Cauchy integral formula with a singularity inside the set

Question

Im struggling a bit with a Cauchy integral formula exercise.

Compute $\int_{C(0,4)} \frac{e^z}{z(z-3)} dz$

In past exercises i've been able to manipulate the integral to not have any singularities for $f(z)$, which is not the case in this one.

I dont really know how to proceed, since no-singularities is one of the criteria of Cauchys integral formula.

Would anyone like to give me a hint?

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$I assume that you haven't learnt the residue theorem yet, and that you mean to use the Cauchy's integral formula:

$$\int_{\gamma}\frac{f(z)}{z-w}\d z=f(w)$$Where $\gamma$ is some suitable closed contour and $f$ is a holomorphic function on $\gamma$'s interior.

Note: $$\frac{e^z}{z(z-3)}=\frac{e^z}{3}\left[\frac{1}{z-3}-\frac{1}{z}\right]$$So, assuming $C(0,4)$ is the boundary of the disk centred at $0$, of radius $4$, you can integrate as follows: $$\int_{C(0,4)}\frac{e^z}{z(z-3)}\d z=\frac{1}{3}\int_{C(0,4)}\frac{e^z}{z}\d z-\frac{1}{3}\int_{C(0,4)}\frac{e^z}{z-3}\d z$$And apply Cauchy's formula individually on the two integrals.

Answer 2

@FShrike's answer using partial fraction decomposition is good.

Another option is to deform the path. Make a new path which goes around a small circle at $0$ starting on the real axis, then moves along the real axis to $3$, then moves around a small circle at $3$, then returns to the starting point.

The integral around $C(0,4)$ is the same as the integral along this new path since they are homotopic in the domain. However the "forwards" and "backwards" portions of the integration along the real axis cancel. So you can split this integral into the sum of the integral around each circle individually. You can then apply Cauchy's theorem to both of these integrals directly, without using partial fraction decomposition.