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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?

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    z=1 is a singularity and lies inside the disc centered at 1/2 with radius 12012-11-12

2 Answers 2

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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.

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    thank you very much for your answer, but at the moment i don't know residue theory, but i'm required to solve it, so i think i should solve in some different way2012-11-12
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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.

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    thank you, now i can solve the exercise2012-11-13