Weird Integral Involving Hermite Polynomials

Question

I have stumbled upon the following integral involving the Hermite polynomials:

$$ I(m) = \int_\mathbb{R} e^{i m x} \left[ e^{-\frac{x^2}{2}} H_m(x) \right] dx \, , \quad m \in \mathbb{N} \cup \{0\} \, , $$

which is rather weird. It came up from trying to obtain an expansion in terms of the harmonic oscillator eigenfunctions from a Fourier series.

I have searched in function manuals and also tried to solve it term by term to find a predictable series, but the coefficients don't make sense to me. Does it look recognizable? Does anyone know how to solve it?

Here are closed expressions for the first six $m$ values:

\begin{align} I(0) &= \sqrt{2 \pi} \\ I(1) &= \sqrt{2 \pi} \left( \frac{2i}{\sqrt{e}} \right) \\ I(2) &= \sqrt{2 \pi} \left( -\frac{14}{e^2} \right) \\ I(3) &= \sqrt{2 \pi} \left( -\frac{180 i}{e^\frac{9}{2}} \right) \\ I(4) &= \sqrt{2 \pi} \left( \frac{3340}{e^8} \right) \\ I(5) &= \sqrt{2 \pi} \left( \frac{80600 i}{e^\frac{25}{2}} \right) \end{align}

There's obviously a rule of type $I(m) \propto \sqrt{2 \pi} e^{-m^2/2} i^m$, but the numerical coefficients follow some non-intuitive rule.

Answer 1

The answer follows from Gradshteyn and Ryzhik, 7.376, $$ I_n(y):=\int_{-\infty}^\infty e^{i\,x\,y}e^{-x^2/2} \, H_n(x) \, dx = \sqrt{2 \pi} e^{-y^2/2} H_n(y) i^n .$$ Just set $n=m$ and $y=m.$ I'd attempt a proof, but I need to what is allowed as a starting point.

EDIT

Formula above is easily derived from generating function $$ \sum_{n=0}^\infty \frac{t^n}{n!} H_n(x) = \exp{(-t^2+2x\,t)} $$ Make a generating formula for $I_n(y),$ interchange integral and sum, and use the previous formula like so: $$\sum_{n=0}^\infty \frac{t^n}{n!} I_n(y) = \int_{-\infty}^\infty e^{i\,x\,y} e^{-x^2/2} \exp{(-t^2+2x\,t)}dx = e^{-t^2} \exp{(\frac{1}{2}(2t+iy)^2 )}\sqrt{2\pi}$$ where the well-known Gaussian integral has been used in the last step. Arrange the exponentials as follows: $$e^{-t^2} \exp{(\frac{1}{2}(2t+iy)^2 )}= e^{-y^2/2} \exp{( -(it)^2 +2(it)y )} $$ Use the generating function again, but now the argument is $it.$ Equate coefficients of $t$ to complete the proof.

Answer 2

By completing the square of the exponential terms it can be seen that \begin{align} \int_{-\infty}^{\infty} e^{i m x} \, e^{-x^2/2} \, H_{m}(x) \, dx &= e^{-m^2/2} \, \int_{-\infty}^{\infty} e^{-(x-i m)^{2}/2} \, H_{m}(x) \, dx \\ &= e^{-m^2/2} \, \int_{-\infty}^{\infty} e^{-u^{2}/2} \, H_{m}(u + i m) \, du \end{align}

Now make use of $$H_{n}(x+y) = \sum_{k=0}^{n} \binom{n}{k} \, H_{k}(x) \, (2 y)^{n - k}$$ in the form $$H_{m}(u + i m) = \sum_{k=0}^{m} \binom{m}{k} \, H_{k}(i m) \, (2 u)^{m-k}.$$ This leads to: \begin{align} \int_{-\infty}^{\infty} e^{i m x} \, e^{-x^2/2} \, H_{m}(x) \, dx &= \sqrt{2} \, e^{-m^2/2} \, \sum_{k=0}^{m} \binom{m}{k} H_{k}(i m) \, 2^{3(m-k)/2} \, \Gamma\left(\frac{m-k+1}{2}\right). \end{align} This series can be reduced, but is sufficient for some calculations.

Answer 3

Let $\psi_m = e^{-x^2/2} H_m(x)$. Applying the Fourier transform to $\psi_0$ and to the relation $\psi_{m + 1} = x \psi_m - \psi_m'$ gives $$\hat \psi_0 = \sqrt {2 \pi} \,\psi_0, \\ \hat \psi_{m + 1} = i x \hat \psi_m - i \hat \psi {}_m', \\ \hat \psi_{m} = C_m \psi_m \supset \hat \psi_{m + 1} = i C_m \psi_{m + 1}, \\ \hat \psi_m = \sqrt {2 \pi} \,i^m \psi_m, \\ I(m) = \hat \psi_m(m).$$