I'm trying to evaluate
$$\int_{\gamma} \frac{e^z}{z^m(1-z)}dz$$
where $\gamma$ is the boundary of $D(\frac{1}{2},1)$.
I can't apply Cauchy's integral formula, since the function has its singularities inside the circle. How can I proceed?
[Math] cauchy’s integral formula with singularities
complex-analysis
Related Solutions
The integral cannot be evaluated in the usual sense when $r=1$ since there is a pole of order $4$ on the contour (at the origin). Remember that the $p$-test says that only poles of order $<1$ are integrable.
The usual method to "assign a value" to divergent integrals like this is the Cauchy principal value method, which in this case would look like
$$ \begin{align*} &\text{PV} \oint_\gamma \frac{e^z-1}{z^5}\,dz \\ &\qquad = -5 \cdot \lim_{\epsilon \to 0^+} \left[ \int_0^{\pi-\epsilon}\frac{\exp\left(1+e^{it}\right)-1}{\left(1+e^{it}\right)^5}\,ie^{it}\,dt + \int_{\pi+\epsilon}^{2\pi} \frac{\exp\left(1+e^{it}\right)-1}{\left(1+e^{it}\right)^5}\,ie^{it}\,dt \right]. \end{align*} $$
Here I've parameterized the circle by $z = 1+e^{it}$ but removed the arc of the circle of length $2\epsilon$ which passes through the pole at $z=0$, as in the picture below.
However, the even order of the pole presents an issue. As $\epsilon \to 0$ we'll be approaching the pole along paths tangent to the imaginary axis, and along this axis the real part of the integrand is even:
$$ \operatorname{Re}\left[\frac{e^{iy}-1}{(iy)^5}\right] = \frac{\sin y}{y^5}. $$
Here's a plot of this:
So, since
$$ \lim_{y \to 0} \frac{\sin y}{y^5} = \infty, $$
no cancellation will occur when calculating the principal value and the result as $\epsilon \to 0$ will be $\infty$.
Best Answer
You do not need the residue theorem. You can decompose your integrand into $$ \int_{\gamma} e^{z} \left( \frac{1}{z} + \frac{1}{z^2} + \ldots + \frac{1}{z^m} + \frac{1}{1 - z} \right) \, dz.$$ To justify the partial fraction decomposition, notice that by adding and subtracting terms, \begin{align} \frac{1}{z^m (1 - z)} & = \frac{(z^{m-1} + \ldots + z + 1)(1 - z) + z^m}{z^m(1 - z)} = \frac{z^{m-1}(1 - z) + \ldots + (1 - z) + z^m}{z^m(1 - z)} \\ & = \frac{1}{z} + \ldots + \frac{1}{z^m} + \frac{1}{1 -z} \end{align} Now, you can calculate each piece of the integral using the generalized Cauchy integral formula, namely that $$ f^{(k)}(z_0) = \frac{k!}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z - z_0)^{k+1}} \, dz$$ for $f(z)$ analytic. I'll leave it to you to take it from here.
EDIT: Justified the partial fraction decomposition.