Sum of $s$-th Power of Reciprocals of Primes We know that for fixed coprime natural numbers $a,m$, we have the estimate 
$$ \sum_{p \equiv a \pmod{m}} p^{-s} = \frac{1}{\phi(m)} \log \frac{1}{s-1} + O_m(1) \hspace{5mm} \forall s \in (1,2)$$
The proof I know only works with real $s$, but I have a feeling that we should be able to say something similar for all complex $s$ with real part $>1$. More precisely, I was wondering if the following holds
$$\sum_{p \equiv a \pmod{m}} p^{-s} = \frac{1}{\phi(m)} \log \frac{1}{s-1} + g(s) \hspace{5mm}\text{for }\Re(s)>1$$
for $\Re(s)>1$ where $g(s)$ is a function which can analytically continued to the region $\Re(s) \geq 1$. I would be really grateful for any proof or reference for the same. Thanks. 
 A: Yes, we do indeed have
$$\sum_{p \equiv a \pmod{m}} p^{-s} = \frac{1}{\phi(m)}\log \frac{1}{s-1} + g(s)$$
for $\Re s > 1$ with $g$ holomorphic on some neighbourhood of $\Re s \geqslant 1$.
Let's first only look at the open half-plane $\Re s > 1$. Expanding $-\log \bigl(1 - \frac{\chi(p)}{p^s}\bigr)$ (the principal branch) using the Taylor series of $-\log(1-z)$ about $0$ and doing a little rearranging we have
$$\sum_{p \equiv a \pmod{m}} p^{-s} = \frac{1}{\phi(m)} \log L(s,\chi_0) + \frac{1}{\phi(m)} \sum_{\chi \neq \chi_0} \overline{\chi(a)} \log L(s,\chi) - \sum_{\substack{p^k \equiv a \pmod{m} \\ k \geqslant 2}} p^{-ks} \tag{1}$$
for $\Re s > 1$. Now let's look whether we have a continuation.
$$L(s,\chi_0) = \zeta(s)\cdot \prod_{p \mid m} \biggl(1 - \frac{1}{p^s}\biggr)\,,$$
whence
$$\log L(s,\chi_0) = \log \frac{1}{s-1} + \log \bigl((s-1)\zeta(s)\bigr) + \sum_{p \mid m} \log \biggl(1 - \frac{1}{p^s}\biggr)\,.$$
Since $s \mapsto (s-1)\zeta(s)$ is holomorphic and nonzero on (a neighbourhood of) $\Re s \geqslant 1$, and $1 - p^{-s}$ is holomorphic and nonzero on $\Re s > 0$ for every $p > 1$ it follows that
$$\log \bigl((s-1)\zeta(s)\bigr) + \sum_{p \mid m} \log \biggl(1 - \frac{1}{p^s}\biggr) = \log L(s,\chi_0) - \log \frac{1}{s-1}$$
is holomorphic on (a neighbourhood of) $\Re s \geqslant 1$.
Since $L(s,\chi) \neq 0$ for $\Re s = 1$ — which we'll assume for the moment — it follows that each $\log L(s,\chi)$ has an analytic continuation to some neighbourhood of $\Re s \geqslant 1$. The final series in $(1)$ converges absolutely for $\Re s > \frac{1}{2}$, hence represents a holomorphic function there.
Putting the parts together, we see that indeed
$$g(s) = \frac{1}{\phi(m)}\Biggl(\log \bigl((s-1)\zeta(s)\bigr) + \sum_{p \mid m} \log \biggl(1 - \frac{1}{p^s}\biggr) + \sum_{\chi \neq \chi_0} \log L(s,\chi)\Biggr) - \sum_{\substack{p^k \equiv a \pmod{m} \\ k \geqslant 2}} p^{-ks}$$
has an analytic continuation to a neighbourhood of $\Re s \geqslant 1$.
What is still missing is the proof that no $L(s,\chi)$ has a zero on the line $\Re s = 1$. The case of a real character and $s = 1$ is special, I'll assume that you already know this from whichever proof(s) of Dirichlet's theorem you've studied. All other cases can be done at once, using an easy modification of Mertens's proof for the $\zeta$-function.

Let $\chi$ be a Dirichlet character modulo $m$, $t \in \mathbb{R}$. Assume that either $\chi^2 \neq \chi_0$ (that is, $\chi$ is a non-real character) or $t \neq 0$. Then $L(1 + it,\chi) \neq 0$.

Proof: Note that $3 + 4\cos \varphi + \cos (2\varphi) = 2(1 + \cos \varphi)^2 \geqslant 0$ for all $\varphi \in \mathbb{R}$. Hence, writing $\chi(p) = e^{i\omega(p)}$,
\begin{align}
3 \log L(\sigma, \chi_0) &{}+ 4\Re \log L(\sigma + it, \chi) + \Re \log L(\sigma + 2it, \chi^2) \\
&= \sum_{p,k} \frac{3 + 4 \cos \bigl(k(\omega(p) - t\log p)\bigr) + \cos \bigl(2k(\omega(p) - t\log p)\bigr)}{k\cdot p^{k\sigma}} \\
&\geqslant 0
\end{align}
for all $\sigma > 1$. Exponentiating,
$$L(\sigma, \chi_0)^3 \cdot \bigl\lvert L(\sigma + it,\chi)\bigr\rvert^4 \cdot \bigl\lvert L(\sigma + 2it,\chi^2)\bigr\rvert \geqslant 1$$
for $\sigma > 1$, and we see that this is only compatible with $L(1 + it, \chi) = 0$ if $L(s,\chi^2)$ has a pole at $1 + 2it$. But this only happens for $t = 0$ and $\chi^2 = \chi_0$, which we have excluded.
