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.
Best Answer
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 $\int_{-1}^1 \sqrt{1-x^2}\,dx=\pi/2$.