The $\psi(x)$ is the second Chebyshev function.

The $\zeta(s)$ is the Riemann zeta function.

Mellin inversion theorem:

$$\varphi(s)=\int_0^\infty f(x)x^{s-1}\,dx$$ $$f(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\varphi(s) x^{-s}\,ds$$

Starting with: $$\tag{1}-\frac{\zeta'(s)}{\zeta (s)} \frac{1}{s}=\int_0^{\infty } \frac{\psi (x)}{x^{s+1}} \, dx$$ Using substitutution $s=-z$: $$\tag{2}-\frac{\zeta'(-z)}{\zeta (-z)} \frac{1}{-z}=\int_0^{\infty } \psi (x) x^{z-1} \, dx$$ Applying inverse Mellin transform: $$\tag{3}\psi(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\left(-\frac{\zeta'(-z)}{\zeta (-z)} \frac{1}{-z}\right) x^{-z}\,dz$$ Reversing substitution $z=-s$ and $\frac{dz}{ds}=-1$: $$\tag{4}\psi(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\left(-\frac{\zeta'(s)}{\zeta (s)} \frac{1}{s}\right) x^{s}\,(-1) ds$$ This simplifies to: $$\tag{5}\psi(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta (s)} \frac{x^{s}}{s}\,ds$$

But the correct formula should be: $$\tag{6}\psi(x)=-\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta (s)} \frac{x^{s}}{s}\,ds$$

I cannot spot a mistake. Can you tell me where I went wrong?

  • 1
    $\begingroup$ I'm working from a "not-so-smart" phone, so apology if I'm not seeing this correctly. It appears the sign error occurs in going from $(3)$ to $(4)$. $\endgroup$ – Mark Viola Nov 14 '17 at 21:45
  • $\begingroup$ @Mark Viola I also thought this step is most likely the source of error, but I looked at it many times and see nothing wrong with it. Maybe I am blind. $\endgroup$ – azerbajdzan Nov 14 '17 at 21:50
  • $\begingroup$ @Mark Viola I got it! I forgot to change integration limits in $(4)$... :-D $\endgroup$ – azerbajdzan Nov 14 '17 at 22:04
  • 1
    $\begingroup$ The Riemann explicit formula is obtained from the residue theorem applied to $$\psi(x)=\lim_{n \to \infty}\frac1{2\pi i}\int_{2-iT_n}^{2+iT_n}+\int_{2+iT_n}^{-\infty+iT_n}+\int_{-\infty-iT_n}^{2-iT_n}\frac{-\zeta'(s)}{\zeta (s)} \frac{x^s}{s}ds$$ The main difficulty is to find a sequence $T_n \to \infty$ such that $\int_{2+iT_n}^{-\infty+iT_n}\frac{-\zeta'(s)}{\zeta (s)} \frac{x^s}{s}ds \to 0$ (which needs a careful analysis of the vertical density of non-trivial zeros). $\endgroup$ – reuns Nov 15 '17 at 21:43
  • 1
    $\begingroup$ You should try first with $\sum_{m=1}^\infty \Lambda(m)e^{-mx}=\lim_{n \to \infty}\frac1{2\pi i}\int_{2-iT_n}^{2+iT_n}+\int_{2+iT_n}^{-\infty+iT_n}+\int_{-\infty-iT_n}^{2-iT_n}\frac{-\zeta'(s)}{\zeta (s)} \Gamma(s)x^sds$ which is much easier because $\Gamma(s)$ is fast decreasing on vertical strips. $\endgroup$ – reuns Nov 15 '17 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.