Does this extend to C?
ζ(x)=∫∞01⌊t⌋xdt, where for 0≤t<1 we say that ⌊t⌋=1.
Answer
For ℜ(x)>1, this integral converges. Further, this integral converges to exactly the regular Riemann zeta function. Thus an analytic continuation of the regular Riemann zeta is an analytic continuation of this function.
[Given the clarification]:
Instead of converging to the regular Riemann zeta function, this zeta function is exactly the regular Riemann zeta function +1 (for the bit between 0 and 1), and thus we still get our continuation from the regular Riemann zeta function.
No comments:
Post a Comment