Alternating sum of primes. 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. 

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
$$ \sum_{p}\frac{\chi(p)}{p}\approx -0.18654766883298535284. $$
