For \(\left\{{a_k}\right\}\) a sequence in \({\mathbf{R}}\), define
\begin{align*}
A(t) := \sum_{0\leq k \leq t} a_k
\end{align*}
For \(\phi\in C^1({\mathbf{R}})\) continuously differentiable,
\begin{align*}
\sum_{x
Fix \(s\in {\mathbf{C}}, a_{k} = 1\) for \(k\geq 1\), and \(\phi(x) := x^{-s}\). Then \(A(t) = {\left\lfloor t \right\rfloor}\), and \begin{align*} \sum_{n=1}^{\lfloor x\rfloor} \frac{1}{n^{s}}=\frac{\lfloor x\rfloor}{x^{s}}+s \int_{1}^{x} \frac{\lfloor u\rfloor}{u^{1+s}} \,du \end{align*} Now take \(\Re(s) > 1\) and \(\lim_{x\to\infty}\) to yield \begin{align*} \zeta(s)=s \int_{1}^{\infty} \frac{\lfloor u\rfloor}{u^{1+s}} \,du \end{align*} Use this to derive Dirichlet’s theorem.md): \zeta(s) has a simple pole at s=1 with $\mathop{\mathrm{Res}}_ \zeta(s: \(/zeta(s)\) has a simple pole at \(s=1\) with \(/Res_{s=1} /zeta(s) = 1\). This works for other Dirichlet series.