# 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

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

2. Hasse's representation yields

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

3. The Laurent series yields

$\zeta\left(z+1\right)-\frac{1}{z}=\sum \limits^{\infty}_{n=0}\dfrac{1}{2^nn!}\left[\sum \limits^{\infty}_{k=0}\dfrac{\gamma_{n+2k}}{2^{2k+1}k!}\right]H_{n}\left(z\right)$

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 $\zeta$?

• 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
• What don't you write the formal manipulations that lead to the 1st formula ? – reuns Oct 25 '17 at 2:40