= Solution
Put
$$
A(x)=\sum_{n\leq x}\Lambda(n)n^{iu}
=\frac{x^{1+iu}}{1+iu}+O\left(x^{1/2}(\log x)^2\right).
$$
For $\Re s>1$, <partial summation> gives
$$
-\frac{\zeta'(s-iu)}{\zeta(s-iu)}
=s\int_1^\infty A(x)x^{-s-1}\,dx
=\frac{s}{(1+iu)(s-1-iu)}
+s\int_1^\infty O\left(x^{1/2}(\log x)^2\right)x^{-s-1}\,dx.
$$
The last integral is holomorphic for $\Re s>1/2$. Thus the logarithmic derivative on the left continues meromorphically to that half-plane with no pole except $s=1+iu$.
A zero $\rho$ of $\zeta$ with $\Re\rho>1/2$ would make $-\zeta'(s-iu)/\zeta(s-iu)$ singular at $s=\rho+iu$, a contradiction unless $\rho=1$, which is a pole rather than a zero. Therefore no nontrivial zero lies to the right of the <critical line>. The <Functional equation of the Riemann zeta function> reflects zeros across that line, so none lies to its left either. Every nontrivial zero lies on the critical line, proving the <Riemann hypothesis>. This is the <Twisted Von Mangoldt estimate implying the Riemann hypothesis>.
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