# The use of limit in Titchmarsh's book “The theory of the Riemann zeta-function” in Theorem $3.13$

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,

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

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 $$n comes from letting the left-most side of the integration rectangle be at $$\Re(w)=-\infty$$ and so $$(x/n)^w\to 0$$ if and only if $$n as $$\Re(w)\to -\infty$$.

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

$$\sum\limits_{n

On the LHS the condition $$n is the condition from Lemma $$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

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 $$x\to\infty$$?

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. $$|\sum\limits_{n for any other half-integer $$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 $$|\sum\limits_{n for $$x>x(\epsilon)$$ is precisely what is needed to conclude that the limit as $$x$$ goes to $$\infty$$ of $$\sum\limits_{n 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

• Awesome, thank You, Conrad! – anonymous Mar 27 '19 at 18:04
• You are welcome – Conrad Mar 27 '19 at 19:03
• $A$ 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 $x,T$ – Conrad Jan 15 '20 at 20:25
• @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.... – Shree Feb 5 '20 at 23:29
• @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 – Conrad Feb 6 '20 at 0:41

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.

• 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! – anonymous Mar 27 '19 at 18:02