It is known that1s−1−ζ(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)=s∫∞1f({t})t−s−1dt,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=1∫n+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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