In this article we show that prime number theorem implies nonvanishing of $\zeta(s)$ on the line $\sigma = 1$ and the argument we present here follows closely the approach taken in Ingham’s book The Distribution of Prime Numbers. The key tool is the formula
\[
-\frac{\zeta’}{\zeta}(s) = s \int_{1}^{\infty} \frac{\psi(x)}{x^{s + 1}}\, dx,
\] which holds for $\sigma > 1$. This formula follows from the following general fact about Dirichlet series: If $\alpha(s) = \sum_{n= 1}^{\infty} a_n n^{-s}$ is a Dirichlet series with abscissa of convergence $\sigma_c$ and $A(s) = \sum_{n \leq x} a_n$, then
\[
\alpha(s) = s \int_{1}^{\infty} \frac{A(x)}{x^{s + 1}} \, dx
\] for $\sigma > \max(\sigma_c, 0)$. Since
\[
\int_{1}^{\infty} \frac{dx}{x^s} = \frac{1}{s-1}
\] for $\sigma > 1$ we obtain
\[
\int_{1}^{\infty} \frac{\psi(x) -x}{x^{s + 1}} \, dx = -\frac{1}{s} \frac{\zeta’}{\zeta}(s)- \frac{1}{s-1} \tag{1}
\] for $\sigma > 1$. It is natural to consider the above integral, which we call $\phi(s)$ for the rest of our discussion, since we want to use $\psi(x) = x + o(x)$ to glean information about zeros of $\zeta(s)$. Let $\varepsilon > 0$ be fixed and take $M > 1$ be such that $|\psi(x)- x| \leq \varepsilon x$ for $x \geq M$. It now follows that
\[
|\phi(s) | \leq \int_{1}^{\infty} \frac{|\psi(x)- x|}{x^{\sigma + 1}} \, dx \leq \int_{1}^{M} \frac{|\psi(x)-x|}{x^2} \, dx + \varepsilon \int_{M}^{\infty} \frac{dx}{x^{\sigma}} \leq I_M + \frac{\varepsilon}{\sigma- 1}.
\] Now let $\delta > 1$ be such that $(\delta- 1) I_M < \varepsilon$. Then for $1 < \sigma < \delta$ we have $|\phi(s)(\sigma- 1)| \leq 2 \varepsilon$. This shows that $\phi(s)(\sigma- 1) \to 0$ as $\sigma \to 1^{+}$.
Suppose for the sake of contradiction that $\zeta(1 + i t_0) = 0$ for some $t_0 \neq 0$. Then $\frac{\zeta’}{\zeta}(s)$ has a simple pole at $s = s_0 = 1 + i t_0$ and so we have
\[
\lim_{\scriptstyle \sigma \to 1^+ \atop \scriptstyle t = t_0} (\sigma- 1) \left( -\frac{1}{s} \frac{\zeta’}{\zeta}(s)- \frac{1}{s-1} \right) = -\frac{1}{ s_0} \underset{s = s_0}{\operatorname{Res}} \frac{\zeta’}{\zeta}(s) \neq 0.
\] But this is a contradiction due to (1).