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Provided that the following generating expression defines the Hermite polynomials $ H_n(x)$

$$ F(x,h)= e^{2hx-h^{2}} = \sum_{n=0}^{\infty} H_n(x) \frac{h^{n}}{n!} $$

Find a recursion relation and solve the following integral

$$ J_n = \int_{-\infty}^{\infty}e^{-\frac{1}{2}x^{2}}H_n(x)dx $$

I've already done the tedious part of solving for the recursion relation among $H_{n-1}, H_n, H_{n+1}$ as

$$ -2\sum_{n=0}^{\infty} \frac {z^{n+1}} {n!}H_n(x) + 2x\sum_{n=0}^{\infty} \frac {z^{n}} {n!}H_n(x) = \sum_{n=0}^{\infty} \frac {z^{n-1}} {(n-1)!}H_n(x) $$ $$ \rightarrow-2\sum_{n=0}^{\infty} \frac {z^{n+1}} {n!}H_n(x) = -2\sum_{n=0}^{\infty} \frac {z^{n}} {(n-1)!}H_n(x) $$ $$ \rightarrow-2\sum_{n=0}^{\infty} \frac {z^{n+1}} {n!}H_n(x) = -2\sum_{n=0}^{\infty}n \frac {z^{n}} {n!}H_{n-1}(x) $$ $$ \rightarrow \sum_{n=1}^{\infty} \frac {z^{n-1}} {(n-1)!}H_n(x) = \sum_{n=0}^{\infty} \frac {z^{n}} {n!}H_{n+1}(x) $$ $$ \rightarrow -2\sum_{n=1}^{\infty} n \frac {z^{n}} {n!}H_{n-1} + 2x\sum_{n=0}^{\infty} \frac {z^{n}} {n!}H_n(x) = \sum_{n=0}^{\infty}\frac {z^{n}} {n!}H_{n+1}(x) $$ $$ \therefore H_{n+1}(x) = 2x H_n(x)-2n H_{n-1}(x) $$

The next part is solving for the integral, which I'm not sure how to approach doing.

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It is $-\frac {1} {2}x^{}2$, yes – Zainy May 5 '13 at 9:21

You really didn't need the recurrence for the Hermites. Rather, note that

$$\int_{-\infty}^{\infty} dx \, e^{-x^2/2} e^{2 h x - h^2} = \sum_{n=0}^{\infty} J_n \frac{h^n}{n!}$$

On the LHS:

$$\begin{align}\int_{-\infty}^{\infty} dx \, e^{-x^2/2} e^{2 h x - h^2} &= e^{-h^2} \int_{-\infty}^{\infty} dx \, e^{-(1/2)(x^2 - 4 h x+4 h^2)} e^{2 h^2} \\ &= e^{h^2} \int_{-\infty}^{\infty} dx \, e^{-(1/2)(x-2 h)^2} \\ &= \sqrt{2 \pi} e^{h^2}\end{align}$$

Set the Taylor expansion of this result to the RHS of the first equation:

$$\sqrt{2 \pi} \sum_{n=0}^{\infty} \frac{h^{2 n}}{n!} = \sum_{n=0}^{\infty} J_n \frac{h^n}{n!}$$

Equating coefficients of $h^n$, I get

$$J_n = \begin{cases} 0 & n \,\text{odd}\\\sqrt{2 \pi} \frac{n!}{(n/2)!} & n \,\text{even} \end{cases} $$

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