# How can I solve this integral equation in terms of Hermite polynomials?

It must be proven that the solution of the integral equation $$f(x)=\int_{-\infty}^{+\infty} e^{-(x-t)^2} g(t)dt$$ is $$g(x)=\frac{1}{\sqrt{}\pi}\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{2^nn!} H_n(x)$$

where the $H_n(x)$ are the Hermite polynomials.

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What is $f$?  – Jeff Dec 18 '11 at 16:51
It looks like a Fredholm equation of the first kind... the sort that's usually solved with a Fourier transform. – J. M. Dec 19 '11 at 3:12

You can start by plugging the series on the right side of the definition of $g(x)$, or rather $g(t)$, into the expression on the right side of $f(x)$. Interchange sum and integral, and you see what looks like the beginning of the Maclaurin series expansion for $f(x)$. (Your notation already assumes $f$ is infinitely differentiable.) It then suffices to show that $$x^n=\frac{1}{2^n\sqrt{\pi}}\int_{-\infty}^\infty\exp(-(x-t)^2)H_n(t)\, dt,$$ as well as justifying the interchange of sum and integral.