Riemann Zeta Function Non-Vanishing on the Line $\mathrm{Re} \; z = 1$ The result quoted in the title is usually a stepping stone in the proof of the prime number theorem and I am familiar with the usual argument for this result. The other day my professor was telling me, however, that the prime number theorem actually implies this result too. He suggested looking at the limit:
$$\lim_{\sigma \to 1} (\sigma-1)\int_1^{\infty}\frac{\psi(x)-x}{x^{\sigma+i\tau + 1}} \mathrm{d}x$$
Apparently by assuming the prime number theorem one can show that this limit is zero and from there deduce that $\zeta(1+i\tau) \neq 0$, but I am having difficulty showing either of these. Can someone help fill in the missing details please?
 A: On the one hand, we have
$$\begin{align}
\int_1^\infty \frac{\psi(x)}{x^{s+1}}\, dx
&= \int_1^\infty x^{-s-1} \sum_{p,k} \log p \cdot \chi_{[p^k,\infty)}(x)\,dx\\
&= \sum_{p,k} \log p \int_{p^k}^\infty \frac{dx}{x^{s+1}}\\
&= \sum_{p,k} \frac{\log p}{s\cdot p^{sk}}\\
&= \sum_p \frac{\log p}{s}\sum_{k} p^{-sk}\\
&= \sum_p \frac{\log p}{s(p^s-1)}\\
&= -\frac{1}{s}\cdot \frac{\zeta'(s)}{\zeta(s)}
\end{align}$$
for $\operatorname{Re} s > 1$, and $\int_1^\infty x^{-s}\,dx = \frac{1}{s-1}$, so writing $\zeta(s) = (s-1)^{-1}\cdot h(s)$ with the known simple pole of $\zeta$ in $1$, we obtain
$$\begin{align}
\int_1^\infty \frac{\psi(x)-x}{x^{s+1}}\,dx &= -\left(\frac{1}{s}\cdot \frac{\zeta'(s)}{\zeta(s)} + \frac{1}{s-1}\right)\\
&= - \left(\frac{(s-1)^{-1}h'(s) - (s-1)^{-2}h(s)}{s\cdot(s-1)^{-1}h(s)} + \frac{1}{s-1}\right)\\
&= - \left(\frac{1}{s}\frac{h'(s)}{h(s)} - \frac{1}{s(s-1)} + \frac{1}{s-1}\right)\\
&= - \frac{1}{s}\left(\frac{h'(s)}{h(s)} + 1\right).\tag{1}
\end{align}$$
If we had $\zeta(1+i\tau) = 0$, then $\frac{h'}{h}$ would have a simple pole in $1 + i\tau$, and we would have
$$\lim_{\sigma\to 1} (\sigma-1)\int_1^\infty \frac{\psi(x)-x}{x^{\sigma+i\tau+1}}\,dx = -\mu \neq 0,$$
where $\mu$ is the multiplicity of the zero of $\zeta$ (or equivalently $h$) in $1+i\tau$.
But the prime number theorem implies that
$$\psi(x) - x \in O\left(\frac{x}{\log x}\right),\tag{2}$$
and therefore
$$\begin{align}
\left\lvert \int_1^\infty \frac{\psi(x)-x}{x^{s+1}}\,dx\right\rvert
&\leqslant \int_1^\infty \frac{\lvert\psi(x)-x\rvert}{x^{\sigma+1}}\,dx\\
&\leqslant K + C\cdot \int_e^\infty \frac{dx}{x^{\sigma}\log x}\,dx\\
&= K + C\cdot \int_1^\infty \frac{e^{-(\sigma-1)t}}{t}\,dt\\
&= K + C\cdot \int_{\sigma-1}^\infty \frac{e^{-t}}{t}\,dt\\
&\leqslant \tilde{K} + C\cdot \log \frac{1}{\sigma-1},
\end{align}$$
which implies
$$\lim_{\sigma\to 1}\quad (\sigma-1)\int_1^\infty \frac{\psi(x)-x}{x^{\sigma+i\tau+1}}\,dx = 0.$$
