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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!

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    $\begingroup$ We know the value of $\zeta(0)$ and $\zeta'(0)$. So just plug them in... $\endgroup$ – user17762 Dec 24 '12 at 21:38
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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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    $\begingroup$ A very beautiful answer one should know! (+1) $\endgroup$ – user 1357113 Dec 25 '12 at 8:44

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