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?

  • $\begingroup$ en.wikipedia.org/wiki/Residue_theorem $\endgroup$ – user17794 Nov 12 '12 at 10:19
  • $\begingroup$ 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. $\endgroup$ – icurays1 Nov 12 '12 at 10:19
  • $\begingroup$ z=1 is a singularity and lies inside the disc centered at 1/2 with radius 1 $\endgroup$ – 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.

  • $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ +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... $\endgroup$ – DonAntonio Nov 12 '12 at 11:25
  • $\begingroup$ 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. $\endgroup$ – bateman Nov 12 '12 at 11:57
  • $\begingroup$ @bateman, careful with the geometric series argument (though heuristically it works), as the geometric series formula for $1/(1 -z)$ only holds for $|z| < 1$. $\endgroup$ – Christopher A. Wong Nov 12 '12 at 12:44
  • $\begingroup$ Don't forget to click the hollow tick mark to accept this answer if you can now solve the problem.... $\endgroup$ – Simon Hayward Nov 12 '12 at 12:44
  • $\begingroup$ 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? $\endgroup$ – bateman Nov 12 '12 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.