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

Use the principal branch of log z $\int_{-1}^{1} log z\, dz$

My attempt was we see how -1 to 1 became pi to 0? That should make the e^whatever terms go away. Then we can do it by parts. $\int_{-1}^{1} log z\, dz$ =$\int_{\pi}^{0} log (e^{i\theta})ie^{i\theta}\, dz=-2+i\pi$

Is this correct result? Could please show me another method of resolution?

Could someone help me through this problem?

share|cite|improve this question
up vote 2 down vote accepted

I would say $$ \int_{-1}^1 \log(z)\,dz = \int_0^1\log(x)\,dx+\int_{-1}^0[\log(-x)+i\pi]\,dx $$ and then do some real integrals.

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.