Why is $\frac{1}{2\pi i} \int_C \left( \frac{x}{n} \right)^s \frac{ds}{s} = \theta(x-n) $?

Question

I'm trying to understand the equation:

$$\frac{1}{2\pi i} \int_C \left( \frac{x}{n} \right)^s \frac{ds}{s} = \theta(x-n).$$

Here $x\in \mathbb{R}, x\geq 0$, and $C = \{s:\operatorname{Re}(s) = \sigma\}$ is a coutour, with fixed $\sigma > 0$ and $t = \operatorname{Im}(s)$ ranging.

This is Eq. 3.26 in Crandall and Pomerance, pp. 159. It's presented as "the Perron formula", although it looks to me to be a particular form of Perron's formula. It seems to me that this can be proven using either a Fourier or Laplace transform, but it's really not obvious to me.

How can I prove the above formula, preferably with elementary methods? Also, is there some intuitive interpretation that I can use to understand this thing?

Answer 1

I assume that $x/n\gt0$ (real implied). If so, we can replace $x/n$ by $x$. So we need to look at $$ \int_Cx^{\large s}\frac{\mathrm{d}s}{s} =\int_Ce^{\large s\log(x)}\frac{\mathrm{d}s}{s} $$ If $x\gt1$, then $\log(x)\gt0$. Then we use a contour that looks like a backwards "D"; going from $\sigma-i\infty$ to $\sigma+i\infty$ then circling counterclockwise around the left half-plane. This encompasses the singularity at $s=0$ with residue $1$. The integral along the curved part of the contour vanishes as it moves to $\infty$. This leaves $$ \frac1{2\pi i}\int_Cx^{\large s}\frac{\mathrm{d}s}{s}=1 $$ If $x\lt1$, then $\log(x)\lt0$. Then we use a contour that looks like a "D"; going from $\sigma-i\infty$ to $\sigma+i\infty$ then circling clockwise around the right half-plane. This encompasses the no singularities. The integral along the curved part of the contour vanishes as it moves to $\infty$. This leaves $$ \frac1{2\pi i}\int_Cx^{\large s}\frac{\mathrm{d}s}{s}=0 $$ If $x=1$, then $$ \begin{align} \frac1{2\pi i}\int_Cx^{\large s}\frac{\mathrm{d}s}{s} &=\frac1{2\pi i}\lim_{t\to\infty}\log\left(\frac{\sigma+it}{\sigma-it}\right)\\ &=\frac1{2\pi i}\pi i\\ &=\frac12 \end{align} $$

From this, I assume that $$ \theta(x)=\left\{\begin{array}{} 1&\text{if }x\gt0\\ \tfrac12&\text{if }x=0\\ 0&\text{if }x\lt0 \end{array}\right. $$

Answer 2

This is a residue problem; the contour can be thought of as the limit of semicircles pointing to the right or left. If $x/n$ is bigger than 1, you complete the contour on the left, so that in the limit the exponential part goes to 0 in the semicircular arc. If its less than 1, complete it to the right. There is a simple pole at 0 with residue 1.