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Problem statement: While I was writing another question and explaining, what I have already tried so far, I came across the following identity: $$\frac{\Gamma\left(\frac{1-2\,s}{2}\right)\,\zeta\left(-2\,s\right)}{\sqrt{\pi}\,s}=-\frac{\pi^{-2s}\,\sec\left(\pi\,s\right)\,\zeta\left(1+2\,s\right)}{\Gamma\left(1-s\right)}$$ $\zeta\left(s\right)$ is the Riemann zeta-function and $\Gamma\left(s\right)$ the gamma function. I assume, that this identity can be proven by the Zeta functional equation, but I haven't figured out how to show that.

EDIT 29.05.21

Inspired by a recent comment of @StevenClark and @metamorphy I finally managed to derivate the reflection functional equation of the Riemann zeta-function starting from the identity above. For the proof only the substituion $s\to -\frac{s}{2}$ is needed.

Attempt at an alternative proof: I found the following alternative proof, which is very interesting. We rewrite the identity in the form: $$-2\frac{\Gamma\left(\frac{1-2\,s}{2}\right)}{\sqrt{\pi}\,s}\sum_{k=1}^{\infty}\left(\frac{1}{k}\right)^{-2s}=\frac{4}{\pi^{\frac{3}{2}}}\frac{\pi^{\frac{3}{2}-2s}\sec\left(\pi\,s\right)}{2\,\Gamma\left(1-s\right)}\sum_{k=1}^{\infty}\left(\frac{1}{k}\right)^{1+2s}$$ Using the Mellin inverse transform on both sides leads to the interesting identity: $$1+2\sum_{k=1}^{\infty}\operatorname{erfc}\left(\frac{k}{\sqrt{z}}\right)=\frac{2\sqrt{z}}{\sqrt{\pi}}+\frac{4}{\pi^{\frac{3}{2}}}\sum_{k=1}^{\infty}\frac{\text{DawsonF}[k\,\pi\,\sqrt{z}]}{k}$$ $\text{DawsonF}\left(z\right)$ is Dawson's integral and $\text{erfc}\left(z\right)$ is the complementary error function. The left- and right-hand side may be written by its integral representation: $$\mathcal{G}\left(z\right)=\frac{1}{\pi}\int_{0}^{1}\frac{\vartheta_{3}\left(0,\exp\left(-\frac{1}{z~\tau}\right)\right)}{\sqrt{\tau\,\left(1-\tau\right)}}=\sqrt{\frac{z}{\pi}}\int_{0}^{1}\frac{\vartheta_{3}\left(0,\exp\left(-\pi^{2}\,z~\tau\right)\right)}{\sqrt{\left(1-\tau\right)}}$$ The equality may be proved by the Jacobi theta functional equation Peter Woit,p.4. The new function $\mathcal{G}\left(z\right)$ , will be subject of a future post of us. It has the known property of the theta function $\vartheta_{3}$, too: the functional equation. The proof of this transformation formula is then based on a result from theory of Fourier series called Poisson Summation Formula. For this the Fourier transform has to be applied, resulting in the last identity.

Gary
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stocha
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    Isn't it just a substitution $s\mapsto 2s+1$ (in the functional equation)? [+ reflection formula for $\Gamma$] – metamorphy May 28 '21 at 18:40
  • @metamorphy: Please descibe more detailed, we might talk about different forms of the Zeta functional equation, – stocha May 28 '21 at 18:49
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    @metamorphy Or try substituting $s\to-s/2$ which leads to $\zeta(s)$ on the left and $\zeta(1-s)$ on the right? – Steven Clark May 28 '21 at 22:53
  • @StevenClark: Thank you ! I finally understand the proof. I only have to substitute $s\to -\frac{s}{2}$. – stocha May 29 '21 at 11:26
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    You're welcome! – Steven Clark May 29 '21 at 15:06
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    It's an interesting identity which I believe is valid for $Re(z)>0$. There seems to be a striking resemblance to my formulas $\text{erfc}\left(\sqrt{z}\right)=\underset{N,f\to\infty}{\text{lim}}\left(2,\pi^{-3/2}\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f,n}\frac{F\left(\frac{k,\pi}{n \sqrt{z}}\right)}{k}\right)$ and $F\left(\sqrt{z}\right)=\underset{N,f\to\infty}{\text{lim}}\left(\frac{1}{2}\pi^{3/2}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f,n}\text{erfc}\left(\frac{k,\pi}{n \sqrt{z}}\right)\right)$ both of which I believe are also valid for $\Re(z)>0$. – Steven Clark May 30 '21 at 00:39
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    Both of my formulas must be evaluated at $M(N)=0$ where $M(N)=\sum\limits_{n=1}^N\mu(n)$ is the Mertens function. My formulas seem much more sensitive to the magnitude of $N$ than they are to the magnitude of $f$ which is assumed to be a positive integer. – Steven Clark May 30 '21 at 00:39
  • That's very interesting: First I wanted to Iook for any special value - similar to [Istvan Mezo] (https://www.ams.org/journals/proc/2013-141-07/S0002-9939-2013-11576-5/home.html) - for the new functions – stocha May 30 '21 at 12:23
  • $\mathcal{H}\left(z\right)=1+2\sum_{k=1}^{\infty}\left(-1\right)^{k}erfc\left(\frac{k}{\sqrt{z}}\right);and;\mathcal{G}\left(z\right)=1+2\sum_{k=1}^{\infty}erfc\left(\frac{k}{\sqrt{z}}\right)$ defined by their integral representations: $\mathcal{H}\left(z\right)=\sqrt{\frac{z}{\pi}}\int_{0}^{1}\frac{\vartheta_{2}\left(0,\exp\left(-\pi^{2}z~\tau\right)\right)}{\sqrt{\left(1-\tau\right)}};and;\mathcal{G}\left(z\right)=\sqrt{\frac{z}{\pi}}\int_{0}^{1}\frac{\vartheta_{3}\left(0,\exp\left(-\pi^{2}z~\tau\right)\right)}{\sqrt{\left(1-\tau\right)}}$ – stocha May 30 '21 at 12:23
  • These functions occur by the treatment of the Duhamel-Integral [stocha] (https://math.stackexchange.com/questions/2082444/closed-form-of-an-integral-involving-a-jacobi-theta-function-int-0t-thet) $\int_{0}^{t}\vartheta_{3}\left(0,\exp\left(-\pi^{2}~\left(t-\tau\right)\right)\right)\vartheta_{2}\left(0,\exp\left(-\pi^{2}~\tau\right)\right)\ d\tau=1$ The new functions, I study have the known property of the theta function, too: the functional equation: – stocha May 30 '21 at 12:27
  • $1+2\sum_{k=1}^{\infty}erfc\left(\frac{k}{\sqrt{z}}\right)=\frac{2\sqrt{z}}{\sqrt{\pi}}+\frac{4}{\pi^{\frac{3}{2}}}\sum_{k=1}^{\infty}\frac{\text{DawsonF}[k,\pi\sqrt{z}]}{k}$ – stocha May 30 '21 at 12:29
  • Let us continue this discussion in chat. – stocha May 30 '21 at 12:30
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    Here are a couple of related formulas which I believe are both valid for $\Re(z)>0$: $\text{erf}\left(\sqrt{z}\right)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f,n}\Gamma\left(0,\frac{k^2 \pi^2}{n^2 z}\right)\right)$ and $e^{-z}=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{2} \sqrt{\frac{\pi }{z}} \sum\limits_{n=1}^N\frac{\mu(n)}{n}\vartheta_3\left(0,e^{-\frac{\pi ^2}{n^2 z}}\right)\right)$. I posted here since chat doesn't render MathJax. – Steven Clark May 30 '21 at 20:20
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    When evaluating these formulas at a finite evaluation limit, $N$ must again be selected such that $M(N)=0$. It's possible to evaluate these formulas at arbitrarily large magnitudes of $N$ since the Mertens function has an infinite number of zeros. – Steven Clark May 30 '21 at 20:21
  • I have another remarkable formula for the dawson function $F[y]$, that could be interesting to you. $\underset{t\rightarrow \infty }{\lim }\sum_{k=1}^{\infty }\frac{1}{k^{4,t+1}}\left( \text{F}[k~\sqrt{y}]\right) =\text{F}[\sqrt{y}]$ I already proofed this formula. – stocha May 30 '21 at 21:20
  • @Gary: Thank you! – stocha May 31 '21 at 11:44

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