Integral Representations of Hermite Polynomial? One of my former students asked me how to go from one presentation of the Hermite Polynomial to another. And I'm embarassed to say, I've been trying and failing miserably. (I'm guessing this is a homework problem that he is having trouble with.)
http://functions.wolfram.com/Polynomials/HermiteH/07/ShowAll.html
So he has to go from the Rodriguez type formula (written as a contour integral) to an integral on the real axis, which is the 3rd formula in the link provided above.  It seems like the hint he was given was to start from the contour integral.
Starting with the contour integral, I tried using different semi-circles (assuming that $z$ was real), but this quickly turned into something weird.
I also tried to use a circle as the contour, then map it to the real line. That was a failure.
I tried working backwards, from the integral on the real axis. Didn't have luck.
The last resort was to show that
1) Both expressions are polynomials.
2) Show that the corresponding coefficients were equal. (That is, I took both functions and evaluated them and their derivatives at 0.)
Even 2), I couldn't see a nice way of showing that
$\int_C \frac{e^{-z^2}}{z^{n+1}}dz = \int_{-\infty}^{\infty} z^n e^{-z^2} dz$ (Up to some missing multiplicative constants.)
I feel like I'm missing something really easy.  If someone could give me some hints without giving away the answer, that would be most appreciated.

\begin{align*}
H_n(x)
&=(-)^n\mathrm{e}^{x^2}\partial_x^n\mathrm{e}^{-x^2} \\
&=(-1)^n\mathrm{e}^{x^2}\partial_x^n\mathrm{e}^{-x^2}\Big(\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}\mathrm{e}^{-(t\pm \mathrm{i}x)^2}\mathrm{d}t\Big) \\
&=(-1)^n\mathrm{e}^{x^2}\frac{1}{\sqrt{\pi}}\partial_x^n\int_{-\infty}^{\infty}\mathrm{e}^{-t^2\mp 2\mathrm{i}xt}\mathrm{d}t \\
&=(-1)^n\mathrm{e}^{x^2}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}\partial_x^n\mathrm{e}^{-t^2\mp 2\mathrm{i}xt}\mathrm{d}t \\
&=(-1)^n\mathrm{e}^{x^2}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}(\mp2\mathrm{i}t)^n\mathrm{e}^{-t^2\mp 2\mathrm{i}xt}\mathrm{d}t \\
&=\frac{(\pm2\mathrm{i})^n}{\sqrt{\pi}}\mathrm{e}^{x^2}\int_{-\infty}^{\infty}t^n\mathrm{e}^{-t^2\mp 2\mathrm{i}xt}\mathrm{d}t \\
&=\frac{(\pm2\mathrm{i})^n}{\sqrt{\pi}}\int_{-\infty}^{\infty}t^n\mathrm{e}^{-(t\pm \mathrm{i}x)^2}\mathrm{d}t
\end{align*}
Average the $+$ and the $-$ situations, we have
$$H_n(x)=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}(2t)^n\mathrm{e}^{x^2-t^2}\cos(2xt-\frac{n\pi}{2})\,\mathrm{d}t .$$
