Could you please tell me, How to evaluate this integral which involve hermite polynomials, $\int_{-\infty}^\infty e^{-ax^2}x^{2q}H_m(x)H_n(x)\,dx=?$ where $H_n$ is the $n$-th Hermite polynomial (Physicist's version) and $q,\,m$ and $n$ are positive integers. If $x^{2q}$ term were absent, I am able to perform the integral by writing the product of the Hermite polynomials into a single Hermite polynomial with a higher degree. Can anybody give me a hint to perform this integral?
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$\begingroup$ But $y^{2q}$ (and $e^{-ay^2}$) is independent of $x$... $\endgroup$ – JohnD Dec 4 '13 at 16:25
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$\begingroup$ sorry man, Now it is corrected in terms of x $\endgroup$ – Sijo Joseph Dec 4 '13 at 16:33
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You say you know how to do it when the $x^{2q}$ term is missing. But the following website explains the effect of multiplying a Hermite polynomial by $x$, and you could just apply this rule $2q$ times.
http://en.wikipedia.org/wiki/Hermite_polynomials#Recursion_relation_2
I know you will end up with a bit of a mess. But at least this should work to give concrete formulae in the case that $q$ is small.
answered Dec 4 '13 at 19:08
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$\begingroup$ Thanks man, that will work. But it will be a series solution, anyway thanks for the hint. Warm welcome to similar and other ideas. $\endgroup$ – Sijo Joseph Dec 5 '13 at 20:24
