Riemann zeta function

Series involving the zeta function, e and pi

We show the proof of this beutiful series posted by @infseriesbot. \[\exp\left(\frac{\zeta(2)}{2}-\frac{\zeta(2)}{3}+\frac{\zeta(4)}{4}-\frac{\zeta(4)}{5}+\frac{\zeta(6)}{6}-...\right)=\sqrt{\frac{2\pi}e}\] To prove this result we use two concepts:

1. The Taylor series expansion of the log-Gamma function $\ln \Gamma(x) $
2. The Euler's reflection formula for the Gamma function $ \Gamma(x)$

Lets start with the left hand side \begin{align*} \frac{\zeta(2)}{2}-\frac{\zeta(2)}{3}+\frac{\zeta(4)}{4}-\frac{\zeta(4)}{5}+\frac{\zeta(6)}{6}-...=&\sum_{n=1}^{\infty} \left(\frac{1}{2n}-\frac{1}{2n+1}\right)\zeta(2n)\\ =&\sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n)(2n+1)} \end{align*} What is the series of the right hand side? This series is related to the expansion series of the log-Gamma function: \[\ln(\Gamma(z))=-\gamma(z-1)+\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k} (z-1)^{k} \quad |z-1|<1\] where $\gamma$ is the Euler-Mascheroni constant. If we let $z=1+t$ and then $z=1-t$ in the expansion series we obtain two equations: \begin{align*} \ln(\Gamma(t+1))&=\gamma t+\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k} t^{k} \\ \ln(\Gamma(1-t))&=-\gamma t+\sum_{k=2}^{\infty}\frac{\zeta(k)}{k} t^{k} \end{align*} Adding both equations: \begin{align*} \ln(\Gamma(t+1))+\ln(\Gamma(1-t))&=\sum_{k=1}^{\infty}\frac{\zeta(2k)}{2k} t^{2k} \label{eq:3} \end{align*} Integrating both sides: \begin{align*} \int_{0}^{1}\left(\ln(\Gamma(t+1))+\ln(\Gamma(1-t))\right)dt &=\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k} \int_{0}^{1}t^{2k} dt \\ \Longrightarrow \frac{1}{2}\int_{0}^{1}\left(\ln(\Gamma(t+1))+\ln(\Gamma(1-t))\right)dt & = \sum_{k=1}^{\infty}\frac{\zeta(2k)}{2k(2k+1)} \tag{1} \end{align*} This is the series we were looking for in terms of a definite integral. Hopefully we can transform it to a maneuverable one. Here is where the Euler's reflection formula for the Gamma function is useful: \begin{align*} \Gamma(t)\Gamma(1-t)=\frac{\pi}{\sin(\pi t)} \quad t\notin \mathbb{Z} \end{align*} Therefore \begin{align*} \frac{1}{2}\int_{0}^{1}\left(\ln(\Gamma(t+1))+\ln(\Gamma(1-t))\right)dt & = \frac{1}{2}\int_{0}^{1}\left(\ln(\Gamma(t+1))(\Gamma(1-t))\right)dt\\ & =\frac{1}{2}\int_{0}^{1}\left(ln(t\cdot\Gamma(t))(\Gamma(1-t))\right)dt\\ & =\frac{1}{2}\int_{0}^{1}\ln\left( \frac{t \cdot \pi}{\sin(\pi t)}\right)dt\\ & = \frac{1}{2}\ln(\sqrt{\pi})-\frac{1}{2}-\frac{1}{2}\int_{0}^{1}\ln(\sin(t\pi))dt \tag{2} \end{align*} We just have to prove that $\int_{0}^{1}\ln(\sin(t\pi))dt=-\ln(2)$ \begin{align*} \int_{0}^{1}\ln(\sin(t\pi))dt& = \int_{0}^{\frac{1}{2}}\ln(\sin(2\pi t))dt\\ & =\ln(2)+ \int_{0}^{\frac{1}{2}}(\sin(\pi t))dt+\int_{0}^{\frac{1}{2}}\ln(\cos(\pi t))dt \quad (\sin(2t)=2\sin(t)\cos(t)) \\ \end{align*} We also know that \begin{align*} \int_{0}^{1}\ln(\sin(t\pi))dt& = 2\int_{0}^{\frac{1}{2}}\ln(\sin(\pi t))dt \quad \text{(using $-\sin(\theta)=\sin(-\theta)$)}\\ &= \int_{0}^{\frac{1}{2}}\ln(\sin(2\pi t))dt \end{align*} Therefore \begin{align*} \int_{0}^{1}\ln(\sin(t\pi))dt& \ln(2)+ \int_{0}^{\frac{1}{2}}\ln(\sin(\pi t))dt+\int_{0}^{\frac{1}{2}}\ln(\cos(\pi t))dt \\ & = \ln(2)+ \int_{0}^{\frac{1}{2}}\ln(\sin(\pi t))dt+\int_{0}^{\frac{1}{2}}\ln(\sin(\pi w))dw \quad \text{($w=\frac{1}{2}-t$)}\\ & = \ln(2)+ 2\int_{0}^{1}\ln(\sin(\pi t))dt+2\int_{0}^{1}\ln(\sin(\pi w))dw \end{align*} \begin{align*} \Longrightarrow \int_{0}^{1}\ln(\sin(t\pi))dt & = \ln(2)+ \int_{0}^{1}\ln(\sin(t\pi))dt+\int_{0}^{1}\ln(\sin(t\pi))dt \\ \Longrightarrow \int_{0}^{1}\ln(\sin(\pi t))dt & = -\ln(2) \\ \tag{3} \end{align*} We conclude from results (1),(2) and (3) our claim: \begin{align*} \frac{\zeta(2)}{2}-\frac{\zeta(2)}{3}+\frac{\zeta(4)}{4}-\frac{\zeta(4)}{5}+\frac{\zeta(6)}{6}-...&= \frac{1}{2}\ln(\sqrt{\pi})-\frac{1}{2}-\int_{0}^{1}\ln(\sin(t\pi))dt\\ =&\frac{1}{2}\ln(\sqrt{\pi})-\frac{1}{2}+\frac{1}{2}\ln(2) \\ \end{align*} \[\boxed{ \exp\left(\frac{\zeta(2)}{2}-\frac{\zeta(2)}{3}+\frac{\zeta(4)}{4}-\frac{\zeta(4)}{5}+\frac{\zeta(6)}{6}-...\right)=\sqrt{\frac{2\pi}e}}\]