Prove that $\int_{\mathbb{R}}x^me^{2ax}e^{-x^2/2}=e^{2a^2}\int_{\mathbb{R}}(x+2a)^me^{-x^2/2}$

Question

As the title indicates, I've been trying for quite some time now to prove that $$\int_{\mathbb{R}}x^me^{2ax}e^{-x^2/2}=e^{2a^2}\int_{\mathbb{R}}(x+2a)^me^{-x^2/2}$$ $\forall m \in \mathbb{N}, \forall a \in \mathbb{R}$. After having derived (using Hermite polynomials) that $$\int_{\mathbb{R}}x^{2n}e^{-x^2/2}=\frac{(2n)!\sqrt{2\pi}}{2^nn!}$$ and expanding $e^{2ax}$ into a power series I've found that $$\int_{\mathbb{R}}x^me^{2ax}e^{-x^2/2}=\int_{\mathbb{R}} \sum_{i=0}^{\infty} \frac{(2a)^ix^{i+m}}{i!}e^{-x^2/2}=\sqrt{2\pi}\sum_{k=\lceil\frac{m}{2}\rceil}^{\infty} \frac{(2a^2)^k}{k!}=\sqrt{2\pi}\left(e^{2a^2}-\sum_{k=0}^{\lfloor\frac{m}{2}\rfloor} \frac{(2a^2)^k}{k!}\right)$$ Despite the somewhat promising $e^{2a^2}$ term though, I'm becoming increasingly convinced that this is not a good way to be going about this. Any and all insights are welcome.

Answer 1

This is easier, just do the change of variables $u=x-2a$. Then $du=dx$ and $$ e^{2ax}e^{-x^2/2}=e^{2a^2}e^{-u^2/2} $$ and we get $$ \int_{\mathbb{R}}x^me^{2ax}e^{-x^2/2}=e^{2a^2}\int_{\mathbb{R}}(u+2a)^me^{-u^2/2}\,du. $$ Finally, just rename the variable $u$ to $x$.