# Tag Archives: Zeta

## Integral representation of the zeta function proof

$$\zeta(s) = \frac{1}{\Gamma(s)}\int^\infty_0\frac{t^{s-1}}{e^t-1}dt$$ $$\textit{proof}$$ Start by the integral representation $$\int^\infty_0 \frac{e^{-t}t^{s-1}}{1-e^{-t}}\,dt$$ Using the power expansion $$\frac{1}{1-e^{-t}} = \sum_{n=0}^\infty e^{-nt}$$ Hence we have $$\int^\infty_0\,e^{-t}t^{s-1}\left(\sum_{n=0}^\infty e^{-nt}\right)\,dt$$ By swapping the series and integral $$\sum_{n=0}^\infty\int^\infty_0\,t^{s-1}e^{-(n+1)t}\,dt = \Gamma(s) \sum_{n=0}^\infty \frac{1}{(n+1)^s}=\Gamma(s)\zeta(s)\,$$

$\forall \,\, n\geq 1$ $$\psi_{n}(z) \, = \, (-1)^{n+1}n!\,\zeta(n+1,z)$$ $$\textit{proof}$$ Use the series representation of the digamma $$\psi_{0}(z) = -\gamma-\frac{1}{z}+ \sum_{n=1}^\infty\frac{z}{n(n+z)}$$ This can be written as the following $$\psi_{0}(z) = -\gamma + \sum_{k=0}^\infty\frac{1}{k+1}-\frac{1}{k+z}$$ By differentiating with respect to … Continue reading
$$\zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$ $$\textit{proof}$$ We start by the product formula of the sine function $$\frac{\sin(z)}{z} = \prod_{n=1}^\infty \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$ Take the logarithm to both sides \log(\sin(z)) – \log(z) = \sum_{n=1}^\infty \log \left(1-\frac{z^2}{n^2 \, \pi^2} … Continue reading