# Why is $\zeta(1+it) \neq 0$ equivalent to the prime number theorem?

Reading through Titchmarsh's book on the Riemann zeta function, chapter 3 discusses the Prime Number Theorem. One way to prove this result is to check the zeta function has no zeros on the line $z = 1 + it,$

$$\zeta(1 + it) \neq 0$$

Indeed the book has $3$ or $4$ proofs of this result. Actually connecting it to the prime number theorem is another matter. One version of the Prime Number Theoriem is:

$$\sum_{n \leq x} \Lambda (n) = x + o(x)$$

involving the van Mangoldt function, but why is this equivalent to the non-vanishing of the Riemann zeta function. I think you can start from Perron's theorem

$$\frac{1}{2\pi i}\int_{1-iT}^{1+iT} \frac{\zeta'(w)}{\zeta(w)} \, \frac{x^w}{w}dw = \sum_{n \leq x} \Lambda (n)$$

and then I don't know how to proceed.

The precise statement of Perron's formula is that for any $c>1$ $$\sum_{n\leq x} \Lambda(n)=\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{\zeta'(w)}{\zeta(w)}\frac{x^w}{w}dw.$$ The reason that the lack of zeros on the $1$-line is equivalent to the prime number theorem comes from evaluating the integral by contour integration. Notice that the integrand $$\frac{\zeta'(w)}{\zeta(w)}\frac{x^w}{w}$$ has simple poles at precisely the zeros and poles of the Riemann zeta function, and so by extending a large contour around the entire plane left of the line $x=c$ we will pick up $$\text{Res}_{w=\rho} \frac{\zeta'(w)}{\zeta(w)}\frac{x^w}{w}=-\frac{x^\rho}{\rho}$$ at every zero of the zeta function, as well as the main term $$\text{Res}_{w=1} \frac{\zeta'(w)}{\zeta(w)}\frac{x^w}{w}=x.$$ Thus, as long as we can bound the edges of this contour and show that they go to zero in the limit (which requires some bounds on the growth of the analytic continuation of the zeta function) we obtain the explicit formula $$\sum_{n\leq x} \Lambda(n) =x -\sum_{\rho:\ \zeta(\rho)=0}\frac{x^\rho}{\rho}-\frac{\zeta'(0)}{\zeta(0)},$$ where this sum is over all zeros of the Riemann zeta function. From this equation we can see where the equivalence comes from: a single term $$\frac{x^\rho}{\rho}$$ will be $o(x)$ if $\text{Re}(\rho)<1$, and it will be $\Omega(x)$ if $\text{Re}(\rho)=1$. Some careful analysis of the zeros allows this intuitive idea to go through and carry over to the infinite series, where we might have to worry about cancellation or divergence of the series.