Chapter 11 Error bounds for primes in arithmetic progressions
Section 11.1 Uniformity in the explicit formula
Put
\begin{equation*}
\psi(x, \chi) = \sum_{n \leq x} \chi(n) \Lambda(n).
\end{equation*}
Then as long as \(x\) is an integer and \(T \leq x\text{,}\) the explicit formula for \(\psi(x, \chi)\) has the form
\begin{equation*}
\psi(x,\chi) = -\sum_{\rho: |\Imag(\rho)| \lt T} \frac{x^\rho}{\rho} - b(\chi)
+ O(xT^{-1} \log^2 (Nx)),
\end{equation*}
where \(b(\chi)\) is defined by
\begin{equation*}
\frac{L'(s,\chi)}{L(s,\chi)} = \begin{cases} s^{-1} + b + O(s)
\amp \chi(-1) = 1 \\
b + O(s) \amp \chi(-1) = -1.
\end{cases}
\end{equation*}
The proof is as for von Mangoldt's formula (Theorem 10.1); I will not redo it here. The point is that everything is uniform in \(N\) except the constant \(b(\chi)\text{.}\)
So to get uniform estimates, one must control \(b(\chi)\text{,}\) which we can do by expressing it in terms of zeroes of \(L(s,\chi)\text{.}\) For this 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*}
If \(a=1\text{,}\) everything is regular 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 \(|\Imag(\rho)| \leq 1\text{.}\)
This means that we 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)).
\end{equation*}
Section 11.2 Controlling the exceptional zeroes
The question that now remains is: how big is the contribution to \(\psi(x,\chi)\) from the sum of \(1/\rho\) over critical zeroes \(L(s,\chi)\) in the range \(|\Imag(\rho)| \lt 1\text{?}\)
To answer this, we need a uniform zero-free region near the real axis. Here's what happens when you try to produce this.
Theorem 11.2.
There is a constant \(c>0\) such that there is no zero of \(L(s,\chi)\) with
\begin{equation*}
|\Imag(\rho)| \lt 1, \qquad \Real(\rho) > 1 - \frac{c}{\log N}
\end{equation*}
except possibly if \(\chi\) is real. In that case, there may be one such zero (counting multiplicity), which is necessarily real.The Riemann Hypothesis for \(L(s,\chi)\) implies that no such oddball zero exists, but to prove unconditional results we must allow for it. In particular, it helps to have a label for the hypothetical zero: we call it an exceptional zero, or Siegel zero, of \(L(s,\chi)\text{.}\) (Note that the criterion for being a Siegel zero is not absolute; it depends on the choice of the cutoff parameter \(c\text{.}\))
If we exclude any exceptional zero \(\beta\) and also its mirror image \(1-\beta\text{,}\) then the sum of \(1/\rho\) over the remaining zeroes in the range \(|\Imag(\rho)| \lt 1\) is \(O((\log N)^2)\text{,}\) since there are \(O(\log N)\) such zeroes and each term contributes \(O(\log N)\) to the sum. We then have
\begin{equation*}
\psi(x,\chi) = -\sum_{\rho: |\Imag(\rho)| \lt T}^{\sim} \frac{x^\rho}{\rho}
- \frac{x^{\beta}}{\beta} - \frac{x^{1-\beta}-1}{1-\beta} +
O(x T^{-1} \log^2 (Nx)),
\end{equation*}
where the tilde means don't count \(\beta\) and \(1-\beta\) as zeroes. The term \((x^{1-\beta}-1)/(1-\beta)\) is \(O(x^c \log x)\text{,}\) but controlling the term \(x^\beta/\beta\) requires preventing the exceptional zero from getting too close to \(1\text{.}\) Here's one way to do that.
Theorem 11.3. Siegel.
For any \(\epsilon > 0\text{,}\) there exists a constant \(c = c(\epsilon)\) with the following property: for any real primitive Dirichlet character \(\chi\) of level \(N\text{,}\) every real zero \(\beta\) of \(L(s,\chi)\) satisfies
\begin{equation*}
\beta \leq 1 - c N^{-\epsilon}.
\end{equation*}
Proof.
This is enough to get the following form of the prime number theorem in arithmetic progressions with error term, called the Siegel–Walfisz theorem.
Definition 11.4.
For \(N\) a positive integer and \(a\) an integer coprime to \(N\text{,}\) put
\begin{align*}
\pi(x, N, a) \amp:= \sum_{p \leq x, p \equiv a (N)} 1\\
\psi(x, N, a) \amp:= \sum_{n \leq x, n \equiv a (N)} \Lambda(n).
\end{align*}
Theorem 11.5. Siegel–Walfisz.
Fix \(\epsilon > 0\) and \(A>0\text{.}\) Then
\begin{align*}
\pi(x, N, a) \amp= \frac{\li(x)}{\phi(N)} + O( x \log^{-A} x)\\
\psi(x, N, a) \amp= \frac{x}{\phi(N)} + O( x \log^{-A} x)
\end{align*}
where the implied constants are explicit and depend only on \(A\) and \(\epsilon\text{,}\) not on \(N\text{.}\)One can improve this error bound a bit unconditionally, but not much. On the other hand, under the Generalized Riemann Hypothesis (i.e., the critical zeroes of every \(L(s,\chi)\) lie on the line \(\Real(s) = 1/2\)), you get errors of \(O(x^{1/2 +\epsilon})\text{.}\)
With the error bound as is, the theorem only has content for \(N\) no bigger than a fixed power of \(\log x\text{.}\) We can prove much better results, say for \(N\) up to \(x^C\) for a fixed \(c \lt 1/2\text{,}\) if you are willing to accept an aggregate statement about the error bounds. More on this when we discuss the Bombieri–Vinogradov theorem (Theorem 19.1).
Section 11.3 Why the exceptional zero?
One can see where the possibility of an exceptional zero arises by beginning to imitate for \(L(s,\chi)\) the proof we gave of the zero-free region for \(\zeta\text{.}\) We have
\begin{equation*}
-\frac{L'(s,\chi)}{L(s,\chi)} = \sum_{n=1}^\infty \Lambda(n)
n^{-\Real(s)} \chi(n) e^{-i \Imag(s) \log n},
\end{equation*}
and using the trigonometric inequality, we have for \(\sigma > 1\)
\begin{equation*}
-3 \frac{L'(\sigma,\chi_0)}{L(\sigma,\chi_0)}
- 4 \Real \frac{L'(\sigma + it, \chi)}{L(\sigma + it,\chi)}
- \Real \frac{L'(\sigma+2it,\chi^2)}{L(\sigma+2it,\chi)} \geq 0.
\end{equation*}
Here \(\chi_0\) is the principal character of the same level as \(\chi\text{.}\)
The argument to exclude zeroes close to the edge of the critical strip proceeds as before if \(\Imag(\rho)\) is bounded away from 0, say \(|\Imag(\rho)| > c/(\log N)\text{.}\) For \(\chi\) nonreal, you do better: \(\chi^2\) is nonprincipal and so \(L(\sigma+2it, \chi^2)\) stays bounded as \(\sigma \to 1^+\text{.}\) So you get an inequality of the form
\begin{equation*}
\frac{4}{\sigma - \Imag(\rho)} \lt \frac{3}{\sigma - 1} + O(\log N
+ \log (|\Imag(\rho)| + 1)),
\end{equation*}
and that gives you a zero-free region all the way down to the real line.
Unfortunately, if \(\chi\) is real, then \(L(\sigma+2it, \chi^2)\) blows up at \(\chi = 1\text{,}\) and our present methods cannot exclude a single zero very close to 1: we only end up with an inequality of the form
\begin{equation*}
\frac{4}{\sigma - \Imag(\rho)} \lt \frac{3}{\sigma - 1} +
\Real \left( \frac{1}{\sigma-1+2i\Imag(\rho)} \right) +
O(\log N
+ \log (|\Imag(\rho)| + 1)).
\end{equation*}
However, we can exclude two such zeroes \(\rho_1, \rho_2\text{,}\) by writing
\begin{equation*}
-\frac{L'(s,\chi)}{L(s,\chi)} \lt - \frac{1}{\sigma - \rho_1}
- \frac{1}{\sigma - \rho_2}
+ O(\log N
+ \log (|\Imag(\rho)| + 1))
\end{equation*}
and so on.
