# Deriving an asymptotic for $\pi(x)$ directly from $\log \zeta(s)$?

Denote by $$\pi(x)$$ the number of primes $$p\leq x$$. We generally give approximations for $$\pi(x)$$ by first approximating $$\psi(x) = \sum_{n\leq x} \Lambda(n)$$. Part of the reason is presumably that, if we apply Perron's formula directly, without going through $$\psi(x)$$, we end up with an expression for $$\pi(x)$$ in terms of an integral involving $$\log \zeta(s)$$. Here we see what the issue is: $$\log \zeta(s)$$ has a branch point at $$s=1$$, and that is unpleasant.

At the same time, it is not that unpleasant; you can do a little loop ("truncated Hankel contour") around $$s=1$$. Is there a standard reference where $$\pi(x)$$ is estimated in this way?

• You mean that $\log\zeta(s)$ has a branch point $s=1$, right? BTW I noticed that you (almost) never use a high-level tag like nt.number-theory. You should, in order for people to see your question! I added this tag now (as I did for many of your posts in the past). Commented Feb 28, 2021 at 23:57
• Koukoulopoulos discusses this approach on p. 51-56 of his book "The Distribution of Prime Numbers," before switching to psi(x) for the reason you mention. He writes that extracting the main term this way "is hard, though certainly not impossible as Riemann himself explained in his 1859 manuscript." I admit I have not read Riemann's original paper, but perhaps you might be interested in p. 4 onward: claymath.org/sites/default/files/ezeta.pdf Commented Mar 1, 2021 at 1:56
• Why $\log \zeta(s)$? That doesn't strike me as useful for this problem since Perron's formula leads to residues, and the residue of $\log \zeta(s)$ at $s=1$ is zero. Commented Mar 2, 2021 at 2:27

A paper of Landau where he uses an integral of $$Z(s) := \log \zeta(s)$$ against $$x^s/s^2$$ (not $$x^s/s$$) was published in 1908 and can be found here. The function $$Z(s)$$ is introduced on p. 753, a funky contour is drawn on p. 754, and lower down on that page he starts integrating $$Z(s)x^s/s^2$$ and he converts things into $$\pi(x)$$ in Section 7.

I found this information in Narkiewicz's book "The Development of Prime Number Theory". Section 6.2 is on Landau's approach to PNT and in part 5 (starting on p. 283) he mentions the above paper:

In the year 1908 Landau gave (Landau 1908d) two new proofs of the Prime Number Theorem in its simplest form, i.e. without giving any evaluation of the error term. The first is similar to the proof presented above but instead of dealing with the the function $$\zeta'(s)/\zeta(s)$$ Landau considers $$Z(s) = \log \zeta(s)$$. This necessitates a small modification of the integration path to take care of the essential singularity of the integrand at $$s=1$$ but the remainder of the argument is carried out along the lines of the preceding proof...

The reference Landau 1908d is the paper I link to above.

• Landau does a similar contour in his Lösung des Lehmer'schen Problems, where he basically has to consider $\zeta(s)^{1/2}$ around $s=1$. jstor.org/stable/2370180 Commented Mar 1, 2021 at 0:54
• Actually, why are arbitrary complex powers $\zeta(s)^z$ (a little) harder? Not much harder, mind you; see Chater II.5 of Tenenbaum. Commented Mar 1, 2021 at 1:25
• Actually, Landau has another 1908 paper (on numbers represented by $a^2+b^2$, without counting a divisor function like in the above cited Lehmer problem) that has this contour. archive.org/details/archivdermathem37unkngoog/page/n324/mode/… Commented Mar 6, 2021 at 17:46
• Can I ask a basic question? Although in these applications (Cauchy's Integral Theorem with branch cut) I feel I do understand the big picture, I still think there are one or two things I'm missing. Could you tell me in these examples, where precisely is he applying Cauchy's Integral Theorem and where exactly are the functions holomorphic? Is he, for example, applying it on $\mathbb C-\mathbb R^{\leq 0}$, where everything is holomorphic? Or is he including $\mathbb R^{\leq 0}$ (once or twice)? I struggle to say exactly where/what/how the conditions of Cauchy's Integral Theorem are satisfied. Commented Nov 20, 2023 at 17:55
• @tomos were you able to look at the paper? Working with the function $\log \zeta(s)$ outside of ${\rm Re}(s) \geq 1$ is delicate. Everything is not holomorphic away from the negative real axis since $\zeta(s)$ certainly has some zeros on the line ${\rm Re}(s) = 1/2$. We can't work with $\log \zeta(s)$ outside a set where $\zeta(s)$ is known to be nonvanishing. It is not proved that $\zeta(s)$ is nonvanishing on ${\rm Re}(s) > 1-\varepsilon$ with $\varepsilon > 0$. The best we could do is use a known zero-free region, which is always asymptotically approaching the line ${\rm Re}(s) = 1$. Commented Nov 20, 2023 at 19:00

Certainly Ayoub does it that way (starting on page 65), considering $$\int\log\zeta(s){x^s\over s}ds$$.

https://archive.org/details/introductiontoan0000ayou

His notes in the Introduction suggest he is following Landau, but I don't think Handbuch does it this way.

When $$T\le\sqrt x$$ and $$x=\frac12+\mathbb Z^+$$, it is possible to show using Perron's formula that

$$\pi(x)={1\over2\pi i}\int_{k-iT}^{k+iT}{x^s\over s}\log\zeta(s)\mathrm ds+\mathcal O\left(x\log x\over T\right)$$

After applying certain analytic properties of $$\zeta(s)$$ in the critical strip, the task turns into evaluating the following residue integral:

$${1\over2\pi i}\oint_{(1+)}{x^s\over s}\log{1\over s-1}\mathrm ds$$

where $$(1+)$$ represents any counterclockwise path containing $$s=1$$ but not $$s=0$$. A possible approach to handle the integral, inspired by Riemann (see chapter 1 of Edwards's 1974 classic Riemann's zeta function), is to introduce a new parameter $$r$$ to the problem:

$$f(r)={1\over2\pi i}\oint_{(r+)}{x^s\over s}\log{1\over s/r-1}\mathrm ds$$

Taking derivatives on both side gives $$f'(r)=x^r/r$$, and integrating back we get

$$f(r)=\operatorname{li}(x^r)+\mathcal O(1)$$

where $$\operatorname{li}(x)$$ is the logarithmic integral function. The reason why there is an O-term is because the sign of $$\Im(r)$$ will introduce certain $$i\pi$$ deviation to the final result. A full derivation of the PNT using this idea can be found here.