Dirichlet L-series associated to periodic sequence Let $\{a_n\}$ be a sequence of complex numbers such that $a_n=a_m $ iff $ n\equiv m \mod q$ for some positive integer $q$. Define the Dirichlet L-series associated to $\{a_n\}$ by
$$L(s)=\sum_{n=1}^{\infty} \frac{a_n}{n^s} \ \ \  \text{  for   Re}(s)>1. $$ 
Also define $$Q(x)=\sum_{m=0}^{q-1}a_{q-m} e^{mx}\ \ \ \text{  with   }\ \  a_0=a_q.$$
I showed that
$$
L(s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{Q(x)x^{s-1}}{e^{qx}-1}dx,  \ \ \text{for   Re}(s)>1 
$$
Now I want to show that $L(s)$ is continuable into the complex plane, with the only
possible singularity a pole at $s=1$.
I follwed the hint in previous asked question, so that $$L(s)=\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{(Q(x)-Q(0))x^{s-1}}{e^{qx}-1}dx+\frac{1}{\Gamma(s)}\int_{0}^{\infty}\frac{Q(0)x^{s-1}}{e^{qx}-1}dx$$
I found that the second term equals to $Q(0)\zeta (s)/q^s$, which is meromorphic except a pole at $s=1$ for $Q(0) \neq 0$. So I want to show that the first term is entire in whole plane. But how can I show it?
 A: If I'm not mistaken, neither integral converges when $\text{Re }s\leq 1$, and should actually be replaced by
$$\frac{1}{(1-e^{2\pi i s})\Gamma(s)} \int_C$$
where $C$ is the keyhole contour. (Regardless of whether I am right about convergence, this is a valid procedure.)  The poles of $\Gamma(s)$ cancel the zeroes of $1-e^{2\pi i s}$ at the negative integers. At the positive integers, the integral over $C$ vanishes because the integrand has no pole at $0$ and the two pieces over the positive real line cancel each other out; so at the positive integers, the poles of $(1-e^{2\pi i s})^{-1}$ are canceled by the vanishing of the integral.  
Thus the question is reduced to showing that the integral over $C$ is holomorphic (forget the factor $\frac{1}{(1-e^{2\pi i s})\Gamma(s)}$). In order to do this, approximate the contour $C$ by a sequence $C_n$ of compact contours (say by chopping off the contour at $n$), and showing that the sequence of holomorphic functions $\int_{C_n}$ converges uniformly to the whole integral $\int_C$. Then use the fact that a uniform limit of holomorphic functions is holomorphic.
