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So I have to solve $$\int_0^1\frac{1}{x^{2/3}(1-x)^{1/3}}dx.$$ To do this I made a branch cut from $z=0$ to $z=1$ and took the bone-shaped contour that straddles the real axis, going clockwise. Now on the line above the real axis, I took the angle with respect to $z=0$ to be $0$ and that with $z=1$ to be $\pi$. Then the corresponding line integral comes out to $e^{-\frac{i\pi}{3}}I$ and so the one below comes out to $-e^{\frac{i\pi}{3}}I$ with the second angle $-\pi$. But this gives me the wrong answer. Most places I found online seem to take the first angle to be $0$ and the second one to go from $0$ to $2\pi$. But why should this make a difference? Why is my way of taking the angles wrong?

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    $\begingroup$ Your angle (branch of arg) has to be continuous along your contours. $\endgroup$ Feb 28, 2021 at 17:18
  • $\begingroup$ @user10354138 could you please explain what you mean by that? $\endgroup$
    – Souroy
    Feb 28, 2021 at 18:10
  • $\begingroup$ Just to add to what @user10354138 wrote: your function has to be single-valued in all complex plane, as soon as you do not cross the cut (relating points 0 and 1) - you should get the same value after a full turn around the cut. If you define, for instance, real function value on the upper bank of the cut $(0,1)$, then making a full turn clockwise around 1 (to move to the lower bank) your function gets the factor $\exp(+\frac{2i\pi}{3})$ -because you turning clockwise (one minus) and negative power ($-\frac{1}{3}$) - another minus.Turning around zero you get a zero phase on the upper bank. $\endgroup$
    – Svyatoslav
    Feb 28, 2021 at 19:29

1 Answer 1

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Note that your integrand here is discontinuous at $x = 0$ and $x=1$.

Thus, if you take the argument for your contour from $ - \pi $ to $ \pi$, then the arc will pass through a discontinuity and your answer may be incorrect by a constant residue.

This problem is not present when taking the angles $0$ to $ 2 \pi$ as there will not be any integration along a discontinuity.

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