# complex analysis – Riemann zeta analytic continuation

$$pi^{-s/2}Gamma(s/2)zeta(s)=int_0^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}, Re(s)>1, where theta(u)=sum_{n=-infty}^{infty}e^{-n^2pi u}$$.

$$pi^{-s/2}Gamma(s/2)zeta(s)=int_0^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$$= $$int_0^1 frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$$+$$int_1^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$$.
Transforming u to 1/u in the first integral and using $$theta(1/u)=u^{1/2}theta(u)$$
We get,
$$pi^{-s/2}Gamma(s/2)zeta(s)=int_1^infty frac{(u^{1/2}theta(u)-1)}{2}u^{(s-1)/2}frac{du}{u}$$+$$int_1^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$$
So the R.H.S. is analytic for all s such that $$0. So is the first integral $$pi^{-s/2}Gamma(s/2)zeta(s)=int_0^infty frac{(theta(u)-1)}{2}u^{(1-s)/2} frac{du}{u}$$ valid for $$0?