How would you prove the following Wallis formula form $ \left(4^{\zeta{(0)}} \cdot e^{-\zeta'{(0)}}\right)^2=\frac{\pi}{2}?$ Thanks in advance!
Prove the Wallis formula form $\left(4^{\zeta{(0)}} \cdot e^{-\zeta'{(0)}}\right)^2=\frac{\pi}{2}$
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4We know the value of $\zeta(0)$ and $\zeta'(0)$. So just plug them in... – 2012-12-24
1 Answers
In this answer, I show that $\zeta(0)=-\frac{1}{2}$, and that $\zeta^{'}(0)=-\frac{1}{2}\log(2\pi ).$ Plugging these in yields $\left(4^{\zeta(0)}e^{-\zeta^{'}(0)}\right)^{2}=4^{-1}e^{\log(2\pi)}=4^{-1}\cdot2\pi=\frac{\pi}{2}.$ Also of interest is part $4$ of my answer regarding the evaluation of $\Gamma\left(\frac{1}{2}\right).$
Proving that $\zeta(0)=-\frac{1}{2}.$
The Riemann zeta function is defined as $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}.$ Writing this as a Riemann Stieltjies integral and applying integration by parts we have that $\zeta(s)=\int_{1}^{\infty}x^{-s}d\left[x\right]=\int_{1}^{\infty}x^{-s}dx-\int_{1}^{\infty}x^{-s}d\left\{ x\right\}$
$=\frac{s}{s-1}-s\int_{1}^{\infty}\left\{ x\right\} x^{-s-1}dx,$ and this holds for all $\text{Re}(s)>0.$ Notice that taking the limit as $s\rightarrow0$ and being careful with the canceling singularities yields $\zeta(0)=-\frac{1}{2}.$
Proving that $\zeta^{'}(0)=-\frac{1}{2}\log(2\pi).$
Recall the functional equation for the zeta function, $\zeta(z)=2^{z}\pi^{z-1}\sin\left(\frac{\pi z}{2}\right)\Gamma\left(1-z\right)\zeta\left(1-z\right).$ Taking the logarithmic derivative, we have that $\frac{\zeta^{'}(z)}{\zeta(z)}=\log2+\log\pi-\frac{\Gamma^{'}\left(1-z\right)}{\Gamma\left(1-z\right)}+\frac{d}{dz}\log\left(\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)\right).$ Since $\zeta(1-z)=-\frac{1}{z}+\gamma+O(z),$ and $\sin\left(\frac{\pi z}{2}\right)=\frac{\pi z}{2}+O\left(z^{3}\right),$ we have that $\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)=\frac{\pi}{2}-\frac{\pi\gamma}{2}z+O\left(z^{2}\right),$
and so $\frac{d}{dz}\log\left(\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)\right)=\frac{-\frac{\pi\gamma}{2}+O\left(z\right)}{\frac{\pi}{2}+O\left(z\right)}=-\gamma+O(z).$ Thus $\frac{\zeta^{'}(0)}{\zeta(0)}=\log2\pi-\Gamma^{'}\left(1\right)-\gamma.$ As $\Gamma^{'}(1)=-\gamma,$ and $\zeta(0)=-\frac{1}{2},$ we conclude that $\zeta^{'}(0)=-\frac{1}{2}\log2\pi.$
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1A very beautiful answer one should know! (+1) – 2012-12-25