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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?

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This is "work work" not "homework". I'm going to be honest here and say that I'm overloaded right now, so I'm asking this question even though I'd probably be able to answer it for myself in a few hours. I'm not sure how this is viewed here, however I'm usually an "answerer" and not an "asker", and I'd really appreciate some help, so I hope it's OK! – Douglas B. Staple Jun 4 '13 at 19:37
up vote 1 down vote accepted

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 $$ \theta(x)=\left\{\begin{array}{} 1&\text{if }x\gt0\\ \tfrac12&\text{if }x=0\\ 0&\text{if }x\lt0 \end{array}\right. $$

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Yes, you are right that $x/n>0$, and that $\theta(x)$ was meant to be the Heaviside step function. What you wrote is basically a much more detailed (and very helpful) explanation of the same thing that Brian said; thanks to both of you. – Douglas B. Staple Jun 5 '13 at 0:53

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.

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Ahh... I start to see it now. It follows directly from the residue theorem. – Douglas B. Staple Jun 4 '13 at 20:07

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