I am trying to evaluate
$$\int_{-\infty}^{\infty} e^{-x^2 / 2} H_n(x) \space dx,$$
where $H_n(x)$ are the Hemite polynomials. I attempted to solve the integral by the using the generating function
$$\sum_{n=0}^{\infty}\frac{z^n}{n!}H_n(x) = e^{-z^2 + 2zx}.$$
Multiplying by $e^{-x^2/2}$ and pulling the summation out from the integral, I found
$$\sum_{n=0}^{\infty}\frac{z^n}{n!}\int_{-\infty}^{\infty}e^{-x^2/2}H_n(x) \space dx = \int_{-\infty}^{\infty}e^{-x^2/2 - z^2 + 2zx}\space dx.$$
Pulling out an $e^{z^2}$, the right side becomes a simple Gaussian integral which together reduces to
$$e^{z^2}\sqrt{2\pi}.$$
But this is where I'm stuck - expressing the RHS in terms of its series expansion, we get $e^{z^2} = \sum_{n=0}^{\infty} \frac{z^{2n}}{n!}$, and now we have a sum with $z^n$ on the LHS and a sum with $z^{2n}$ on the RHS. Is there a way to shift indices and equate the coefficients on both sides, or another way to go about this problem? Otherwise, I'm left with the integral evaluating to $z^n \sqrt{2\pi}$, which doesn't make much sense to me.
Note
$$\sum_{n = 0}^\infty \frac{z^{2n}}{n!} = \sum_{n = 0}^\infty \frac{(2n)!}{n!}\frac{z^{2n}}{(2n)!} = \sum_{m = 0}^\infty c_m \frac{z^m}{m!}$$ where $$c_m = \begin{cases}\dfrac{m!}{(m/2)!},&\text{if $m$ is even}\\0,&\text{if $m$ is odd}\end{cases}$$
Thus $\int_{-\infty}^\infty e^{-x^2/2}H_n(x)\, dx = c_n\sqrt{2\pi}$.