Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Could you please tell me, How to evaluate the integral, $\int_{-\infty}^\infty\int_{-\infty}^\infty\dfrac{e^{-a(x^2+y^2)}}{\sqrt{k^2+\beta^2(x^2-y^2)^2}}dx~dy$

I already have obtained a series solution in terms of gamma function, if anybody can write this as a single function, it would be really appreciated. my answer is, $\int_{-\infty}^\infty\int_{-\infty}^\infty\dfrac{e^{-a(x^2+y^2)}}{\sqrt{k^2+\beta^2(x^2-y^2)^2}}dx~dy=\dfrac{\pi}{\beta}\sum\limits_{k=0}^\infty\dfrac{(\beta/a)^{2k+1}}{(k!)^2}\dfrac{\Gamma(k+1/2)\Gamma(2k+1)}{\Gamma(1/2-k)} .$ Can anybody relate this answer to a single function?

share|cite|improve this question

Assuming $a,k,\beta > 0$, I believe the answer is $$ \dfrac{\pi^2}{4 \beta} \left( J_0\left(\dfrac{ak}{2\beta}\right)^2 + Y_0\left(\dfrac{ak}{2\beta}\right)^2\right) $$ where $J_0$ and $Y_0$ are Bessel functions of the first and second kinds.

Maple (after changing to polar coordinates and some simplification) got a similar result but with $Y_0(-ak/(2\beta))$ which is obviously not quite right (the result would be complex).

share|cite|improve this answer
Hi Robert Israel, thank you very much for the result. Could you please enlighten me by providing the way that you got the answer ? I have tried the integral by simply taking the series expansion of the polynomial in the denominator and then changing the things to polar coordinate, I got a series of gamma function. But I don't like it since it is a series expansion. This sounds pretty good, if somebody provide me the details I will be happy. – Sijo Joseph Nov 7 '13 at 19:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.