I have been reviewing primes in Arithmetic progression from Apostol's Introduction to analytic number theory book. It is given that:

$$ \pi_a(x) \sim \frac{x}{\log(x)} \cdot \frac{1}{\phi(k)} $$

Where $ \pi_a(x) = \sum_{n \leq x\\ n \cong a (mod \text{ } k)} 1 $.

However, I am interested in the error terms as well. Can someone please provide an asymptotic formula with the error terms as well?

Any help is highly appreciated.


1 Answer 1


Siegal-Walfisz theorem asserts that

$$ \psi(x;q,a)=\sum_{\substack{n\le x\\n\equiv a\pmod q}}\Lambda(n)={x\over\varphi(q)}+\mathcal O(xe^{-C_N\sqrt{\log x}}) $$

wherein $(a,q)=1$, $C_N$ is a constant only dependent on the real number $N$, and $q=q(x)\le(\log x)^N$, and the O constant is independent of $q$. The drawback of this theorem is that $C_N$ cannot be effectively computed.

If we set

$$ \Delta(x;q)=\max_{(a,q)=1}\sup_{y\le x}\left|\psi(x;q,a)-{x\over\varphi(q)}\right| $$

then Bombieri-Vinogradov's theorem states that

$$ \sum_{q\le Q}\Delta(x;q)\ll_A x(\log x)^{-A}+\sqrt xQ(\log xQ)^4 $$

For more details about Bombieri-Vinogradov, see this document.

  • $\begingroup$ What about the $O$ constant? For any $A$: $\psi(x;q,a)=\frac{x}{\varphi(q)}+O(\frac{x}{\log^A x})$ where the $O$ constant depends on $A$ and $q$. $\endgroup$
    – reuns
    Apr 11, 2021 at 12:20
  • $\begingroup$ @reuns Just edited $\endgroup$
    – TravorLZH
    Apr 11, 2021 at 12:27

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