2

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

Lord_Farin
  • 17,743
JFNJr
  • 1,055
  • Expand out the binomial term and express the double integral as a sum of double integrals, each of which can be expressed as a product of two integrals, one w.r.t. $x$ and the other w.r.t $Y$. Then work out each integral or look up the answer in a table since each integral is an even moment of an $\mathcal N(\mu,\sigma^2)$ random variable, and there is a well-tabulated (if not well-known) expression for the $n$-th moment. – Dilip Sarwate May 15 '13 at 18:28
  • @DilipSarwate I did not get it. Can you be more specific? How can I express it as the sum of double integrals? – JFNJr May 15 '13 at 18:30
  • Example: $(x^2+y^2)^2 = x^4y^4+2x^2y^2+y^4$. Your integral is thus the sum of three double integrals, the middle of which is $$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}dxdy\quad 2x^2y^2\cdot e^{-\frac{(x-x_0)^2+(y-y_0)^2}{a^2}}=2\int_{-\infty}^{+\infty}x^2e^{-(x-x_0)^2/a^2},dx \int_{-\infty}^{+\infty}y^2e^{-(y-y_0)^2/a^2},dy.$$ Can you relate each integral on the right to a similar integral that computes the second moment of a $\mathcal N(\mu,\sigma^2)$ normal random variable for certain choices (that you need to figure out) for the values of $\mu$ and $\sigma^2$? – Dilip Sarwate May 15 '13 at 18:38

1 Answers1

0

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: $$\mathrm{I}=\pi \left(\frac{a^2}{2}\right)^{k+1}\sum_{n=0}^k{k \choose n}(2n-1)!!(2k-2n-1)!!$$

Caran-d'Ache
  • 3,564