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

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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

1 Answer 1

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 $\frac{2}{3} + \frac{2}{3}i$.

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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

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