}.\end{align*}$$ 
