Analytic continuation of Riemann zeta function using integral formula

Question

From the integral formula $$\zeta(s)=s/(s-1)-s\int_1^\infty \{x\}x^{-s-1}dx$$ one can show that $\zeta$ has an analytic continuation for $\sigma(=\Re(s))>0$. Can I also derive the analytic continuation for $\sigma>-1$?

Answer 1

Yes, substract $1/2$ (the mean value of $\{x\}$) to get a conditionally convergent integral $$\zeta(s)=\frac{s}{s-1}-\frac12-s\int_1^\infty (\{x\}-\frac12)x^{-s-1}dx \qquad (\Re(s) > -1)$$ Then integrate by parts if you want an absolutely convergent integral.

Repeating the same process $k$ times, you'll get the continuation to $\Re(s) > -k$ and the value of $\zeta(-n)$, $n > k$.

Also note that $\zeta(s) = -s\int_0^\infty (\{x\}-1/2) x^{-s-1}dx$ for $\Re(s) \in (-1,0)$ is a way to prove the functional equation, using the Fourier series $\{x\}-1/2 = -\sum_{n=1}^\infty \frac{\sin(2\pi n x)}{\pi n}$ and $\int_0^\infty \sin(2\pi nx)x^{-s-1}dx = n^{s-1}\int_0^\infty \sin(2\pi x)x^{-s-1}dx$