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In an engineering application, I've been trying to calculate the following integral involving Laguerre polynomials:

$\int_{-\infty}^\infty dx L_n(x^2+\beta^2) e^{-x^2/2+i x \alpha}$,

where $n=0,1,2,3,\ldots$ and $\alpha,\beta$ are real.

For $\beta=0$ the integral can be found in Gradshteyn: $\sqrt{2\pi}/(2^n n!)e^{-\alpha^2/2}[H_n(\alpha/\sqrt{2})]^2$, where $H_n$ is the Hermite polynomial.

However, I am not sure what to do in the case $\beta\neq 0$ -- any thoughts?

Note: Mathematica numerically evaluates the integral for small $n$ (and general $\beta\neq 0$), but I am looking for the solution in a closed form for any $n$.

Thanks alot in advance!

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Interestingly, mathematica keeps telling me that the integral diverges! (even when I change the integral to the integral from $0$ to $\infty,$ and replace your $e^{i a x}$ by $\cos(a x).$ Weird. – Igor Rivin Dec 19 '11 at 21:28
    
Yes, it reports that if you keep $n$ as a parameter... however, if you plug in specific values of $n$ like 0,1,2,3,4,..., you get a well-defined answer. – Mark M Dec 19 '11 at 23:00
2  
What are you trying to do with these integrals? For many purposes a generating function is good enough, and here the generating function $\sum_{n=0}^\infty L_n(x) z^n = (z-1)^{-1} \exp(-xz/(z-1))$ yields a closed form for the generating function of your integrals, which is $$ \sqrt{\frac{2\pi}{(z-1)(3z-1)}} \phantom. \exp\left(-\frac{(\alpha^2+6\beta^2)z^2-2(\alpha^2+\beta^2)z + \alpha^2}{2(z-1)(3z-1)}\right) $$ if I calculated (and TeXed) correctly. – Noam D. Elkies Dec 19 '11 at 23:21

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