As part of the proof of the Prime Number Theorem in my online notes, we are told to show the following identity:

For $y>0$, $c>0$ show that

$ \int _{c-i \infty}^{c+i \infty} \dfrac{y^s}{s(s+1)}ds$

is $0$ if $ y \leq 1$ and $ 1-y^{-1}$ if $y > 1$.

Now I can see where the residues play a role in the second result - the poles are at $0$ and $1$ with residues $1$ and $-y^{-1}$ respectively. My question is, should the answers have conditions $c>1$ and $ c \leq 1$ rather than on $y$. Also, I'm having trouble determining an appropriate contour for integration - I had been looking at something rectangular with corners at $ \pm c \pm i N $ then letting $N \rightarrow \infty$. The only thing is it along the vertical sections they don't match up so well.

Thanks for any help you can give!


Consider the following contour integral:

$$\oint_C dz \frac{e^{z t}}{z (z+1)} $$

where $C$ is the contour including the line segment from $c-i R$ to $c+i R$ and a circular arc of radius $R$ closed to the left. Note that $y=e^t$; I made this substitution to help us see things easier.

The contour integral is equal to

$$\int_{c-i R}^{c+i R} ds \frac{e^{s t}}{s (s+1)} + i R \int_{\pi/2}^{3 \pi/2} d\theta \, e^{i \theta} \frac{e^{R t \cos{\theta}} e^{i R t \sin{\theta}}}{R e^{i \theta} (R e^{i \theta}+1)}$$

The second integral vanishes as $R \to \infty$ when $t \gt 0$. To see this, we may provide a bound on its magnitude as it is less than:

$$\frac1{R-1} \int_{\pi/2}^{3 \pi/2} d\theta \, e^{R t \cos{\theta}} = \frac{2}{R-1} \int_{0}^{\pi/2} d\theta \, e^{-R t \sin{\theta}} \le \frac{2}{R-1} \int_{0}^{\pi/2} d\theta \, e^{-2 R t \theta/\pi} \le \frac{\pi}{R-1}$$

The contour integral is then simply equal to the original integral. However, the contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles $z=0$ and $z=-1$. As these poles are simple, the residue calculation is straightforward. Thus, the integral is equal to, for $t \gt 0$:

$$i 2 \pi \left (1-e^{-t} \right ) $$

For $t \lt 0$, the second integral does not converge and we may not use the residue theorem for the integral over $C$ as defined. Rather, we need to consider another contour $C'$ over which the integral converges. In this case, $C'$ merely differs by closing the circular arc to the right. In this case, there are no poles within $C'$ and thus the contour integral is zero for $t \lt 0$.

Translating this result back to the original notation $y=e^t$, we see that

$$\int_{c-i \infty}^{c+i \infty} ds \frac{y^s}{s (s+1)} = i 2 \pi \left (1-\frac1{y} \right ) u(y-1) $$

where $u$ is the Heaviside step function.

  • $\begingroup$ Thanks! That makes it so much clearer. I haven't done these in years...I've embarrassingly forgotten. $\endgroup$ – user204388 Jan 2 '15 at 16:14

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