Here is the Wikipedia article on it:
https://en.wikipedia.org/wiki/Schwarz%E2%80%93Christoffel_mapping
I feel it doesn't make sense.
The integration produces real values (on the real line), so f(x) traces a curve on the real line to start. Then, at the crossing of a branch point, from left to right, one of the factors in the denominator experiences a sign change, so we pick up a factor of -1.
But -1 = $e^{i\pi}$ or $e^{-i\pi}$. So I feel the exponents for the mapping that traces out an equilateral triangle, which has interior angles = 60 degrees and 3 edges of equal length, should be 1/3, so that at the moment of the sign change, we pick up $e^{\frac{i\pi}{3}}$, which will show that the integrand turns by 60 degrees. But, according to the formula, the exponent should be 1-(alpha/pi), which would equal to 2/3, which doesn't make sense.
Any thoughts or suggestions would be greatly appreciated.
EDIT, so it turns out that the exponents should correspond to exterior angles of the image polygon - but why? For an equilateral triangle, the edge clearly has to turn 60 degrees, in order to get a 60 degree interior angle.
EDIT 2: I got it now, I think :-). But if anyone feels like writing up an answer, you're more than welcome!
Thanks,
You can visualizes this as follows. Consider the real line in the complex plane with $n$ marked points. Traverse the real line from left to right. As you cross the first marked point, we rotate the rest of the line after the mark by your exterior angle. The same for the next marked point, and so on. We eventually form the desired polygon after all these rotations of the line at these marks are done.
What's going on is that after $z > z_1$, the corresponding factor in $(z-z_1)^{\alpha_1} \cdots (z-z_n)^{\alpha_n}$ no longer contributes an angle, so we have an angle change of $\alpha_1$.
There's a very clear explanation of this idea in Brown and Churchill in the section on Schwarz-Christoffel mappings.