I wish to prove the following equality
$$\sum_{p \le x, p \equiv 3 \bmod 10} \frac{1}{p} = \frac{1}{4} \log\log(x)+A+O(\frac{1}{\log x})$$
For some constant A
Own work: Let $$A(x) = \sum_ {p \le x} \frac{\log p}{p} $$
and let: $a(n)= 1$, when $n$ is prime, and let $a(n)=0$ otherwise $$\sum_{p \le x} \frac{1}{p} = \sum_{n \leq x} \frac{a(n)}{n}$$ and $$A(x) = \sum_{n \leq x} \frac{a(n)}{n} \log n$$
If $f(t) = \frac{1}{\log t}$, and since $A(t) = 0$ for $t<2$
$$\sum_{p \le x} \frac{1}{p} = \sum_{n \leq x} \frac{A(n)}{\log x} + \int_{2}^{x} \frac{A(t)}{t\log^2 t} dt$$
I know that:
$$\sum_ {p \le x} \frac{\log p}{p} = \log x + O(1)$$
Which can be plugged in to obtain:
$$\sum_{p \le x} \frac{1}{p} = \frac{\log x + O(1)}{\log x} + \int_{2}^{x} \frac{\log t + R(t)}{t\log^2 t} dt$$
Which will be:
$$\sum_{p \le x} \frac{1}{p} = 1 + O(\frac{1}{\log x}) + \int_{2}^{x} \frac{1}{t\log t} dt +\int_{2}^{x} \frac{ R(t)}{t\log^2 t} dt $$
But I think, I think I didn't start correct. I think I have to do more with $p \equiv 3 \bmod 10$
Here is a heuristic proof.
Start with $\sum_{p < x} \dfrac1{p} =\ln \ln x + O(1) $.
Then, mod any positive integer $m$, the primes in the $\phi(m)$ congruence classes $p \equiv k \bmod m$ for $(k, m) = 1, 1\le k \lt m$ are approximately equally distributed, so $\sum_{p < x, p \equiv k \bmod m} \dfrac1{p} =\dfrac1{\phi(m)}\ln \ln x + O(1) $.
Since $\phi(10)=4$, all you have to do is make this rigorous.
Note: Look up "prime number races."