The problem statement reads:
This function $\large (−z)^\frac{2}{3} (resp., (1−z)^\frac{2}{3})$ is determined as to be real and positive when $z=x<0$ (resp. when $z=x<1$) and analytic in the open upper half plane.
What is the image of the upper half-plane under the map $\large Φ:w→∫_w^{−∞}\frac{dz}{(−z)^\frac{2}{3}(1−z)^\frac{2}{3}}$
the integral being taken in the upper half plane, starting from $−∞=−∞+i0$
The hint given is: It is helpful to show that $Φ(∞+i0)=∞$ and to express the image in terms of the constant $\large C=∫_{−∞}^0 \frac{dz}{(−z)^\frac{2}{3}(1−z)^\frac{2}{3}}=∫_0^{∞}x^{-\frac{2}{3}}(1+x)^{−\frac{2}{3}}$
Any hints or suggestions on how to get started on this problem are welcome.
I've read a bit on Wikipedia about the Schwarz-Christoffel mapping but am not sure of how to apply it to this problem.
Thanks,
The image is an equilateral triangle with one vertex at 0, and another at $C$.
If you want to learn about Schwarz-Christoffel map, it is much better to use a good book, rather than Wikipedia. The standard undergraduate text by Saff and Snider is OK to begin with, and it has some interesting examples. A more advanced source is Hurwitz-Courant but you have to read German or Russian, there is no English translation.
EDIT. To address your comment. Ahlfors is fine. Formula (3) on page 236. Here $n=2$, $w_1=0$, $w_2=1$, $\beta_1=\beta_2=-2/3$, $(1/3-1=2/3)$, the third vertex corresponds to $\infty$. Infinity is mapped to the vertex at $0$. The triangle with angles $\pi/3$ is equilateral. The ray $(-\infty,0)$ is mapped on a side. If you know one (oriented ) side, you know the equilateral triangle.