Prove that zeta function is continuable in the complex plane with only singularity a simple pole at s = 1.
Required to use the formula $ζ(s) = \frac{1}{Γ(s)} \int_0^1\frac{x^{s-1}}{e^x-1}dx+\frac{1}{Γ(s)} \int_1^\infty\frac{x^{s-1}}{e^x-1}dx$.
I used some methods from functional analysis, but I imagine that there is a short solution.
Best Answer
Expand $\frac{x}{e^x-1}$ in power series in the $\int_0^1$ integral and use the theorem that $\Gamma(s)$ has no zero and simple poles at negative integers (which follows for example from that $\Gamma(s)\Gamma(1-s)-\pi/\sin(\pi s)$ is entire and bounded)