# cauchy's integral formula with singularities

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?

-
en.wikipedia.org/wiki/Residue_theorem – user17794 Nov 12 '12 at 10:19
Make sure you're reading Cauchy's integral formula carefully. Does the function $$f(z)=\frac{e^z}{z-1}$$ have singularities in that disc? You might also find Cauchy's integral formula for derivatives useful. – icurays1 Nov 12 '12 at 10:19
z=1 is a singularity and lies inside the disc centered at 1/2 with radius 1 – bateman Nov 12 '12 at 10:59

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.

-
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 way – bateman Nov 12 '12 at 11:01

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.

-
+1 For the very clever way to decompose the integrand function, yet I'm not sure whether the OP can use Cauchy's Integral Formula if he/she can't use the Residue Theorem... – DonAntonio Nov 12 '12 at 11:25
thank you very much, the decomposition follows from the geometric series, where terms are divided by z^m, so we have denominators for the first m terms, then i can apply Cauchy for derivatives, taking z_0=1 for the first m integrals and z_0=1 for the last. – bateman Nov 12 '12 at 11:57
@bateman, careful with the geometric series argument (though heuristically it works), as the geometric series formula for $1/(1 -z)$ only holds for $|z| < 1$. – Christopher A. Wong Nov 12 '12 at 12:44
Don't forget to click the hollow tick mark to accept this answer if you can now solve the problem.... – Simon Hayward Nov 12 '12 at 12:44
so i can't use geometric series, because the disc is not contained in |z|<1. How can i prove the decomposition of integrand? by induction on m maybe? – bateman Nov 12 '12 at 18:36