Let $$Li(x)=\int_2^x\frac{dt}{\log t},$$ the logarithmic integral and we define for $\Re s>1$ the complex function $$f(s)=s\int_2^\infty\frac{Li(u)}{u(u^s-1)}du.$$

Question. I say that previous expression defines $f(s)$ as an analytic function for $\Re s>1$. Can you discuss if the analytic continuation of $f(s)$ has pole/s, and what is its residue? (Please if my question isn't well possed explain why.)

There is in the literature the corresponding identity defining $\log\zeta(s)$, with the prime-counting function $\pi(u)$ in the integrand.

Thanks in advance.

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    $\begingroup$ $Li(u) = \mathcal{O}(u)$ and $\frac{1}{u^s-1} - u^{-s} = \mathcal{O}(u^{-2s})$ so $\int_2^\infty \frac{Li(u)}{u(u^s-1)} du = \int_2^\infty Li(u)u^{-s-1} du + \int_2^\infty \mathcal{O}(u^{-2s})du$. the second integral is analytic for $Re(s) > 1/2$, while the first can be integrated by parts $\int_2^\infty Li(u)u^{-s-1} du = \frac{1}{s} \int_2^\infty \frac{u^{-s}}{\log u} du$ having a (logarithmic) singularity at $s=1$ $\endgroup$ – reuns Sep 20 '16 at 10:32
  • $\begingroup$ Now I take notes from your valuable response, feel free to add an answer. Many thanks @user1952009 $\endgroup$ – user243301 Sep 20 '16 at 12:12
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    $\begingroup$ note that $F(s) = \int_2^\infty \frac{u^{-s}}{\log u} du$ means $F'(s) = -\int_2^\infty u^{-s} du = \frac{2^{1-s}}{1-s}$ hence $F(s) =C+ \int \frac{2^{1-s}}{s-1}ds$ (having clearly a logarithmic singularity at $s=1$) $\endgroup$ – reuns Sep 20 '16 at 12:24
  • $\begingroup$ Great explanation user1952009 for the first integral, now I try think what about it $\endgroup$ – user243301 Sep 20 '16 at 12:49

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