Residue of a Dirichlet Series at $s=1$

Question

I have encountered this problem of determining the leading term in the Laurent expansion of a Dirichlet series. Let $d(n)$ be integers and consider the Dirichlet series $$D(s)=\sum_{n=1}^{\infty}\frac{d(n)}{n^s}$$ Suppose that $D(s)$ has a simple pole at $s=1$. Then in a neighborhood of $1$, we can write $$D(s)=\frac{C}{s-1}+\sum_{n=0}^{\infty}C_n(s-1)^n$$ I want to determine $C$ when we only know the $d(n)$. Any help would be appreciated. Also, what if $D(s)$ is given by only an Euler product? Can we compute the limit $\lim_{s\to1}(s-1)D(s)$?

Answer 1

Let $S(x) = \sum_{n \leq x} d(n)$. Assuming the abscissa of convergence of $D(s)$ is $1$, summation by parts yields, for $\mathfrak{Re}(s) > 1$, $$D(s) = s \int_1^{+\infty} \frac{S(x)}{x^{s+1}} \, \mathrm{d} x.$$ Now let $R(x) = S(x) - Cx$ where $C$ is the residue you are looking for. Plugging $S(x) = R(x) + Cx$ in the above formula, we get \begin{align}D(s) &= s \int_1^{+\infty} \frac{R(x)}{x^{s+1}} \, \mathrm{d} x + Cs \int_1^{+\infty} \frac{1}{x^{s}} \, \mathrm{d} x\\ &= \frac{Cs}{s-1} + s \int_1^{+\infty} \frac{R(x)}{x^{s+1}} \, \mathrm{d} x\\ &= \frac{C}{s-1} + C + s \int_1^{+\infty} \frac{R(x)}{x^{s+1}} \, \mathrm{d} x.\end{align}

The hypothesis that $D$ has a simple pole at $1$ with residue $C$ implies that the last integral remains bounded as $s \to 1$. This already provides some information about what this number $C$ should be, it's the only complex number such that $$\int_1^{+\infty} \frac{R(x)}{x^{s+1}} \, \mathrm{d} x$$ is bounded in the neighborhood of $1$.

In particular, if you manage to prove something along the line of $$S(x) = Cx + O(x^{1-\varepsilon})$$ for some $\varepsilon > 0$, then your residue is this particular $C$.

Conversely, the Wiener-Ikehara tauberian theorem (see Theorem 2 from this link) implies that $S(x) \underset{x \to +\infty}{\sim} Cx$ if the coefficients $d(n)$ are non-negative.