# Branch cut integral $\int_{-1}^1\left(1-x^2\right)^{\frac{1}{2}} d x$

Define the branch of $$f(z)=\left(1-z^2\right)^{\frac{1}{2}}$$ by the branch cut $$(-\infty,-1] \cup [1,\infty), f(0)=1$$.

Use this branch and a suitably chosen semi-circular contour (with finite radius $$R$$ greater than 1) in the upper half plane to evaluate $$\int_{-1}^1\left(1-x^2\right)^{\frac{1}{2}} d x$$

I am thinking of this contour

where $$\gamma_3,\gamma_4$$ should cancel and $$\gamma_1,\gamma_2$$ evaluates to $$0$$. I stuck at $$\gamma_R$$ since there seems to be no place to use the Residue theorem.

Any help is appreciated.

• there is no singularity in the integrand you can't apply any kind of residue Commented Feb 27 at 14:23
• @ThetaOmega I don't know, to me the exercise doesn't make sense. The integral indeed is trivial with the substitution $x=\sin t$ Commented Feb 27 at 15:24
• @xpaul It's good to see you back now. You are of course correct; one not need to use contour integration. But, if one chooses to do so, then the posted solution might be useful. ;-) Commented Feb 28 at 21:45
• @Masacroso: The integral is even easier if you recognize its geometric meaning of "What's the area of half of a circle with radius 1?"
– Dan
Commented Feb 28 at 21:53
• @Dan The question is on the use of contour integration to evaluate the integral. It is NOT simply "What is an easy way to evaluate the integral?" Commented Feb 29 at 22:44

The integral over $$\gamma_R$$, $$R>1$$ can be evaluated by noting that $$\int_0^\pi e^{i2\phi}\,d\phi = 0$$ and that $$\sqrt{1-z^2}=iz\sqrt{1-1/z^2}$$ for the selected branch cuts and branch. Therefore, we find that
\begin{align} \int_0^\pi \sqrt{1-R^2e^{i2\phi}}\,iRe^{i\phi}\,d\phi&=\int_0^\pi \left(-R^2e^{i2\phi}\right)\sqrt{1-\frac1{R^2e^{i2\phi}}}\,d\phi\\\\ &=\int_0^{\pi} \left(-R^2e^{i2\phi}\right)\left(1-\frac1{2R^2e^{i2\phi}}+O\left(\frac1{R^4}\right)\right)\,d\phi\\\\ &=\frac\pi 2+O\left(\frac1{R^2}\right) \end{align}
Letting $$R\to \infty$$, we find that $$\int_{-1}^1 \sqrt{1-x^2}\,dx=\pi/2$$.