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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?

Community
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Gobi
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1 Answers1

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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.

Bruno Joyal
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