Given what we now know, this is a straightforward adaptation of our proof of the prime number theorem (
Chapter 1). We thus skip over some details which are not much changed, in order to focus on the new ingredients.
For \(\chi\) a Dirichlet character of level \(N\text{,}\) define
\begin{equation*}
\vartheta_\chi(x) := \sum_{p \leq x} \chi(p) \log p\text{.}
\end{equation*}
Given a choice of \(m\) coprime to \(N\text{,}\) put
\begin{align*}
\vartheta_m(x) \amp := \sum_\chi \overline{\chi(m)} \vartheta_\chi(x)\\
\amp = \phi(N) \sum_{p \leq x\colon p \equiv m\,(N)} \log p
\end{align*}
where the last equality follows from
Theorem 4.10. As in
Lemma 1.7, the desired result is equivalent to
\(\vartheta_m(x) \sim x\text{.}\)
As in
Section 1.3, it will suffice to show convergence of the improper integral
\begin{equation}
\int_1^\infty \frac{\vartheta_m(x) - x}{x^2}\,dx\text{.}\tag{4.4.1}
\end{equation}
We will again use the Tauberian argument (
Section 1.4) to reduce this to an application of
Theorem 1.8. Keeping in mind
(4.3.1), define the function
\begin{align}
\Phi(s) \amp:= \sum_\chi - \overline{\chi(m)}\frac{L'(s,\chi)}{L(s,\chi)}\notag\\
\amp= \sum_\chi \sum_p \sum_{n=1}^\infty \overline{\chi(m)} \chi(p^n) (\log p) p^{-ns}\tag{4.4.2}
\end{align}
in which the sum runs over the Dirichlet characters of level
\(N\text{.}\) For
\(\chi\) principal,
\(L(s,\chi)\) differs from
\(\zeta(s)\) by finitely many Euler factors, none of which has a zero or pole at
\(s=1\text{;}\) hence
\(-L'(s,\chi)/L(s,\chi)\) again has a simple pole at
\(s=1\) with residue
\(1\) and no other poles in
\(\Real(s) \geq 1\text{.}\) Meanwhile, for
\(\chi\) nonprincipal,
\(L(s,\chi)\) is actually holomorphic for
\(\Real(s) \geq 1\) (by
Theorem 3.5) with no zeroes in this region (by
Theorem 3.8,
Theorem 3.10, and
Theorem 3.11); hence
\(-L'(s,\chi)/L(s,\chi)\) is holomorphic for
\(\Real(s) \geq 1\text{.}\)
We deduce from the above that
\(\Phi(s+1) - \frac{1}{s} \) extends holomorphically at
\(s=0\text{.}\) Now applying
Theorem 1.8 as in
Section 1.4, we deduce that the value of this function at
\(s=0\) computes
\(\int_0^\infty (\vartheta_m(e^t)e^{-t} - 1)\,dt\) plus an integral which converges absolutely (corresponding to the summands with
\(n>1\) in
(4.4.2)). Substituting
\(x = e^t\) yields
(4.4.1) as required.