Cómo se probaría el Wallis forma fórmula $$ \left(4^{\zeta{(0)}} \cdot e^{-\zeta'{(0)}}\right)^2=\frac{\pi}{2}?$ $ gracias de antemano!
Respuesta
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En esta respuesta, me muestran que la $\zeta(0)=-\frac{1}{2}$, $$\zeta^{'}(0)=-\frac{1}{2}\log(2\pi ).$$ Enchufar estos en los rendimientos de $$\left(4^{\zeta(0)}e^{-\zeta^{'}(0)}\right)^{2}=4^{-1}e^{\log(2\pi)}=4^{-1}\cdot2\pi=\frac{\pi}{2}.$$ Also of interest is part $4$ of my answer regarding the evaluation of $\Gamma\left(\frac{1}{2}\right).$
Demostrando que $\zeta(0)=-\frac{1}{2}.$
La Riemann zeta función se define como: $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}.$ Escribir esto como una Riemann Stieltjies integral y aplicando integración por partes tenemos que $$\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,$$ and this holds for all $\texto{Re}(s)>0.$ Notice that taking the limit as $s\rightarrow0$ and being careful with the canceling singularities yields $$\zeta(0)=-\frac{1}{2}.$$
Demostrando que $\zeta^{'}(0)=-\frac{1}{2}\log(2\pi).$
Recordemos la ecuación funcional para la función zeta, $$\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),$$
y así $$\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.$$
