Tauberian theorem with better error term

Question

This is a fairly vague question.

Suppose we have a sequence of positive numbers $(c_n)_n$ and we want to find an asymptotic formula for $S(x) = \sum_{n \leq X} c_n$. In favorable circumstances, standard Tauberian theorems, e.g. the one in Appendix A of Chambert-Loir and Tschinkel's paper "Fonctions zeta des hauteurs des espaces fibres", available at http://www.math.nyu.edu/~tschinke/papers/yuri/01zeta/zeta.pdf, give an asymptotic formula of the form $X^a P(\log X) + O(X^{a - \delta})$, with $P$ a polynomial of degree $b-1$, with $b$ the order of pole at $a$ ($a>0$ is the right most pole). The leading term of the polynomial $P$ is explicit.

Question. Can one explicitly compute the polynomial $P$?

In my particular problem, I have a zeta function $Z(s)= \sum_n c_n.n^{-s}$ which I understand fairly well. One can show that $Z(s) = f(s) + h(s)$ with $f$ written in terms of Eisenstein series with a pole at $a=1$ of order $b=2$, and $h(s)$ harmless. In the Chambert-Loir–Tschinkel theorem $\delta = 1/2$ works, and I have an asymptotic formula of the form $(3/\pi) X \log X + B X + O(X^{1/2})$. Now I'm trying to find the value $B$.

Answer 1

If your zeta function $Z(s)$ has analytic continuation to the left of your pole $a$ (with some reasonable bounds etc.), then you can just use an inverse Mellin transform, no? The main term is the residue of $X^s Z(s)$ at $s=a$, so all you need is the Taylor series expansion of $X^s$ at $s=a$ (which gives the powers of $\log X$) and the Laurent series expansion of $Z(s)$ at $s=a$. Multiply the two expansions and pick the coefficient of $(s-a)^{-1}$.

Answer 2

Explicit expressions for such lower order terms can be found in Chapter II.5 (Selberg-Delange method) of the book

Tenenbaum - Introduction to analytic and probabilistic number theory.

See in particular Theorem 3.

The expression contains things like the coefficients of the higher order terms of the Taylor expansion of your function at the singularity.

I agree Alexey however that they are usually quite ugly.