# Dirichlet series 'shifted' by a polynomial

Let $F(x) \in \mathbb{Z}[x]$ and $$\xi(s) = \sum^\infty_{n=1}g(n)n^{-s}$$ be the Dirichlet series associated an arithmetic function $g(n)$. Define a new Dirichlet series $$\xi_F(s) = \sum^\infty_{n=1}g(n)F(n)^{-s}.$$ I call $\xi_F(s)$ the Dirichlet series obtained from $\xi(s)$ by shifting by $F(x)$. (Maybe this is known by another name.) I will start with a broad question: what can one say about such Dirichlet series? I would be greatful for any references.

Now I will be more specific.

1) Can one write down the coefficients of the Dirichlet series $\xi_F(s)$ (in terms of g(n) and F(n)) ?

2) It seems clear that if $\xi(s)$ converges for $\Re(s) > k$ then $\xi_F(s)$ converges for $\Re(s) > k/\deg(F)$. Suppose that $\xi(s)$ has meromorphic continuation to $\Re(s)> k+\epsilon$, for $\epsilon > 0$. Does it follow that $\xi_F(s)$ also has meromorphic continuation?

3) If 2) is true and the continued function $\xi(s)$ has a pole of order $d$ at $s=k$ does it follow that the continued function $\xi_F(s)$ has a pole of order $d$ at $s=k/\deg(F)$? or does it have a pole of some other order?

Perhaps, some of these question can be answered for special cases. For example, if $F(x) = x^p$ then $\xi_F(s)=\xi(ps)$ and all of the above is true.

• Is there a particular motivating example for this? Because the nice thing about Dirichlet series is the relative ease of the product of two such $\xi$. The only $F$ for which $\xi_F$ will have this nice property are $F(x)=x^p$. In particular, the strong reason we set the condition that $g$ is multiplicative is so that we can write $\xi$ as a product. Again, you won't be able to do that for $\xi_F$, for general $F$. – Thomas Andrews Apr 3 '12 at 13:16
• Possibly relevant: Lerch transcendent: mathworld.wolfram.com/LerchTranscendent.html where $F(n)=a+n$ (except the numerators are $z^n$) – graveolensa Apr 3 '12 at 13:28
• The example which lead me to ask this question is: $\phi(n)$ (Euler totient) and $x^2-x$. ie $\sum \phi(n) (n^2-n)^{-s}$. – tipper Apr 3 '12 at 14:04

## 1 Answer

When $g\equiv 1$, this is a special case of the Shintani zeta function. (I'm using the notation from the wikipedia article for the remainder of this post.) If you factor your polynomial as $F(T) = \prod_{j=1}^k(T-\alpha_i)$, and define the linear form $L_j$ in $m=1$ variable by $L_j(n) = n-\alpha_i$, then the Shintani zeta function associated to this data is

$$L(s_1, \dots, s_k) = \sum_{n\geq 0} \prod_{j=1}^kL_j(n)^{-s_j} = \sum_{n\geq 0} \prod_{j=1}^k(n-\alpha_j)^{-s_j}.$$

Your zeta function can be obtained as the restriction of the above Shintani zeta function along the diagonal $s_1 = \dots = s_k = s$, because then

$$L(s, \dots, s) = \sum_{n\geq 0} \prod_{j=1}^kL_j(n)^{-s} = \sum_{n\geq 0} F(j)^{-s}.$$

You can read about the Shintani zeta function in Hida's book "Elementary theory of $L$-functions and Eisenstein series".