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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?

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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).

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  • $\begingroup$ 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. $\endgroup$ – Sijo Joseph Nov 7 '13 at 19:55

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