Proving that $\int_{-\infty}^{\infty}x^{m}e^{-x^2}H_n(x)dx=0$ for $m$ an integer and $0\leq m\leq n-1$ I find myself trying to prove that $\int_{-\infty}^{\infty}x^{m}e^{-x^2}H_n(x)dx=0$ for $m$ an integer and $0\leq m\leq n-1$
I tried to solve it by the following method:
Using the Rodrigues formula for Hermite polynomials:
$H_n(x)=(-1)^{n}e^{x^2}\cfrac{d^{n}}{dx^n}e^{-x^2}=(-1)^{n}e^{x^2}\cfrac{n!}{2\pi i}$
And substituting this equality in the integral that we want to demonstrate, we have:
\begin{align*}
\int_{-\infty}^{\infty}x^{m}e^{-x^2}H_n(x)dx&=\int_{-\infty}^{\infty}x^{m}e^{-x^2}(-1)^{n}e^{x^2}\cfrac{n!}{2\pi i}dx\\ 
&=(-1)^{n}\cfrac{n!}{2\pi i}\int_{-\infty}^{\infty}x^{m}e^{-x^2}e^{x^2}dx\\
&=(-1)^{n}\cfrac{n!}{2\pi i}\int_{-\infty}^{\infty}x^{m}e^{-x^2+x^2}dx\\
&=(-1)^{n}\cfrac{n!}{2\pi i}\int_{-\infty}^{\infty}x^{m}e^{0}dx\\
&=(-1)^{n}\cfrac{n!}{2\pi i}\int_{-\infty}^{\infty}x^{m}(1)dx\\
&=(-1)^{n}\cfrac{n!}{2\pi i}\int_{-\infty}^{\infty}x^{m}dx\\
\end{align*}
As we know that the integrals in symmetric intervals of odd functions are 0, for the $ m $ odd (where $ x ^ {m} $ is an odd function) the integral will always remain as 0 therefore:
$\text{If $m$ is odd } \Longrightarrow \int_{-\infty}^{\infty}x^{m}e^{-x^2}H_{n}(x)dx=0$
But in the case that $ m $ is even I cannot ensure that the integral is 0, can someone tell me how to justify the case where $ m $ is even?
Because perhaps the proof using orthogonality is simpler. Thanks in advanced!
 A: The $n^{\textrm{th}}$ order Hermite polynomial is orthogonal to all polynomials of degree less than $n$.  You just have to write the $x^m$ term as a sum of Hermite polynomials, and use orthogonality.
I will use the probabilist's Hermite polynomials, so I need to rewrite your integral a little bit:
$$
I_{nm}=\frac{1}{(\sqrt{2})^{n+m+1}}\int_{-\infty}^\infty y^m \operatorname{He}_n(y)e^{-\frac{y^2}{2}}dy 
$$
The Hermite inversion formula
$$
y^m = m!\sum_{k=0}^{\left \lfloor \frac{m}{2}\right \rfloor} \frac{1}{2^k k!(m-2k)!} \operatorname{He}_{m-2k}(y), 
$$
can be found on the wiki and proven here, can be used to transform the integral to
$$
I_{nm}=\frac{m!}{(\sqrt{2})^{m-n+1}}\sum_{k=0}^{\left \lfloor \frac{m}{2}\right \rfloor}\frac{1}{2^k k!(m-2k)!}\int_{-\infty}^\infty \operatorname{He}_{m-2k}(y)\operatorname{He}_n(y)e^{-\frac{y^2}{2}}dy.
$$
Using the orthogonality condition
$$
\int_{-\infty}^\infty \operatorname{He}_{\alpha}(y)\operatorname{He}_{\beta}(y)e^{-\frac{y^2}{2}}dy=\sqrt{2\pi}\alpha!\delta_{\alpha\beta}
$$
and the fact that $m<n$ proves that your integral is zero.
We can find the general solution: $I_{nm}$ is only non-zero whenever $m$ and $n$ have the same parity $\operatorname{mod}(n,2)=\operatorname{mod}(m,2)$ (ie they are both even or both odd) and $m\ge n$, and the general formula is
$$
I_{nm} = \frac{m!\sqrt{\pi}}{2^{m-n}\left(\frac{m-n}{2}\right)!}.
$$
