# Has the Weierstass transform been used to give Hermite series representations of the Riemann zeta function?

The inverse of the Weierstrass transform expands a function as a series of Hermite polynomials $H_{n}$. There are several ways to invert the Weierstrass transform which led me to the following expansions:

1. Along the critical line

$\zeta\left(\tfrac{1}{2}+ix\right)=\overset{\infty}{\underset{m=0}{{\textstyle \sum}}}\left[\overset{\infty}{\underset{n=1}{{\textstyle \sum}}}\dfrac{1}{n^{1/2+\ln n/4}}\left(\frac{-i\ln n}{2}\right)^{m}\right]\frac{H_{m}\left(x\right)}{m!}\text{.}$

2. Hasse's representation yields

$\zeta\left(z\right)\left(1-2^{1-z}\right)=\overset{\infty}{\underset{m=0}{{\textstyle \sum}}}\left[\overset{\infty}{\underset{n=0}{{\textstyle \sum}}}\overset{n}{\underset{k=0}{{\textstyle \sum}}}\frac{\left(-1\right)^{k}\left(\ln\left(k+1\right)\right)^{m}}{2^{n+1}}\binom{n}{k}e^{\left[\ln\left(k+1\right)\right]^{2}/4}\right]\frac{\left(-1\right)^{m}H_{m}\left(z\right)}{2^{m}m!}\text{.}$

3. The Laurent series yields

$\zeta\left(z+1\right)-\frac{1}{z}=\overset{\infty}{\underset{n=0}{{\textstyle \sum}}}\tfrac{1}{n!2^{n}}\left[\underset{k=0}{\overset{\infty}{{\textstyle \sum}}}\frac{\gamma_{n+2k}}{2^{2k+1}k!}\right]H_{n}\left(z\right)$

where the $\gamma_{n}$ are the Stieltjes constants.

Et cetera. The calculations were all formal, and I've mostly ignored convergence, assuming this territory is well-trod. Has this been explored, or is there some flaw in this approach?

Also, is there some better way to find a Hermite expansion of $\zeta$?

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What exactly is your question? – Qiaochu Yuan Feb 25 '10 at 22:30
I think he means to ask if these are actually correct, and what other similar expansions (in terms of Hermite polynomials) are there which are connected to $\zeta$. – Jacques Carette Feb 26 '10 at 0:35
Newly edited for clarification. – Craig Calcaterra Feb 26 '10 at 6:06