# Statement Equivalent to the Riemann Hypothesis

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.

• Is my answer clear to you – reuns Feb 26 at 14:58

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 $$T \to \infty$$.