integral of hermite polynomials Objective:
Show that
$$
\int^{\infty}_{-\infty} x e^{-x^2} H_n(x) H_m(x) dx = \pi^{1/2} 2^{n-1} n! \delta_{m,n-1} + \pi^{1/2} 2^n (n+1)! \delta_{m,n+1}
$$
My attempt at this is:

\begin{eqnarray*}
\sum^\infty_{n=0} \frac{2^n s^n u^n}{n!} (s+u) \sqrt{\pi} &=& \sum^\infty_{m,n=0}\frac{s^m u^n}{m! n!} \int x H_m(x) H_n(x) e^{-x^2} dx \\
\sqrt{\pi}\bigg( \sum^\infty_{n=0} \frac{2^n s^{n+1} u^n}{n!} +\sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!} \bigg) &=& \sum^\infty_{m,n=0}\frac{s^m u^n}{m! n!} \int x H_m(x) H_n(x) e^{-x^2} dx 
\end{eqnarray*}
I'm reasonably confident up to here.
\begin{eqnarray*}
\sqrt{\pi}\bigg( \sum^\infty_{n=0} \frac{2^n s^{n+1} u^n}{n!} +\sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!} \bigg) \delta_{m,n+1}&=& \delta_{m,n+1}\sum^\infty_{m,n=0}\frac{s^m u^n}{m! n!} \int x H_m(x) H_n(x) e^{-x^2} dx 
\end{eqnarray*}
\begin{eqnarray*}
\sqrt{\pi}\bigg( \sum^\infty_{n=0} \frac{2^n s^{n+1} u^n}{n!}\big(\delta_{m,n+1}\frac{m! n!}{s^m u^n}\big) +\sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!}\big(\delta_{m,n+1}\frac{m! n!}{s^m u^n}\big)  \bigg) \delta_{m,n+1}&=&  \int x H_m(x) H_n(x) e^{-x^2} dx
\end{eqnarray*}
Then I thought maybe I could try re-indexing the n and m in the second $\big(\delta_{m,n+1}\frac{m! n!}{s^m u^n}\big)$ term.
\begin{eqnarray*}
\sqrt{\pi}\bigg( \sum^\infty_{n=0} \frac{2^n s^{n+1} u^n}{n!}\big(\frac{(n+1)! n!}{s^{n+1} u^n}\big) +\sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!}\big(\delta_{m,n+1}\frac{n! m!}{s^n u^m}\big)  \bigg) \delta_{m,n+1}&=&  \int x H_m(x) H_n(x) e^{-x^2} dx \\
\bigg(\sqrt{\pi} 2^n (n+1)!  + \sqrt{\pi} \sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!}\big(\delta_{m,n+1}\frac{n! m!}{s^n u^m}\big) \bigg) \delta_{m,n+1}&=&  \int x H_m(x) H_n(x) e^{-x^2} dx \\
\sqrt{\pi} 2^n (n+1)! \delta_{m,n+1} + \sqrt{\pi} \sum^\infty_{n=0} \frac{2^n s^n u^{n+1}}{n!}\big(\frac{n! (n+1)!}{s^n u^{n+1}}\big)  \delta_{m,n+1}&=&  \int x H_m(x) H_n(x) e^{-x^2} dx \\
\sqrt{\pi} 2^n (n+1)! \delta_{m,n+1} + \sqrt{\pi} 2^n  (n+1)! \delta_{m,n+1}&=&  \int x H_m(x) H_n(x) e^{-x^2} dx \\
\sqrt{\pi} 2^{n+1} (n+1)! \delta_{m,n+1} &=&  \int x H_m(x) H_n(x) e^{-x^2} dx
\end{eqnarray*}

Obviously the wrong answer. I got the following hint from a friend: "You should take d^n/du^n d^m/ds^m (n-th and m-th derivatives) of the second line, answer has two different Kronecker deltas."
Can anyone specifically help me figure out how to get this $\delta_{m,n-1}$ factor? I can't figure out that one.
 A: This is related to the question Problem proving $\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx$, so the answer is very similar.
I prefer to work with the probabilists Hermite polynomials, which can be related to the physicists (which appear in this question) via
$$
He_{\alpha}(x) = 2^{-\frac{\alpha}{2}}H_\alpha\left(\frac{x}{\sqrt{2}}\right)
$$
and the desired integral becomes
$$
I_{nm} = \int_{-\infty}^\infty xe^{-x^2}H_n(x)H_m(x)dx \\
= 2^{\frac{n+m-1}{2}}\sqrt{\pi}\int_{-\infty}^\infty He_1(y)He_n(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}},
$$
where the identity $x=He_1(x)$ has been used.
The integral can be turned into a sum of products of Hermite polynomials by the linearization theorem
$$
He_\alpha(x)He_\beta(x)=\sum_{k=0}^{\min(\alpha,\beta)}{\alpha \choose k}{\beta \choose k}k!He_{\alpha+\beta-2k}(x)
$$
with $\alpha=1$ and $\beta=n$:
$$
He_1(x)He_n(x) = He_{n+1}+nHe_{n-1}(x).
$$
The integral is now in a form where orthogonality can be used
$$
I_{nm}= 2^{\frac{n+m-1}{2}}\sqrt{\pi}\left\{\int_{-\infty}^\infty He_{n+1}(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}} + n\int_{-\infty}^\infty He_{n-1}(x)He_m(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}\right\}.
$$
The probabilists orthogonality condition is
$$
\int_{-\infty}^\infty He_{\alpha}(x)He_\beta(x)e^{-\frac{x^2}{2}}\frac{dx}{\sqrt{2\pi}}=\alpha!\delta_{\alpha,\beta}
$$
leading to
$$
I_{nm}= 2^{\frac{n+m-1}{2}}\sqrt{\pi}\left\{(n+1)!\delta_{n+1,m} + n(n-1)!\delta_{n-1,m} \right\}.
$$
which leads directly to your desired result.
