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let $s,z$ be two complex parameters. Consider the integral : $$I(s,z)=s\int_{1}^{\infty}x^{z}\zeta^{'}(sx)dx$$ Computing the integral for $\Re(s)>1$ is easily done, using the Dirichlet series representation of the zeta function. Namely : $$I(s,z)=-s\int_{1}^{\infty}x^{z}\left(\sum_{n=2}^{\infty}\log (n)n^{-sx}\right)dx=\sum_{n=2}^{\infty}\left(s\log n\right)^{-z}\Gamma\left(1+z,s\log n\right)$$ Where $\Gamma(\cdot,\cdot)$ is the incomplete gamma function. It seems that $I(s,z)$ is entire in the $z$ plane, but not defined for $\Re(s)<1$ -as expected-. How to derive the analytic continuation for $I(s,z)$ in $s$ ?

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For $s > 1$ with the change of variable $x= y/s$ $$I_z(s) = s \int_1^\infty x^z \zeta'(sx)dx=s^{-z} \int_s^\infty y^z \zeta'(y)dy$$ and by analytic continuation for every $s \in \mathbb{C} \setminus (-\infty,1]$.

$ I_z(s)$ has two branch points : at $s=0$ and another one at $s=1$ caused by the pole of $\zeta'(s)$.

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  • $\begingroup$ I am sorry, but i don't understand this part of your answer "and by analytic continuation for every $s\in \mathbb{C}\setminus(-\infty,1]$". How did you arrive at this result ? say i want to compute $I_{z}\left(\frac{1}{2}+i\epsilon\right)$, what formula should i use ? $\endgroup$ – Mohammad Al Jamal Oct 31 '17 at 8:58
  • $\begingroup$ @MohammadAlJamal Don't you see the RHS is analytic on $ \mathbb{C}\setminus(-\infty,1]$ ? ($\int_s^\infty$ is a contour integral for $s$ complex) $\endgroup$ – reuns Oct 31 '17 at 19:34

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