Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Calculate $$\int_\gamma \frac{1+\sin(z)}{z} dz$$

where $\gamma$ is the circle of centre 0 and radius $\log(\sqrt2)$ oriented counter clockwise.

Well there is a singularity at 0. The fact that the radius is $\log(\sqrt2)$ has no real relevance as this function fails to be analytic at any circle centered at 0.

I used the Cauchy integral formula to calculate the integral and got $2\pi i$. Does that look correct?

share|cite|improve this question
yes your are right. – noname1014 Apr 18 '12 at 0:24
up vote 1 down vote accepted

$$ \int_\gamma \frac 1 z\,dz + \int_\gamma\frac {\sin z}{z}\,dz. $$ The second integral is zero because the singularity at $z=0$ is removable. You can see that by noticing that the power series for $\sin z$ has no constant term, so you can divide the whole thing by $z$ without getting a $1/z$ term.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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