I am asking this question here, since my comment on this post on Math Overflow got no responses.
I need an explicit (computable) bound on the $O(1)$ term in the following asymptotic formula:
$$\sum_{\substack{p\le x \\p\equiv a\;(\text{mod} \; q)}}\frac{\log(p)}{p}=\frac{\log(x)}{\phi(q)}+O(1),$$
where $\gcd(a,q)=1$. Specifically for the cases $q=4$ and $a\in\{1,2\}$. However, using the prime number theorem for arithmetic progressions, I have not been able to get the error better than $O(\log(\log(n)))$. This is what I have done so far:
let $\pi(x;q,a)$ denote the number of primes less than or equal to $x$ that are congruent to $a$ modulo $q$. By Abel's summation formula, we have
\begin{equation}
\sum_{\substack{p\le x \\ p\equiv a (\text{mod}\; q)}}\frac{\log(p)}{p}=\frac{\log(x)}{x}\pi(x;q,a)+\int_{2}^{x}\frac{\log(t)-1}{t^2}\pi(t;q,a)dt
\hspace{8mm}(1)\end{equation}
Let $\theta(x;q,a):=\sum_{\substack{p\le x \\ p\equiv a (\text{mod}\; q)}}\log(p)$. Then
\begin{align*}
\pi(x;q,a)=\frac{\theta(x;q,a)}{\log(x)}+\int_{2}^{x}\frac{\theta(t;q,a)}{t\log^2(t)}dt
\end{align*}
by the result on page 142 here. Therefore:
\begin{align*}
\frac{\log(x)}{x}\pi(x;q,a)=\frac{\theta(x;q,a)}{x}+\frac{\log(x)}{x}\int_{2}^{x}\frac{\theta(t;q,a)}{t\log^2(t)}dt.
\end{align*}
The term $\tfrac{\theta(x;q,a)}{x}=\frac{1}{\phi(q)}+o(1)$, while
$$\frac{\log(x)}{x}\int_{2}^{x}\frac{\theta(t;q,a)}{t\log^2(t)}dt=\frac{\log(x)}{x}O(\frac{x}{\log^2(x)})=O(\frac{1}{\log(x)})=o(1).$$
Thus, it remains to consider the integral in $(1)$. We have
\begin{align*}
\int_{2}^{x}\frac{\log(t)-1}{t^2}&\pi(t;q,a)dt=\int_{2}^{x}\frac{\log(t)}{t^2}\pi(t;q,a)dt-\int_{2}^{x}\frac{\pi(t;q,a)}{t^2}dt. \hspace{6mm} (2)
\end{align*}
Let $\varepsilon$ be defined by $\tfrac{\log(t)\pi(t;q,a)}{t}=\frac{1}{\phi(q)}+\varepsilon(t)$. Then the first integral in the right hand side of $(2)$ can be written as
\begin{align*}
\int_{2}^{x}\frac{\log(t)\pi(t;q,a)}{t^2}dt=&\int_{2}^{x}\frac{1}{t}\cdot \frac{\log(t)\pi(t;q,a)}{t}dt=\int_{2}^{x}\frac{1}{t}\cdot (\frac{1}{\phi(q)}+\varepsilon(t))dt\\[1mm]
=&\frac{\log(x/2)}{\phi(q)}+\int_{2}^x\frac{\varepsilon(t)}{t}dt.
\end{align*}
The second integral in the right hand side of $(2)$ can be written as
\begin{align*}
\int_{2}^{x}\frac{\pi(t;q,a)}{t^2}dt=&\int_{2}^{x}\frac{1}{t\log(t)}\cdot \frac{\log(t)\pi(t;q,a)}{t}dt\\
=& \int_{2}^{x}\frac{1}{t\log(t)}\cdot (\frac{1}{\phi(q)}+\varepsilon(t))dt\\
=&\frac{\log(\log(x))-\log(\log(2))}{\phi(q)}+\int_{2}^x\frac{\varepsilon(t)}{t\log(t)}dt.
\end{align*}
Thus, we have:
\begin{align*}
\sum_{\substack{p\le x \\ p\equiv q (\text{mod}\; q)}}\frac{\log(p)}{p}=\frac{1}{\phi(q)}+\frac{\log(x/2)}{\phi(q)}-\frac{\log(\tfrac{\log(x)}{\log(2)})}{\phi(q)}+\int_{2}^x\frac{\varepsilon(t)}{t}dt-\int_{2}^x\frac{\varepsilon(t)}{t\log(t)}dt+o(1)
\end{align*}
Thanks to the result on page 6 here, if $1\le q\le 1200$ is an integer, and $a$ is coprime to $q$, then
$$\frac{x}{\phi(q)\log(x)}<\pi(x;q,a)<\frac{x}{\phi(q)\log(x)}(1+\frac{5}{3\log(x)})$$
for all $x\ge 50q^2$. Consequently,
$$0<\varepsilon(x)<\frac{5}{2\phi(q)\log(x)}$$
for $x \ge 50q^2$. For our purposes we have $q=4$, so the bound holds for $x\ge 800$. Assuming $x\ge 800$, we then have
\begin{align*}
\int_{2}^x\frac{\varepsilon(t)}{t}dt=\int_{2}^{800}\frac{\varepsilon(t)}{t}dt+\int_{800}^x\frac{\varepsilon(t)}{t}dt,
\end{align*}
where
\begin{align*}
0<\int_{800}^x\frac{\varepsilon(t)}{t}dt<\frac{5}{2\phi(q)}\int_{800}^x\frac{1}{t\log(t)}dt=\tfrac{5}{2\phi(q)}\log(\log(x))-\tfrac{5}{2\phi(q)}\log(\log(800)).
\end{align*}
Moreover,
\begin{align*}
\int_{2}^x\frac{\varepsilon(t)}{t\log(t)}dt=&\int_{2}^{800}\frac{\varepsilon(t)}{t\log(t)}dt+\int_{800}^x\frac{\varepsilon(t)}{t\log(t)}dt,
\end{align*}
where
\begin{align*}
0<\int_{800}^x\frac{\varepsilon(t)dt}{t\log(t)}<\frac{5}{2\phi(q)}\int_{800}^x\frac{dt}{t\log^2(t)}=\frac{5}{1600\phi(q)\log(800)}-\frac{5}{2\phi(q)\log(x)}.
\end{align*}
As a conclusion, I get that
\begin{align*}
\sum_{\substack{p\le x \\ p\equiv a (\text{mod}\; q)}}\frac{\log(p)}{p}=const.-\frac{\log(\frac{\log(x)}{\log(2)})}{\phi(q)}+\frac{\log(x)}{\phi(q)}+\int_{800}^x\frac{\varepsilon(t)}{t}dt-\int_{800}^x\frac{\varepsilon(t)}{t\log(t)}dt+o(1).
\end{align*}
However, for the error term(the stuff between $conts.$ and $o(1)$, call it $r(x)$), the best I get is
\begin{align*}\frac{5}{2\phi(q)\log(x)}-\frac{5}{1600\phi(q)\log(800)}-\frac{\log(\frac{\log(x)}{\log(2)})}{\phi(q)}<r(x)<\frac{2\log(\log(x))}{3\phi(q)}+\frac{\log(\log(2))}{\phi(1)},\end{align*}
which I cannot see is better than $O(\log(\log(x)))$.
It have tried using the Siegel–Walfisz theorem, but so far been unsuccessful. I know the constant in the theorem is not effectively computable, so I don't see how I can get the bound explicit using the theorem.
Best Answer
If $k=\mathcal{O}(\log^\alpha x)$, then by the Siegel–Walfisz theorem $$ \theta (x;k,\ell ) = \frac{x}{{\phi (k)}} + \mathcal{O}(xe^{ - c(\alpha )\sqrt {\log x} } ). $$ Thus, by partial summation, \begin{align*} \sum_{\substack{p\le x \\p\equiv \ell \,(\text{mod} \, k)}}\frac{\log p }{p} & = \frac{{\theta (x;k,\ell )}}{x} + \int_1^x {\frac{{\theta (t;k,\ell )}}{{t^2 }}dt} \\ & = \frac{{\theta (x;k,\ell )}}{x} + \frac{1}{{\phi (k)}}\int_1^x {\frac{{dt}}{t}} + \int_1^{ + \infty } {\frac{1}{{t^2 }}\left[ {\theta (t;k,\ell ) - \frac{t}{{\phi (k)}}} \right]dt} \\ &\quad\; - \int_x^{ + \infty } {\frac{1}{{t^2 }}\left[ {\theta (t;k,\ell ) - \frac{t}{{\phi (k)}}} \right]dt} \\ & = \frac{{\log x}}{{\phi (k)}} + \frac{1}{{\phi (k)}} + \int_1^{ + \infty } {\frac{1}{{t^2 }}\left[ {\theta (t;k,\ell ) - \frac{t}{{\phi (k)}}} \right]dt} \\ & \quad\; - \mathcal{O}(1)\int_x^{ + \infty } {\frac{1}{t}e^{ - c(\alpha )\sqrt {\log t} } dt} + \mathcal{O}(e^{ - c(\alpha )\sqrt {\log x} } ) \\ & = \frac{{\log x}}{{\phi (k)}} + \frac{1}{{\phi (k)}} + \int_1^{ + \infty } {\frac{1}{{t^2 }}\left[ {\theta (t;k,\ell ) - \frac{t}{{\phi (k)}}} \right]dt} + o(1). \end{align*} The error term may be improved from $o(1)$ to $$ \mathcal{O}(\sqrt {\log x} \,e^{ - c(\alpha )\sqrt {\log x} } )=\mathcal{O}(e^{ - d(\alpha )\sqrt {\log x} } ). $$