We now ask the question: to what extent can we use
Theorem 11.2 to obtain an estimate for the prime number theorem in arithmetic progressions with some degree of uniformity in the modulus?
To begin with, recall that the proof of
Theorem 10.9 included a step where we shifted the value of
\(T\) to avoid getting too close to the imaginary part of a zero of
\(\zeta\) in the critical strip. A similar step occurs in the proof of
Theorem 11.2, so we must make sure that this adjustment can be made uniformly in
\(N\text{;}\) this follows from the following lemma.
The other remaining issues lie in the shape of the main terms of
Theorem 11.2. One of these is the quantity
\(b(\chi)\) which had no analogue for
\(\zeta\text{.}\) To describe it, we go back to the product expansion:
\begin{equation*}
\frac{L'(s,\chi)}{L(s,\chi)} = -\frac{1}{2} \log (N/\pi) - \frac{1}{2}
\frac{\Gamma'(s/2 + a/2)}{\Gamma(s/2 + a/2)} + B(\chi) +
\sum_{\rho} \left( \frac{1}{s-\rho} + \frac{1}{\rho} \right),
\end{equation*}
where \(a=0\) if \(\chi\) is even and \(a=1\) if \(\chi\) is odd. The constant \(B\) here is not the same as \(b\) (it includes the contribution from the exponential part of the Hadamard product expansion), but no matter; we can eliminate it by comparing a given \(s\) with \(s=2\text{.}\) Hence
\begin{equation*}
\frac{L'(s,\chi)}{L(s,\chi)}
= O(1) - \frac{1}{2} \frac{\Gamma'(s/2 + a/2)}{\Gamma(s/2 + a/2)} +
\sum_\rho \left( \frac{1}{s-\rho} - \frac{1}{2-\rho} \right),
\end{equation*}
where the implied constant in \(O(1)\) is absolute (independent of \(N\)). If \(a=1\text{,}\) everything is holomorphic near \(s=0\text{;}\) if \(a=0\text{,}\) the two log derivatives both have a simple pole at \(s=0\text{,}\) and the residues match. We can thus equate the constant terms of the expansions around \(s=0\text{,}\) to obtain
\begin{equation*}
b(\chi) = O(1) - \sum_\rho \left( \frac{1}{\rho} + \frac{1}{2-\rho} \right).
\end{equation*}
As happened with \(\zeta\text{,}\) we can bound the contribution of the zeroes with \(|\Imag(\rho)| \geq 1\) by \(O(\log N)\text{.}\) The same goes for the term \(1/(2-\rho)\) when \(|\Imag(\rho)| \leq 1\text{.}\)
Putting this together, we now have
\begin{equation}
\psi(x, \chi) = -\sum_{\rho: |\Imag(\rho)| \lt T} \frac{x^\rho}{\rho}
+ \sum_{\rho: |\Imag(\rho)| \lt 1} \frac{1}{\rho} + O(x T^{-1} \log^2 (Nx)).\tag{11.2.1}
\end{equation}