Integral Formula for Particular Dirichlet Series
The following is a formula for a Dirichlet series where \(s \in \mathbb{R}\) and we satisfy a particular constraint on the sequence \(a_n\).
Theorem
If \(A(x) = \sum_{n \leq x} a_n = O(x^\delta)\). For \(s > \delta\),
\[ \sum_{n = 1}^\infty \frac{a_n}{n^s} = s \int_1^\infty \frac{A(t)}{t^{s + 1}} \,\mathrm{d}t.\]
Proof
We use Abel summation with \(f(n) = \frac{1}{n^s}\) to deduce that
\[\begin{align*}
\sum_{n \leq x} \frac{a_n}{n^s} &= \frac{A(x)}{x^s} + s \int_1^x \frac{A(t)}{t^{s + 1}} \,\mathrm{d}t \\
\lim_{x \to \infty} \sum_{n \leq x} \frac{a_n}{n^s} &= \lim_{x \to \infty} \left(\frac{A(x)}{x^s} + s \int_1^x \frac{A(t)}{t^{s + 1}} \,\mathrm{d}t\right) \\
\sum_{n = 1}^\infty \frac{a_n}{n^s} &= \lim_{x \to \infty} \left(\frac{A(x)}{x^s}\right) + s \int_1^\infty \frac{A(t)}{t^{s + 1}} \,\mathrm{d}t. \tag{1} \\
\end{align*}\]
Then because \(A(x) = O(x^\delta)\) by assumption, there exists an \(x_0\) and constant \(C\) such that
\[\begin{align*}
x > x_0 &\implies |A(x)| < C |x^\delta| \\
&\implies \left|\frac{A(x)}{x^s}\right| < C \left|\frac{x^\delta}{x^s}\right| = C|x^{\delta - s}|.
\end{align*}\]
Because \(s > \delta\), \(\delta - s < 0\) and hence as \(x \to \infty\), \(|x^{\delta - s}| \to 0\). Therefore we have by the pinching theorem that
\[ \lim_{x \to \infty} \left|\frac{A(x)}{x^s}\right|
= \lim_{x \to \infty} \frac{|A(x)|}{x^s} = 0.\]
Therefore (1) becomes
\[ \sum_{n = 1}^\infty \frac{a_n}{n^s} = + s \int_1^\infty \frac{A(t)}{t^{s + 1}} \,\mathrm{d}t.\]