Contour for calculating this complex integral 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!
 A: 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.
