How to compute the inner product of a Hermite polynomial with itself I am trying to prove the well-documented fact that if $H_{n}(x)$ is the $n$th Physicists Hermite polynomial then:
$$\left\langle H_{n} \middle| H_{m} \right\rangle=\int_{-\infty}^{\infty}H_{n}^{*}(x)H_{m}(x)e^{-x^{2}}\:\mathrm{d}x=\delta_{nm} 2^{n} n! \sqrt{\pi}$$
I know the formula for a general formula for $H_{n}$:
$$H_{n}(x)\triangleq (-1)^{n}e^{x^{2}}\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}e^{-x^{2}}$$
But I'm not sure how I can use this to prove the above identity. I can prove the Kronecker delta, but not the actual value of $\langle H_{n} \mid H_{n}\rangle$. I assumed that I could use integration by parts, but I have a squared derivative term so I wasn't sure what to use.

Have you tried the book of Gabor Szego on Orthogonal Polynomials, there is a Hermite Polynomial section 5.5, page 105ff.
See, e.g. http://books.google.com
