Integral involving gaussian function I would like to calculate the following integral:
$$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\quad (x^2+y^2)^k\exp\left(-\dfrac{(x-x_0)^2+(y-y_0)^2}{a^2}\right)\,\mathrm dx\,\mathrm dy$$
Any clue on how to proceed?
Thanks
 A: You can use the fact that 
$$(x^2+y^2)^k = \sum_{n=0}^k {k \choose n}x^{2n}y^{2(k-n)}$$
So
$$\mathrm{I}=\int_{\!-\infty}^{\!+\infty}\!\!\!\int_{\!-\infty}^{+\infty}dxdy(x^2\!\!+\!y^2)^k\! e^{-\frac{(x-x_0)^2+(y-y_0)^2}{a^2}}\!\!=\!\sum_{n=0}^k{k \choose n}\int_{-\infty}^{+\infty}\!\!\!x^{2n}e^{-\frac{(x-x_0)^2}{a^2}}dx  \!\!\int_{-\!\infty}^{\!+\infty}y^{2(k-n)}e^{-\frac{(y-y_0)^2}{a^2}}dy$$
One can work with it like with scaled plane moments of the normally distributed random variable. It is well known that $$\operatorname{E} \left[ X^p \right] =a^p \cdot (-i\sqrt{2}\mathrm{sgn}x_0)^p \; U\left( {-\frac{1}{2}p},\, \frac{1}{2},\, -\frac{1}{2}(x_0/a)^2 \right)$$ for a probability function $ \frac{1}{a\sqrt{2\pi}} e^{ -\frac{(x-x_0)^2}{2a^2} }$ but we have a scaled version (up to the multiplyer $\frac{1}{a\sqrt{\pi}}$).
So
$$\int_{-\infty}^{+\infty}\!\!\!x^{2n}e^{-\frac{(x-x_0)^2}{a^2}}dx=a\sqrt{\pi}\operatorname{E} \left[ X^{2n} \right]$$
and $$\int_{-\infty}^{+\infty}\!\!\!y^{2(k-n)}e^{-\frac{(y-y_0)^2}{a^2}}dx=a\sqrt{\pi}\operatorname{E} \left[ Y^{2(k-n)} \right]$$
So one can obtain:
$$\mathrm{I}=(-1)^k (a^2)^{k+1}\pi \sum_{n=0}^k{k \choose n}(\mathrm{sgn}x_0)^{2n}(\mathrm{sgn}y_0)^{2(k-n)}U\left(\!\!-\!n\!,\!\frac{1}{2}\!,\!-\frac{x_0^2}{a^2} \right)U\left(\!\!n\!-\!k\!,\!\frac{1}{2}\!,\!-\frac{y_0^2}{a^2} \right)$$
If $x_0=y_0=0$ then   $$\mathrm{E}\left[X^{2n}\right] =  \left(\frac{a}{\sqrt{2}}\right)^{2n}\,(2n-1)!! $$ and  $$\mathrm{E}\left[Y^{2(k-n)}\right] = \left(\frac{a}{\sqrt{2}}\right)^{2(k-n)}\,(2(k-n)-1)!! $$ So everything gets even simpler:
$$\mathrm{I}=\pi \left(\frac{a^2}{2}\right)^{k+1}\sum_{n=0}^k{k \choose n}(2n-1)!!(2k-2n-1)!!$$
