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 $\zeta$?

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 $\zeta$?

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 lead 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.

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

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 $\zeta$?