How can I get an explict representation for the dirichlet series, $$\sum_{k=0}^\infty\frac{\sigma_x(a k+b)}{(ak+b)^s} $$

Possibly in terms of hurrwitz zeta functions and other elementary functions

Where $\sigma_x(n)$ is the divisor function, counting the $x^{th}$ powers of the divisors of n


I'll assume $(a,b) = 1$ for simplicity.

Let $\chi$ denote a Dirichlet character modulo $b$, i.e. a function from $\mathbb{N}$ satisfying 3 conditions: $\chi(st)=\chi(s)\chi(t), \chi(s+b)=\chi(s), \chi(s)\neq 0$ iff $(b,s)>1$.

For example, when $m=4$, there are 2 characters: $\chi_{1}(n) = \begin{cases} 1 &&n\equiv 1(2) \\ 0 && 2|n\end{cases}$ and $\chi_{2}(n) = \begin{cases} 1 &&n\equiv 1(4) \\ -1 && n \equiv 3(4) \\ 0 && 2|n\end{cases}$.

Some properties of those:

  1. A product of 2 characters is a character. The inverse of a character $\chi$ is $\overline{\chi}$ and is also character. There's always the trivial character $\chi(n)=1_{(n,b)=1}$.
  2. $\chi(1)=1$
  3. If $(n,b)=1$ then $\chi(n)$ is a root of unity of order $\phi(n)$ (Hint: Euler's Theorem)
  4. There are $\phi(b)$ characters. Try to prove it at least for $b$ an odd prime power - in this case, the group $(\mathbb{Z}/b\mathbb{Z})^{\times}$ is cyclic, i.e. generated by some single element $g$, which can be sent to any root of unity of order $\phi(b)$ which determines the character.
  5. Orthogonality - $(s,b)=(t,b)=1 \implies \sum_{\chi} \chi(s)\overline{\chi(t)} = \phi(b)1_{s \equiv t \mod b}$. Try to prove it.

The last part shows that in general, $\sum_{n\equiv a \mod b} \frac{a_n}{n^s}$ equals $\frac{1}{\phi(b)} \sum_{\chi} \sum_{n} \frac{a_n \chi(n)}{n^s}$.

This reduces your problem to simplifying $\sum_{n} \frac{\sigma_x(n) \chi(n)}{n^s}$. Since $\sigma_x$ and $\chi$ are both multiplicative, so is their product, hence $\sum_{n} \frac{\sigma_x(n) \chi(n)}{n^s} = \prod_{p \text{ is prime}} (\sum_{k \ge 0} \sigma_x(p^k)p^{-ks})$. We'll understand the general $p$-term: $$\sum_{k \ge 0} \sigma_x(p^k)p^{-ks} = \sum_{k \ge 0} \frac{p^{(k+1)x}-1}{p^x-1} \frac{\chi^{k}(p)}{p^{ks}}$$ Some algebra shows that this is $(1-\chi(p)p^{-s})^{-1}(1-\chi(p)p^{x-s})^{-1}$. Some familiarity with Dirichlet's L-functions, $L(s,\chi)=\sum_{n} \frac{\chi(n)}{n^s} = \prod_{p} (1-\chi(p)p^{-s})^{-1}$ shows that $\sum_{n} \frac{\sigma_x(n) \chi(n)}{n^s} = L(s,\chi)L(s-x,\chi)$.

All in all, your expression is $$\frac{1}{\phi(b)} \sum_{\chi} L(s,\chi)L(s-x,\chi)$$

All this and more is found in Chapter 16 of Ireland and Rosen's "A Classical Introduction to Modern Number Theory".


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