Alternating sum of primes.

Question

Do we know the sum, $$\sum_p \frac{\chi\left(p\right)}{p}\textrm{?}$$ for $\chi\left(n\right)$ being $1$ if $n=1\textrm{ mod }4$, $-1$ if $n=3\textrm{ mod }4$ and zero otherwise.

I'm looking for an exact value, if there is one, and a derivation.

Answer 1

Since $|\chi(p)|\leq 1$, we may approximate $\frac{\chi(p)}{p}$ with $-\log\left(1-\frac{\chi(p)}{p}\right)$ and deduce that $$ \sum_{p}\frac{\chi(p)}{p} \approx \log\prod_{p}\left(1-\frac{\chi(p)}{p}\right)^{-1}=\log L(1)=\log\frac{\pi}{4} $$ where $L(s)$ stands for the Dirichlet $L$-function $L(s)=\sum_{n\geq 1}\frac{\chi(n)}{n^s}$. With the same mechanism leading to a series representation for the prime $\zeta$ function (i.e. Moebius' inversion formula) we may state that

$$ \sum_{p}\frac{\chi(p)}{p} = \sum_{n\geq 1}\mu(n)\frac{\log L(n)}{n} $$ where the series on the RHS has a reasonable convergence speed, due to $L(n)\approx 1-\frac{1}{3^n}\Rightarrow\log L(n)\approx -\frac{1}{3^n}$. Numerically we have $$ \sum_{p}\frac{\chi(p)}{p}\approx -0.18654766883298535284. $$