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

How can evaluate $\int_{-1}^{1} z^{\frac{1}{2}}\, dz$ with the main branch of $z^{\frac{1}{2}}$? Thanks for your help

share|cite|improve this question
Perhaps split it into $\int_{-1}^0$ and $\int_0^1$? – anon May 7 '12 at 14:05
It is being evaluated in the complex plane – Miguel Mora Luna May 7 '12 at 14:25
up vote 1 down vote accepted

This is an expansion on anon's comment above.

Caveat: I'm not 100% certain what the "main branch" is supposed to do to the negative real axis, but I am going to assume it maps to the positive imaginary axis.

To integrate from $0$ to $1$, that's no problem, that's an old-school integral of a real-valued function on the real line, and we get 2/3.

From $-1$ to $0$, we have a complex valued function. I think the easiest way to do this one is to let $t = -z$. Now, because you're working with the main branch, $\sqrt{-t} = i\sqrt{t}$ for $t$ a positive real number - note, confusingly, that this identity isn't true for all complex numbers, moreover, a different choice of branch cut of the square root function can make it false. $$ \int_{-1}^0 z^{\frac{1}{2}}dz = -i\int_1^0 t^{\frac{1}{2}}dt $$ This latter integral is $\frac{2}{3}i$ so the final answer is $\frac{2}{3} + \frac{2}{3}i$.

share|cite|improve this answer
Agreed. Unless a specific curve is given, use this simplest one. For a more complex (ha ha) problem, use a curve that starts at $-1$, wraps around $0$ some number of times, and ends at $1$. Of course the "main branch" is discontinuous when your curve crosses the cut. – GEdgar May 7 '12 at 14:43

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.