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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.

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  • $\begingroup$ @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
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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} $

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