As noted in
Remark 10.3, by
(10.1.2) and
Lemma 10.2, we can interpret
\(R(x,T)\) as the limit, as
\(U\) runs over all odd positive integers, of the integral of
\(- \frac{\zeta'(s)}{\zeta(s)} \frac{x^s}{s} ds\) over the contour
(10.2.1). Note that we are free to choose any
\(c>1\text{;}\) it will keep the notation simple to take
\(c = 1 + (\log x)^{-1}\text{.}\) Note that then
\(x^c = ex = O(x)\text{.}\)
Moreover, by
Lemma 10.4, if we shift
\(T\) by a bounded amount, we may ensure that the difference between
\(T\) and the imaginary part of any zero of
\(\zeta\) is at least some constant times
\((\log T)^{-1}\text{,}\) at the price that the sum over zeroes may change by the presence or absence of
\(O(\log T)\) terms each of size
\(O(x T^{-1} \log T)\text{.}\) This is subsumed by our proposed error bound; we are thus free to assume hereafter that such a shift has been made.
It thus remains to bound the contribution of each segment of the contour
(10.2.1). We analyze this contour by separating it into three terms:
the infinite segments \(c-i\infty \to c-iT\) and \(c+iT \to c+i\infty\text{;}\)
the horizontal segments \(c-iT \to -1-iT\) and \(-1+iT \to c+iT\text{;}\)
the remaining segments \(-1-iT \to -U-iT \to -U+iT \to -1+iT\text{.}\)
We treat these three contributions in reverse order. For the remaining segments
\(-1-iT \to -U-iT \to -U+iT \to -1+iT\text{,}\) we apply
Lemma 10.8 to bound the horizontal contributions by
\begin{equation*}
O \left( \int_1^\infty (\log s + \log T) x^{-s}/T \,ds \right)
\leq
O\left( \frac{1}{T x \log^2 x} + \frac{\log T}{T x \log x} \right),
\end{equation*}
which is subsumed by our error bound. We bound the vertical contribution in the limit as \(U \to 0\) by
\begin{equation*}
O \left( \frac{T \log U}{U x^U} \right),
\end{equation*}
which tends to zero.
For the horizontal segments
\(c-iT \to -1-iT\) and
\(-1+iT \to c+iT\text{,}\) we combine
Lemma 10.4 and
Lemma 10.6 to see that for
\(s\) on these segments,
\begin{equation*}
\frac{\zeta'(s)}{\zeta(s)} = O(\log^2 T).
\end{equation*}
We can thus bound these contributions by
\begin{equation*}
O \left( \log^2 T \int_{-1}^c |x^s/s|\,ds \right)
\leq O\left( \frac{x \log^2 T}{T \log x} \right),
\end{equation*}
which is subsumed by our proposed error bound.
This leaves the infinite segments
\(c-i\infty \to c-iT\) and
\(c+iT \to c+i\infty\text{.}\) By
Lemma 10.5 applied with
\(y = x/n\text{,}\) the contour integral is dominated by
\begin{equation*}
\sum_{n=1, n \neq x}^\infty
\Lambda(n) \left( \frac{x}{n} \right)^c \min\{1, T^{-1} |\log (n/x)|^{-1}\}
+ c T^{-1} \Lambda(x).
\end{equation*}
To estimate the summand, it helps to distinguish between terms where \(\log (n/x)\) is close to zero, and those where it is bounded away from zero. For the latter, the quantity \(|\log (n/x)|^{-1}\) is bounded above; so the summands with, say, \(|n/x - 1| \geq 1/4\text{,}\) are dominated by
\begin{equation*}
O\left( x T^{-1} \left( -\frac{\zeta'(c)}{\zeta(c)} \right) \right)
= O(x T^{-1} \log x).
\end{equation*}
For the former, consider values \(n\) with \(3/4 \lt n/x \lt 1\) (the values with \(1 \lt n/x \lt 5/4\) are treated similarly, and \(n/x = 1\) contributes \(O(\log x)\)). Let \(x'\) be the largest prime power strictly less than \(x\text{;}\) then the summands \(x' \lt n \lt x\) all vanish. In particular, it is harmless to assume \(x' > 3x/4\text{,}\) since otherwise the summands we want to bound all vanish.
We now separately consider the summand \(n = x'\text{,}\) and all of the summands with \(3/4 \lt n \lt x'\text{.}\) The former contributes
\begin{equation*}
O\left( \log(x) \min\left\{1, \frac{x}{T(x-x')}\right\}\right).
\end{equation*}
For each term of the latter form, we can write \(n = x' - m\) with \(0 \lt m \lt x/4\text{,}\) and
\begin{equation*}
\log \frac{x}{n} \geq - \log \left(1 - \frac{m}{x'} \right)
\geq \frac{m}{x'},
\end{equation*}
so these terms contribute
\begin{equation*}
O \left( x T^{-1} (\log x)^2 \right).
\end{equation*}
All of this is subsumed by the proposed error bound, completing the proof.
