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?
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?
You can use the Residue theorem.
For every $a \in D(\frac{1}{2},1)$ we have $\int_\gamma f(z) ~dz= 2\pi i\sum \text{Res}(f, a_k),$
Where the $a_k$ are the residues of the poles of $f$ inside $\gamma$. The residues of $f$ are given by the coefficients of $z^{-1}$ in the Laurent Series expansion of $f(z)$ at each of the singularities.
In this case we have two singularities, a simple pole at $z=1$ and a pole of order 5 at $z=0$.
There are in fact closed form identities for determining residues of $f$ at these sorts of poles.
For a simple pole we have,
$\text{Res}(f, a) = \lim_{z \to a}(z-a)f(z),$
and for the pole of order n we have,
$\text{Res}(f, a) = \frac{1}{(n-1)!}\lim_{z \to a}\frac{d^{n-1}}{dz^{n-1}}((z-a)^nf(z))$
You should now be able to be able to compute the integral, by substituting in the derived values for the residues.
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.