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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.)

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.

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This is an older problem so I will give a fairly complete solution.

We use the physicist's convention for the Hermite polynomials. There are many ways to show the two representations are the same. Let's expand the integrals in small $x$ and show that they have the same series. We start from the contour integral, $$\begin{eqnarray*} H_n(x) &=& \frac{n!}{2\pi i} \oint_\gamma d t\, \frac{e^{2tx -t^2}}{t^{n+1}} \\ &=& \frac{n!}{2\pi i} \sum_{k=0}^\infty \frac{(2x)^k}{k!} \sum_{l=0}^\infty \frac{(-)^l}{l!} \oint_\gamma d t\, t^{2l+k-n-1}. \end{eqnarray*}$$ But the contour integral is zero unless $2l+k-n = 0$, otherwise it is $2\pi i$. Notice that $n-k$ must be even. Thus, $$H_n(x) = n! \sum_{k=0 \atop n-k\ \mathrm{even}}^\infty (-)^{\frac{n-k}{2}} \frac{(2x)^k}{k!(\frac{n-k}{2})!}.$$ The sum will terminate at $k=n$ since the gamma function diverges at $0,-1,-2,\ldots$. Letting $l = (n-k)/2$ we find $$\begin{equation} H_n(x) = n! \sum_{l=0}^{\lfloor \frac{n}{2}\rfloor} (-)^l \frac{(2x)^{n-2l}}{l!(n-2l)!},\tag{1} \end{equation}$$ the standard series representation of the Hermite polynomials.

Now for the other integral representation. Expand the binomial series,
$$\begin{eqnarray*} H_n(x) &=& \frac{2^n}{\sqrt{\pi}} \int_{-\infty}^\infty dt\, e^{-t^2}(x+it)^n \\ &=& \frac{2^n}{\sqrt{\pi}} \sum_{k=0}^n {n\choose k} x^k i^{n-k} \int_{-\infty}^\infty dt\, e^{-t^2} t^{n-k}. \end{eqnarray*}$$ The integral is zero unless $n-k$ is even. Therefore, $$\begin{eqnarray*} \int_{-\infty}^\infty dt\, e^{-t^2} t^{n-k} &=& 2\int_{0}^\infty dt\, e^{-t^2} t^{n-k} \\ &=& \Gamma\left(\frac{n-k}{2}+\frac{1}{2}\right) \\ &=& \sqrt{\pi} \frac{(n-k)!}{2^{n-k}(\frac{n-k}{2})!}. \end{eqnarray*}$$ (Change variables, let $z = t^2$, to see the connection to the gamma function.) Substitute this into our series for $H_n(x)$. After the dust settles, we let $l = (n-k)/2$ and arrive back at equation (1) above.

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The contour integral, of course, is the result of applying Cauchy's differentiation formula to the generating function for the Hermite polynomials. – J. M. Apr 15 '12 at 15:00
@J.M.: Do you mean Cauchy's integral formula? Yes, we just fold each side of the formula for the generating function with $\frac{n!}{2\pi i}\oint_\gamma dt\, t^{-n-1}$. I chose to go directly to the series and not use the properties of the generating function. The presentation is self-contained. I'm glad someone is reading! – user26872 Apr 15 '12 at 17:43


$=(-)^n\mathrm{e}^{x^2}\partial_x^n\mathrm{e}^{-x^2}\Big(\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}\mathrm{e}^{-(t\pm \mathrm{i}x)^2}\mathrm{d}t\Big)$

$=(-)^n\mathrm{e}^{x^2}\frac{1}{\sqrt{\pi}}\partial_x^n\int_{-\infty}^{\infty}\mathrm{e}^{-t^2\mp 2\mathrm{i}xt}\mathrm{d}t$

$=(-)^n\mathrm{e}^{x^2}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}\partial_x^n\mathrm{e}^{-t^2\mp 2\mathrm{i}xt}\mathrm{d}t$

$=(-)^n\mathrm{e}^{x^2}\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}(\mp2\mathrm{i}t)^n\mathrm{e}^{-t^2\mp 2\mathrm{i}xt}\mathrm{d}t$

$=\frac{(\pm2\mathrm{i})^n}{\sqrt{\pi}}\mathrm{e}^{x^2}\int_{-\infty}^{\infty}t^n\mathrm{e}^{-t^2\mp 2\mathrm{i}xt}\mathrm{d}t$

$=\frac{(\pm2\mathrm{i})^n}{\sqrt{\pi}}\int_{-\infty}^{\infty}t^n\mathrm{e}^{-(t\pm \mathrm{i}x)^2}\mathrm{d}t$


Average the $+$ and the $-$ situations, we have



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Start from $H_n(x)=(-1)^ne^{x^2}\dfrac{d^n}{dx^n}e^{-x^2}$ :

$H_n(x)=(-1)^ne^{x^2}\dfrac{d^n}{dx^n}e^{-x^2}=\dfrac{(-1)^ne^{x^2}}{\sqrt\pi}\dfrac{d^n}{dx^n}\int_0^\infty e^{-\frac{t^2}{4}}\cos xt~dt=\dfrac{(-1)^ne^{x^2}}{\sqrt\pi}\int_0^\infty t^ne^{-\frac{t^2}{4}}\cos\left(xt+\dfrac{n\pi}{2}\right)dt$

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