Sunday, 10 April 2016

Analytic continuation of zeta

It is known that1s1ζ(s)s=1x[x]xs+1dx for Re(s)>0.



The last integration can provide the analytic continuation for ζ.



I encouter one generalization of the situation :




Let f be a continuous function on [0,1]. If F(s)=s1f({t})ts1dt,s=σ+it,σ>0 then F(s) has the analytic continuation to the entire C, where{a}=a[a].



This general case is for Riemann zeta if f(x)=x.



I want to prove this theorem, and it states just set A=10f(u)du,f1(u)=u0(f({v})A)dv.
Then induction and integration by parts will yields the result, but I do not understand. At least, F has different integration (1=n=1n+1n) from A and f1.



It sounds like it is a well-known theorem, but most of the time I notice that ζ analytic continuation method is done using gamma function Γ or Bernoulli polynomial.




I cannot find a reference with good detail to follow the steps in the proof.



Any help please ?

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