I am told that the Riemann Hypothesis is equivalent to the condition: $\psi(x) = x + O(x^{1+o(1)})$, and asked to prove this in the forward direction. (Here $\psi(x)$ is the Chebyshev Function).

Given the context of my notes, I am aware that I am expected to do this using a contour integral.

I believe from a converse of Perron's Formula, we get that $\psi(x) = \int_{\sigma - i\infty}^{\sigma + i\infty}\frac{\zeta'(s)}{\zeta(s)}\cdot \frac{x^s}{s}ds$, provided $\sigma > 1$ where we write $s = \sigma + it$ for a general complex number.

Given this, and the knowledge that the Riemann Hypothesis tells us where the roots of $\zeta(s)$ and thus the poles of $\frac{\zeta'(s)}{\zeta(s)}$ are, I expect that we are supposed to take a contour integral extending to the right of the vertical line $\left[\sigma_0 - in, \sigma_0 + in \right]$, that stops before the line $\sigma = \frac{1}{2}$. Thus, by the Residue Theorem, we conclude that the contour integral evaluates to $0$ for all $n \in \mathbb N$, and this allows us to find the desired form of $\psi(x)$.

However, where I am stuck now is evaluating the integral over the other parts of the contour.

For example suppose we have the contour $C_n = [\frac{3}{2} - in, \frac{3}{2} + in] \cup [\frac{3}{2} + in, \frac{3}{4} + in] \cup [\frac{3}{4} + in, \frac{3}{4} - in] \cup [\frac{3}{4} - in, \frac{3}{2} - in]$, and label each straight line $C_n^1, C_n^2, C_n^3, C_n^4$ in order, then how can I compute the integral over $C_n^2$, say?

Is there a nice form for $\frac{\zeta'(s)}{\zeta(s)}$ in this range that may make the integration any nicer?

Perhaps am I supposed to take a rectangular contour that gets thinner as it gets taller? If I were to do that, would I be able to justify bounding the integral by some constant, which may be absorbed by the error term.

I am quite confused by this question and would appreciate any help you may be able to offer, thank you.

  • $\begingroup$ Is my answer clear to you $\endgroup$
    – reuns
    Feb 26, 2019 at 14:58

1 Answer 1


Let $\psi(x) =\sum_{n \le x} \Lambda(n),\psi_1(x) = \int_1^x \psi(y)dy$.

$\frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1}$ is analytic for $\Re(s) > a$ iff for every $\sigma > a$, $ \psi_1(x) =\frac{x^2}{2}+ O(x^{\sigma+1})$.

One direction is obvious : if $\psi_1(x) - \frac{x^2}{2} = O(x^\sigma)$ then $s(s+1) \int_1^\infty (\psi_1(x)-\frac{x^2}{2})x^{-s-2}dx= -\frac{\zeta'(s)}{\zeta(s)}-\frac{s(s+1)/2}{s-1}$ is analytic for $\Re(s) >\sigma$.

The other direction needs a lot of estimates specific to $\zeta(s)$, in particular that the density of zeros implies $\frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1}=C+\sum_\rho \frac{1}{s-\rho}+\frac{1}{\rho} = O(\Im(s)^\epsilon)$ for $\Re(s) \ge \sigma$ which implies the absolute convergence of

$$\psi_1(x)-\frac{x^2}{2} = \frac{-1}{2i\pi}\int_{\sigma - i\infty}^{\sigma + i\infty}\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s+1}}{s(s+1)}ds= O(x^{\sigma+1})$$

To translate to $\psi(x)$ you'll look at $\sum_{n \le x} \Lambda(n) (1-\frac{\log n}{\log x}) = x-\frac{1}{2i\pi}\int_{\sigma - i\infty}^{\sigma + i\infty}\frac{\zeta'(s)}{\zeta(s)}\frac{x^{s}}{s^2}ds= x+O(x^{\sigma})$

Applying the residue theorem to obtain the explicit formula for $\psi_1$ as a series over the poles of $\zeta'/\zeta$ needs even more estimates, crossing the critical strip, then using the functional equation to evaluate the $\int_{-\infty+iT}^{-1+iT} \frac{\zeta'(s)}{\zeta(s)} \frac{x^s}{s(s+1)}ds$ part, for the explicit formula for $\psi$ there is the additional problem of the convergence as $T \to \infty$.


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