Hermite polynomial relations How one can prove the relation on Hermite polynomial given as $$\int_{-\infty}^\infty H_n\left(x+\frac{x_0}{2}\right)e^{^{-\frac{x^2}{2}}}dx=\sqrt{\pi}x_0^n$$I also didn't understand the meaning of $\displaystyle H_n\left(x+\frac{x_0}{2}\right)$, what does that signify, I know by the way that $H_n(x)$ represents the Hermite polynomial but what is the meaning of $\displaystyle H_n\left(x+\frac{x_0}{2}\right)$? Please explain this alongwith.
Thanks.
 A: This convolution integral is proven in Theorem 2.7 in a recent preprint "A General Expression for Hermite Expansions with Applications" found here: http://dx.doi.org/10.13140/RG.2.2.30843.44325
Define the integral you are looking for
$$
I_n = \int_{-\infty}^\infty H_n\left(x+\frac{x_0}{2}\right)\exp\left(-\frac{x^2}{2}\right)dx
$$
In order to prove this, we Taylor expand the Hermite polynomial in $x_0$
$$
I_n= \sum_{m=0}^n {n \choose m} \left(\frac{x_0}{2}\right)^m\int_{-\infty}^\infty H_{n-m}(x)\exp\left(-\frac{x^2}{2}\right)dx.
$$
This integral is only zero for $n=m$ by the orthogonality of Hermite polynomials
$$
\int_{-\infty}^\infty H_\alpha(x)H_\beta(x)\exp\left(-\frac{x^2}{2}\right)dx=c_n \delta_{\alpha\beta}
$$
where $c_n$ depends on your weighting function and normalization.  Here it is $c_n = \sqrt{2\pi}n!$.
Plugging this into the sum results in
$$
I_n=\left(\frac{x_0}{2}\right)^n\int_{-\infty}^\infty \exp\left(-\frac{x^2}{2}\right)dx
$$
leading to (almost) your result
$$
I_n = \sqrt{2\pi}\left(\frac{x_0}{2}\right)^n.
$$
