Hermite Polynomials: Rodrigues to Integral Representation I would like to go from this representation of the Hermite polynomials:
$$H_n(z)=(-1)^ne^{z^2}\frac{d^n}{dz^n}e^{-z^2} \tag{1}$$
to this representation
$$H_n(z)=\frac{2^n}{\sqrt{\pi}}\int_{-\infty}^{\infty}(z+is)^n\,e^{-s^2}\,ds \tag{2}$$
I have tried to use Cauchy's theorem to give the $n^{\textrm{th}}$ derivative term in terms of a contour integral, and then contort the integrand and contour so that we're left with an integral over the real line, but I just can't find a way to do it. I've also read this question, but all of answers go through a different representation/route than the one I desire. Could you help me out?
EDIT: To shed some light on what exactly I desire, I will present the first few steps of my attempted path. For an analytic function $f(z)$, we can express its $n^{\textrm{th}}$ derivative by a contour integral.
$$\frac{d^n}{dz^n}f(z)=\frac{n!}{2\pi i}\oint_{\Gamma} \frac{f(w)}{(w-z)^{n+1}} dw $$
Where $\Gamma$ is a simple closed curve that encircles the point $z$ in the complex plane. Using this we can rewrite $(1)$ as
$$H_n(z)=(-1)^ne^{z^2}\frac{n!}{2\pi i}\oint_{\Gamma} \frac{e^{-w^2}}{(w-z)^{n+1}} dw$$
I don't know where to go from here. If I make the substitution $s=w-z$, I end up going in a circle showing that $H_n(z)=H_n(z)$. Moreover, I can't find a simple way to reduce that integral to an integral over $\mathbb{R}$ (due to the gaussian term in the integral).
 A: Note that we can write $e^{-z^2}$ as the integral
$$e^{-z^2}=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-s(s+2iz)}\,ds \tag 1$$
Taking the $n$'th derivative of $(1)$ with respect to $z$ yields
$$\begin{align}
\frac{d^n}{dz^n}\left(e^{-z^2}\right)&=\frac{1}{\sqrt{\pi}}\int_{-\infty}^\infty (i2s)^n\,e^{-s(s+2iz)}\,ds\\\\
&=\frac{2^n}{\sqrt{\pi}}e^{-z^2}\int_{-\infty}^\infty (-is)^n\,e^{-(s+iz)^2}\,ds\\\\
&=\frac{2^n}{\sqrt{\pi}}e^{-z^2}\int_{-\infty}^\infty (-i(s-iz))^n\,e^{-s^2}\,ds\\\\
&=\frac{(-1)^n\,2^n}{\sqrt{\pi}}e^{-z^2}\int_{-\infty}^\infty (z+is)^n\,e^{-s^2}\,ds\tag 2
\end{align}$$
Multiplying both sides of $(2)$ by $(-1)^n\,e^{z^2}$ reveals 
$$\begin{align}
H_n(z)&=(-1)^n\,e^{z^2}\,\frac{d^n}{dz^n}\left(e^{-z^2}\right)\\\\
&=\frac{\,2^n}{\sqrt{\pi}}\int_{-\infty}^\infty (z+is)^n\,e^{-s^2}\,ds
\end{align}$$
as was to be shown!

ORIGINAL POST
Although I am not sure exactly what you are seeking, I wanted to present a way to show that the integral expression in the OP is indeed the $n$'th order Hermite Polynomial.  
Proceeding, we use the binomial theorem to write the integral of interest as
$$\begin{align}\frac{2^n}{\sqrt{\pi}}\int_{-\infty}^\infty (z+is)^n e^{-s^2}\,ds&=\frac{2^n}{\sqrt{\pi}}\sum_{k=0}^n\binom{n}{k}i^kz^{n-k}\int_{-\infty}^\infty s^ke^{-s^2}\,ds \\\\
&=\frac{2^n}{\sqrt{\pi}}\sum_{k=0}^{n}\binom{n}{k}i^kz^{n-k}\left(\frac{1-(-1)^k}{2}\right)\Gamma\left(\frac{k+1}{2}\right) \\\\
&=\frac{2^n}{\sqrt{\pi}}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}(-1)^kz^{n-2k}\Gamma\left(k+\frac12\right) \\\\
&=\frac{2^n}{\sqrt{\pi}}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}(-1)^kz^{n-2k}\left(\sqrt{\pi}\,\frac{(2k)!}{4^k\,k!}\right) \\\\
&=2^nn!z^n\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\left(\frac{1}{2z}\right)^{2k}\frac{1}{k!(n-2k)!} \tag 3
\end{align}$$
where $(3)$ is the explicit expression for the $n$'th order Hermite Polynomial found here. 
