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 $P$? 
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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 $B$. 



Tauberian theorem with better error term
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 $(s-a)^{-1}$.


