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In Titchmarsh's book "The theory of the Riemann Zeta-function" his Lemma $3.12$ is one of the main tools. Lemma $3.12$ is a version of Perron formula. Lemma $3.12$ starts by observing that, with $c>0$ and with $n<x$,

$$ \frac{1}{2\pi i}\int\limits_{c-iT}^{c+iT}\left(\frac{x}{n}\right)^w \frac{\mathrm dw}{w}=1+ O\left( \frac{(x/n)^c}{T\log (x/n)} \right) $$

This stems from the fact that $c>0$, and so if one completes the integration contour to the left by encircling the point $w=0$ completely in the complex $w$ plane by a rectangle whose right-most side is the original integration line $(c-iT,c+iT)$, the residue is simply $1$. This is the reason for the requirement $c>0$. The requirement $n<x$ comes from letting the left-most side of the integration rectangle be at $\Re(w)=-\infty$ and so $(x/n)^w\to 0$ if and only if $n<x$ as $\Re(w)\to -\infty$.

And so, in Theorem $3.13$ the starting point is the application of Lemma $3.12$,

$$ \sum\limits_{n<x} \frac{\mu(n)}{n^s}=\frac{1}{2\pi i}\int\limits_{c-iT}^{c+iT} \frac{x^w \mathrm dw}{w\zeta(s+w)}+ O\left( \frac{x^c}{cT} \right)+ O\left( \frac{\log x}{T} \right) $$

On the LHS the condition $n<x$ is the condition from Lemma $3.12$, and $c$ is still required to obey $c>0$.

Now Titchmarsh enlarges the contour of integration and encloses the $w=0$ completely, as earlier. Do notice that $c>0$ should still hold, since $c$ is nothing else but $\Re(w)$ on the right-most side the integration contour. No matter how the original line $(c-iT,c+iT)$ is deformed, the deformed closed integration contour must enclose the point $w=0$, and hence $c>0$ on the RHS part of the contour that intersects the real axis in $w$ plane.

With $c>0$ the pole at $w=0$ produces the residue $1/\zeta(s)$ and the result is now

$$\sum\limits_{n<x} \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}+O$$

The $O$ term tends to $0$ in the end, after Titchmarsh adjusts $T$ and $x$ appropriately.

And so then Titchmarsh requires

$$c=\frac{1}{\log x}$$

Additionally, he then lets $x\to\infty$.

But in this limit, $c\to 0$. And this makes the original starting equation useless, since the pole of the integrand at $w=0$ is no longer enclosed in the integration countour, but lies on the integration contour instead, if one interprets the limit as $\lim\limits_{x\to\infty}x=\infty$.

On the other hand, if one interprets the limit $\lim\limits_{x\to\infty}x$ in the way that $x$ grows large but never hits the point at infinity, then $c$ never reaches $0$, and this would do. But aren't there problems with this interpretation of a limit in standard mathematics?

Maybe Titchmarsh uses the Sokhotski–Plemelj theorem, but then the result misses the summand $\frac{1}{2\zeta(s)}$...

Or maybe I'm missing some detail completely here...

So my question is:

What exactly enables Titchmarsh to take the limit $x\to\infty$?

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Let $\epsilon >0$; by the above (standard stuff you seem to agree with), there is a large enough half-integer $x(\epsilon)$ (depending on $s=1+it, \epsilon$), s.t. $|\sum\limits_{n<x} \frac{\mu(n)}{n^s}-\frac{1}{\zeta(s)}| < \epsilon$ for any other half-integer $x \geq x(\epsilon)$ - as usual there is no restriction in using half-integers (i.e. $odd/2$) beacause the Dirichlet sum is constant between integers and the jump goes to zero since $s=1+it$

(choosing in Perron, $c=\frac{1}{\log x}, \log T= (\log x)^{\frac{1}{10}}, \delta = A(\log T)^{-9}=A(\log x)^{-\frac{9}{10}},x$, hence $T$ large enough, $A$ positive absolute constant coming from zero-free regions of $\zeta$)

But then the relation $|\sum\limits_{n<x} \frac{\mu(n)}{n^s}-\frac{1}{\zeta(s)}| < \epsilon$ for $x>x(\epsilon)$ is precisely what is needed to conclude that the limit as $x$ goes to $\infty$ of $\sum\limits_{n<x} \frac{\mu(n)}{n^s}$ is precisely $\frac{1}{\zeta(s)}$, by the usual definition of limit.

So the point is that you apply Perron with large but finite $x$ and then in "shorthand" you let $x$ go to $\infty$ meaning the above $\epsilon - x(\epsilon)$ limit relation, which has nothing to do with making $x$ infinite as a number

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  • $\begingroup$ Awesome, thank You, Conrad! $\endgroup$ – anonymous Mar 27 '19 at 18:04
  • $\begingroup$ You are welcome $\endgroup$ – Conrad Mar 27 '19 at 19:03
  • $\begingroup$ $A$ is a constant that comes from the zero-free region - it probably has a value depending on an inferior bound on $x$ but not sure what is though if you allow large enough $x$ you can probably make it as big as you want (eg if you use $x >1000$ then the choice of $T$ gives $T > T_0=(\log 1000)^{.01}$ and that gives a zero free region constant $A$ valid for those $T$). In general $A$ is irrelevant being just a constant that depends only on your intiial starting point but has no bearing on large values of $x,T$ $\endgroup$ – Conrad Jan 15 '20 at 20:25
  • $\begingroup$ @Conrad, what if we set $c = \eta + 1/\log(x)$ where $\eta$ is another small positive number. We will need to change $T = x$ So that all terms in RHS go to zero. And we don’t have to worry about $c$ going to zero. $\frac{x^c}{Tc} = \frac{x^{(\eta+1/\log(x))}}{ (x\eta+ x/\log(x))}$ goes to 0 as $x \rightarrow \infty$ etc.... $\endgroup$ – Shree Feb 5 '20 at 23:29
  • $\begingroup$ @shree the question makes sense only in the context of titchmarsh book as there is more going on and one needs to move the integral through the pole at $0$ coming from $1/w$ and still stay in a non zero area of zeta which is a function of $T$ so this determines $\delta(T)$ which in turn forces what $x$ can be in terms of $T$ etc; we then reverse engineer and get $T$ in terms of $x$ so they are not free parameters $\endgroup$ – Conrad Feb 6 '20 at 0:41
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To see the idea of the proof, let $M(x) = \sum_{n \le x} \mu(n)$ and we'll look instead at $f(x) = \int_1^x M(y)dy$ to obtain absolutely convergent integrals.

Then Titchmarsh shows a lower bound $|\zeta(1+it)|\ge B/\log(2+| t|)$ plus an upper bound $|\zeta'(s)|\le B\log(2+| t|)$ to obtain some bound $|\zeta(s)|>1/(A\log(2+|t|)) ,|\frac1{\zeta(s)}|<\log(2+|t|)$ for $s=\sigma+it, \sigma\ge 1-\frac{1}{A\log^2(2+ |t|)}$.

And this is what we need to conclude $$f(x) = \frac{1}{2i\pi} \int_{2-i\infty}^{2+i\infty} \frac{1}{\zeta(s)} \frac{x^{s+1}}{s(s+1)}ds= \frac{1}{2i\pi} \int_{\Re(s) = 1-\frac{1}{A\log^2 (2+|\Im(s)|)}} \frac{1}{\zeta(s)} \frac{x^{s+1}}{s(s+1)}ds \\= O(\int_{-\infty}^\infty \frac{\log (2+|t|)}{1+t^2 }x^{2-\frac{1}{A \log^2(1+ |t|)}}dt)\\ = O(\int_0^T \frac{x^{2-\frac{1}{A \log^2(1+ |T|)}}\log (2+|t|)}{1+t^2 }dt)+O(\int_T^\infty \frac{x^{2}\log T}{1+t^2 }dt)\\ =O(x^{2-\frac{1}{A \log^2(1+ |T|)}}) + O(\frac{x^2 \log T}{T})\\=O(x^{2-\frac{1}{A \log^2(1+ e^{\log^{1/4} x})}}) +O(\frac{x^2\log^{1/4} x}{e^{\log^{1/4} x}})=O(\frac{x^2}{e^{\log^{1/8} x}})=o(x)$$

If $M(x)> cx$ infinitely often, as $M(x+y)\ge M(x)-y$ then $|f(x+cx/2)-f(x) | \ge \sum_{n=0}^{cx/2} (cx-n) \ge x^2 c^2/8$ infinitely often, contradicting that $f(x)=o(x^2)$. Thus we proved $M(x)=o(x)$, the PNT.

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  • $\begingroup$ An interesting proof of an equivalent of PNT, thanks! There's a typo at the end of the long equation, the end result is $o(x^2)$. It's all visible though, no problems here. Thanks again! $\endgroup$ – anonymous Mar 27 '19 at 18:02

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