Riemann was able to establish the following link between the Riemann zeta function and the weighted prime counting function $J(x)$. $$\ln(\zeta(s))=s\int_1^\infty J(x)x^{-s-1}dx$$ Using the Mellin inverse transform, he then obtained the following: $$J(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\ln(\zeta(s))}{s} x^{-s}ds$$ The problem with the above equation is that it evaluates $\zeta(s)$ at complex values. However, he was able to surmount that problem by using the Zeta functional equation, which states $$\pi^{-s / 2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right) = - \frac 1 {s \left({1 - s}\right)} + \int_1^\infty \left({x^{s / 2 - 1} + x^{- \left({s + 1}\right) / 2} }\right) \omega \left({x}\right)dx$$ I understand how this formula is obtained, but cannot grasp this next step. From combining the functional equation and the result from the inverse Mellin transform, he obtains $$J(x)=Li(x)-\sum_\rho Li(x^\rho)-\ln(2)+\int_x^\infty\frac{dt}{t(t^2-1)\ln(t)}$$ How did the functional equation result in this simplification? How did Riemann simplify the complex integral? a step by step explanation from the Mellin integral to the final result above would be very helpful!

NOTE: I am not very familiar with integration over the complex plane, but very willing to learn.

  • $\begingroup$ What is the exact question? Is this a historical question about Riemann or a question about proofs of the Prime Number Theorem? Anyway, the question seems somewhat based on a false premise. (What is your source for the claims you make.) Most any book on analytic number theory should be able to clear this up. $\endgroup$
    – quid
    Jul 7, 2015 at 22:53
  • $\begingroup$ @quid thanks for pointing that out - just edited the question $\endgroup$
    – Romain S
    Jul 7, 2015 at 22:56
  • 1
    $\begingroup$ start with $\frac{\zeta'(s)}{s \zeta(s)}$ you'll get with the residue theorem (and the functionnal equation to allow applying residue theorem on the contour $\Re(s)$ going to $-\infty$) a sum of $x^\rho$ for $\varphi(x)$, then remark that $\zeta'(s)/\zeta(s)$ is the derivative of $\ln\zeta(s)$ so that $J(x) = \int \varphi'(x)/\ln x$ and you'll get the $Li(x^\rho)$ terms... the integral in the right of the $Li(x^\rho)$ is a sum of $Li(x^{-2k})$ for the "trivial" zeros of zeta. my advice : read wikipedia, the Tichmarsh's book "the theory of the riemann zeta function" and forums and pdfs. $\endgroup$
    – reuns
    Jul 18, 2015 at 8:10

1 Answer 1


pleas look this page 4 may you get what you want (P. Bourgade and J.P. Keating Seminaire Poincare)


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