# What's wrong with my method for this integral?

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?

• Your angle (branch of arg) has to be continuous along your contours. Feb 28, 2021 at 17:18
• @user10354138 could you please explain what you mean by that? Feb 28, 2021 at 18:10
• 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. Feb 28, 2021 at 19:29

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.