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Is there any formula which represent the sinc function $\operatorname{\rm sinc}(x)=\dfrac{\sin(\pi x)}{\pi x}$ (its expansion) in terms of the Chebychev-Hermite function?

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2 Answers 2

Denote the Chebyshev-Hermite functions as $$ \varphi_n(x) = \frac{(-1)^n }{\sqrt{2^n n! \sqrt{\pi}}} \mathrm{e}^{\frac{x^2}{2}} \frac{\mathrm{d}^n}{\mathrm{d} x^n} \mathrm{e}^{-x^2} = \frac{1}{\sqrt{2^n n! \sqrt{\pi}}} H_n(x) \mathrm{e}^{-\frac{x^2}{2}} $$ where $H_n(x)$ denote the $n$-th Hermite orthogonal polynomial.

Because $\varphi_n(x)$ form complete orthonormal system, i.e. $\int_{-\infty}^\infty \varphi_n(x) \varphi_m(x) \mathrm{d} x = \delta_{n,m}$ we have $$ \operatorname{\mathrm{sinc}}(x) = \sum_{n=0}^\infty c_n \varphi_n(x) $$ where $c_n = \int_{-\infty}^\infty \varphi_n(x) \operatorname{\mathrm{sinc}}(x) \mathrm{d} x$. Now, due to parity $c_{2m+1} = 0$.

Notice that $c_n = \int_{-\infty}^\infty \hat{\varphi}_n(\omega) \hat{\operatorname{\rm sinc}}(\omega) \mathrm{d} \omega $. Since $$ \hat{\varphi}_{2n}(\omega) = (-1)^n \varphi_{2n}(\omega) \qquad \hat{\operatorname{\rm sinc}}(\omega) = \frac{1}{\sqrt{2 \pi}} [ -\pi < \omega < \pi ] $$ we have $$ c_{2n} = \frac{(-1)^n}{\sqrt{2\pi}} \int_{-\pi}^\pi \varphi_{2n}(\omega) \mathrm{d} \omega = (-1)^n \sqrt{\frac{2}{\pi}} \int_{0}^\pi \varphi_{2n}(\omega) \mathrm{d} \omega $$ Now, using the explicit expression for the Hermite polynomials: $$ \varphi_{2n}(\omega) = \frac{\sqrt{(2n)!}}{2^n \pi^{1/4}} \mathrm{e}^{-\frac{\omega^2}{2}} \sum_{k=0}^n (-1)^{n-k} \frac{(2\omega)^k}{(2k)! (n-k)!} $$ we have $$ c_{2n} = \sqrt{\frac{2}{\pi}} \frac{\sqrt{(2n)!}}{2^n \pi^{1/4}} \sum_{k=0}^n \frac{(-1)^k}{(2k)!(n-k)!} 2^{3 k - 1/2} \gamma\left(k+\frac{1}{2}, \frac{\pi^2}{2} \right) $$ where $ \gamma\left(k+\frac{1}{2}, \frac{\pi^2}{2} \right)$ stands for the incomplete gamma function.

Here is a verification in Mathematica:

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Copy-paste ready input:

In[77]:= Table[
  Sqrt[2/Pi] Sqrt[(2 n)!]/(2^n Pi^(1/4))
     Sum[(-1)^k/((2 k)! (n - k)!) 2^(3 k - 1/2)
       Gamma[k + 1/2, 0, Pi^2/2], {k, 0, n}] - 
   Integrate[(HermiteH[2 n, x] E^(-(x^2/2)))/Sqrt[
     2^(2 n) (2 n)! Sqrt[\[Pi]]] Sinc[Pi x], {x, -Infinity, 
     Infinity}], {n, 0, 4}] // FullSimplify

Out[77]= {0, 0, 0, 0, 0}
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What you have called $\Gamma_{k+1/2}\left(0,\frac{\pi^2}{2} \right)$ I think is more commonly denoted as $\gamma\left(k+\frac12,\frac{\pi^2}{2}\right)$. As you know, Mathematica only supports the lower incomplete gamma function implicitly... –  J. M. Dec 15 '11 at 17:50
@J.M. Yes, you have a point. I will edit the post to change that. –  Sasha Dec 15 '11 at 18:01
Thank you very much –  David Dec 16 '11 at 14:07

$$\frac{\sin (\pi x)}{\pi x}=\sum _{n=0}^{\infty } \frac{\left((-1)^n e^{-\frac{1}{2} i \pi n} \left(1+e^{i \pi n}\right) \Gamma \left(\frac{n}{2}+\frac{1}{2},0,\frac{\pi ^2}{4}\right)\right) H_n(x)}{2 \pi n!}$$

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