In Titchmarsh's book "The theory of the Riemann zeta-function" there's theorem 14.25(A) on page 369 of the second edition where a summand $1/\zeta(s)$ appears out of the blue, so it seems... Oh, I do know I'm missing something here. The book can be read online here.

The equation at hand appearing in the book proving theorem 14.25(A) goes as follows:

$\sum\limits_{n<x}\frac{\mu(n)}{n^s}=\frac{1}{2\pi i}\int\limits_{2-iT}^{2+iT}\frac{x^w}{\zeta(s+w)}\frac{dw}{w}+O\left( \frac{x^2}{T} \right)\\ =\frac{1}{2\pi i}\left( \int\limits_{2-iT}^{1/2-\sigma+\delta-iT} + \int\limits_{1/2-\sigma+\delta-iT}^{1/2-\sigma+\delta+iT} + \int\limits_{1/2-\sigma+\delta+iT}^{2+iT} \right) \frac{x^w}{\zeta(s+w)}\frac{dw}{w}+ \frac{\mbox{$\bf{1}$}}{\mbox{$\bf{\zeta(s)}$}} +O\left( \frac{x^2}{T} \right) $

Here, $\sigma=\Re(s)$, $\sigma>1/2$ and $0<\delta<\sigma-1/2$. The integrals are all taken along paths to the right of the critical line $\Re(s+w)=1/2$ and one assumes the truth of RH.

So, Titchmarsh simply deforms the contour of integration in the region free of singularities of the integrand. And then adds $1/\zeta(s)$...

My question is:

$\textbf{Where did the summand }\frac{1}{\zeta(s)}\textbf{ come from?}$ How does one justify the introduction of the summand $1/\zeta(s)$ in the second line of this Titchmarsh's proof of theorem 14.25(A)?

I believe the explanation is trivial, but nonetheless it escapes me completely at the moment. Any suggestion is welcome.

  • $\begingroup$ @Xander Henderson Hi. The first integral is a finite version of the Mellin transform, it integrates along a single finite straight line from 2-iT to 2+iT. Here T is finite. So Titchmarsh just deformed the contour without intersections... The three integrals just follow a different contour connecting the same 2 points. I don't see how could one produce a closed contour here, and how it could then introduce $1/\zeta(s)$... Help please! :D $\endgroup$ – anonymous Sep 24 '18 at 1:09
  • $\begingroup$ The deformed contour runs to the left of $w=0$, so it seems $1/\zeta(s)$ is the residue at $w=0$? $\endgroup$ – anonymous Sep 24 '18 at 1:22
  • $\begingroup$ If $1/\zeta(s)$ is the residue leftover from the closed contour integration around $w=0$ then the 3 integrals above appear with the wrong sign, don't they? $\endgroup$ – anonymous Sep 24 '18 at 1:29
  • $\begingroup$ Oh the integration goes in the reverse order, clock-wise! Yes, you were right. OK, I see it now. It's the residue. Thanks Xander! Cheers! :D Titchmarsh writes in the really condensed manner... Not bothered to explain steps either :D $\endgroup$ – anonymous Sep 24 '18 at 1:31

I'll answer my own question. As suggested by Xander, the term $1/\zeta(s)$ is the residue. Consider the following result:

$\frac{1}{2\pi i}\left( \int\limits_{2-iT}^{1/2-\sigma+\delta-iT} + \int\limits_{1/2-\sigma+\delta-iT}^{1/2-\sigma+\delta+iT} + \int\limits_{1/2-\sigma+\delta+iT}^{2+iT}+ \int\limits_{2+iT}^{2-iT} \right) \frac{x^w}{\zeta(s+w)}\frac{dw}{w} =-\frac{1}{\zeta(s)}$

Here, the integration runs along the closed contour clock-wise, hence the minus sign on the right. The term $1/\zeta(s)$ is simply the residue of the integrand at $w=0$, because the contour does run around the point $w=0$.

From the above equation one finds trivially

$ \frac{1}{2\pi i}\left( \int\limits_{2-iT}^{1/2-\sigma+\delta-iT} + \int\limits_{1/2-\sigma+\delta-iT}^{1/2-\sigma+\delta+iT} + \int\limits_{1/2-\sigma+\delta+iT}^{2+iT} \right) \frac{x^w}{\zeta(s+w)}\frac{dw}{w}+\frac{1}{\zeta(s)} =\\ -\frac{1}{2\pi i}\int\limits_{2+iT}^{2-iT}\frac{x^w}{\zeta(s+w)}\frac{dw}{w}= \frac{1}{2\pi i}\int\limits_{2-iT}^{2+iT}\frac{x^w}{\zeta(s+w)}\frac{dw}{w} $


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.