I'm looking for a closed form expression for the indefinite integral involving the product of a Gaussian with a Hermite polynomial. The integral is given by:
$I_a(x):=\int e^{-ax^2} H_k(x) dx $
where $a>0$ and $k$ is the index of the (physicist's) Hermite polynomial. In particular, I only need this for $a=1,2$ and $3$. Mathematica was able to do this for $a=1$ (it gives $I_1(x)=2^k\sqrt{\pi}\left[\frac{_1F_1\left(\frac{k+1}{2};\frac{3}{2};-y^2\right)y}{\Gamma((1-k)/2)}+\frac{_1F_1\left(\frac{k}{2};\frac{1}{2};-y^2\right)-1}{k\Gamma(k/2)}\right]$), but gave up for larger $a$.
For $a=1$, using the definition of the Hermite polynomial ($H_k(x):=(-1)^ke^{x^2}\frac{d^k}{dx^k}e^{-x^2}$), I was able to obtain a simple form:
$I_1(x)=-e^{-x^2}H_{k-1}(x)$
but I'm not sure how to extend this for $a=2$ or $3$. Integration by parts did not seem to help. Can a closed form be found for $I_a(x)$ (preferably in terms of elementary/special functions)?
PS: I'm actually interested in the definite integral from $x=a$ to $b$ ($0<a<b$), and so I was looking for a closed form for the indefinite integral.
I am assuming you are using the physicist version of Hermites since you used $H$ and not $He$.
We just have to massage into a form Mathematica can do.
You can compute this first by a substitution $y = \sqrt{a} x$, $\exp(-ax^2) \rightarrow \exp(-y^2)$ and $H_k(x) \rightarrow H_k(y/\sqrt{a})$, so that $$ \int_u^v dx H_k(x) \exp(-ax^2) = \frac{1}{\sqrt{a}}\int_{\sqrt{a}u}^{\sqrt{a}v} dy H_k(y/\sqrt{a}) \exp(-y^2). $$ Then the scaling identity can be utilized: $$ H_n(\lambda u)=n ! \sum_{k=0}^{[n / 2]} \frac{\left(2\lambda^2-1\right)^k (\sqrt{2}\lambda)^{n-2 k}}{k !(n-2 k) !} H_{n-2 k}(u/\sqrt{2})=n ! \sum_{k=0}^{[n / 2]} \frac{\left(2\lambda^2-1\right)^k (\sqrt{2}\lambda)^{n-2 k}}{k !(n-2 k) !} \times\sum_{m=0}^{n-2 k-1} H_{n-2 k-m}(0)H_m(u) $$ where in the last step one uses the addition theorem see here, (54). This identity with $\lambda = 1/\sqrt{a}$ yields
$$ \frac{1}{\sqrt{a}}\int_{\sqrt{a}u}^{\sqrt{a}v} dy H_k(y/\sqrt{a}) \exp(-y^2) = n ! \sum_{k=0}^{[n / 2]} \frac{\left(2/a-1\right)^k (\sqrt{2/a})^{n-2 k}}{k !(n-2 k) !} \sum_{m=0}^{n-2 k-1} H_{n-2 k-m}(0)\frac{1}{\sqrt{a}}\int_{\sqrt{a}u}^{\sqrt{a}v} dy H_m(y)\exp(-y^2) $$
This last integral yields to Mathematica: $$ \int dy H_m(y) \exp(-y^2) = -\frac{\Gamma \left(\frac{1}{2}-\frac{m}{2}\right) \left( _1F_1\left(\frac{m}{2};\frac{1}{2};-y^2\right)-1\right)+m y \Gamma \left(-\frac{m}{2}\right) _1F_1\left(\frac{m+1}{2};\frac{3}{2};-y^2\right)}{2 \Gamma (1-m)} $$
With a bit more algebra, one can show these $_1F_1$ functions themselves are Hermite polynomials, due to the second argument being a $\frac{1}{2}$ or a $\frac{3}{2}$, see here. You can plug in your desired integral bounds, and obtain your answer as a double sum over Hermite polynomials.
This should be correct, up to algebraic errors and signs, which you should verify.
Where did you come across this integral?