Relation to the Gamma Function 5 Click on the image for a detailed proof \zeta(s)=\frac{1}{2}+\frac{1}{s-1}+\frac{1}{\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}\right)\frac{x^{s-1}}{e^x}dx \zeta(s)=\frac{1}{2}+\frac{1}{s-1}+\frac{1}{\Gamma(s)}\int_{0}^{\infty}\left(\frac{1}{e^x-1}-\frac{1}{x}+\frac{1}{2}\right)\frac{x^{s-1}}{e^x}dx This Relation connecting the Riemann Zeta and the Gamma Function has the benefit of making explicit the pole of the Zeta Function at s=1.This demonstration is based on the Relation to the Gamma Function 1. Any questions? Feel free to contact our authors at: [email protected], you will be put directly in touch with the person who wrote the text. Didn’t find what you were looking for? Keep looking through all our formulas regarding the Riemann Zeta Function! First Functional EquationLaurent ExpansionEuler's Product Formula First DerivativeRelation to the Gamma Function 2Analytic continuation for 0<Re s<1 See all Formulas for the Riemann Zeta… X-twitter Pinterest