Integrals and Miscellaneous 20

$f(z):=(1-z^3)^{-1/2}$ , $C$ is the contour which consists of two lines and two arcs. One of the lines is indented at $z=1$ by an arc of radius $1/R$. The radius of the other arc is $R$.

From the Cauthy's theorem we find $\oint_C f(z)dz=0$. The integral round the arcs tend to zero.\begin{eqnarray*}\oint f(z)dz &=&\int_0^1(1-x^3)^{-\frac{1}{2}}dx+\int_1^\infty\left\{(x-1)e^{-\pi i}\right\}^{-\frac{1}{2}}(1+x+x^2)^{-\frac{1}{2}}dx+i\int_\infty^0(1+ix^3)^{-\frac{1}{2}}dx\\&=&\int_0^1(1-x^3)^{-\frac{1}{2}}dx+i\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx\\&&\quad-i\int_0^\infty(1+ix^3)^{-\frac{1}{2}}dx\\&=&\int_0^1(1-x^3)^{-\frac{1}{2}}dx+i\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx\\&&\quad-ie^{-\frac{\pi}{6}i}\int_0^\infty(1+y^3)^{-\frac{1}{2}}dx\end{eqnarray*}Hence,\begin{eqnarray*}\int_0^1(1-x^3)^{-\frac{1}{2}}dx&=&\frac{1}{2}\int_0^\infty(1+x^3)^{-\frac{1}{2}}dx=\frac{1}{2}\int_{-\infty}^0(1-x^3)^{-\frac{1}{2}}dx\\\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx&=&\frac{\sqrt{3}}{2}\int_0^\infty(1+x^3)^{-\frac{1}{2}}dx\end{eqnarray*}Therefore,\begin{eqnarray*}\int_{-\infty}^1(1-x^3)^{-\frac{1}{2}}dx&=&\frac{3}{2}\int_0^\infty(1+x^3)^{-\frac{1}{2}}dx\\\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx&=&\frac{\sqrt{3}}{2}\int_0^\infty(1+x^3)^{-\frac{1}{2}}dx\end{eqnarray*}$$\therefore\quad\int_{-\infty}^1(1-x^3)^{-\frac{1}{2}}dx=\sqrt{3}\int_1^\infty(x^3-1)^{-\frac{1}{2}}dx$$