In studying the Schwarz-Christoffel mapping, I have come across certain integrals that I want to numerically integrate. For example, one such integral is:
$\int_{0}^{1} (z^2 - z)^{-2/3} dz$
A cursory look at a graph of the integrand by Desmos shows this:
The integral is symmetric about $x = 1/2$; however, it has singularities at both endpoints. Moreover, on a glance it seems that these endpoints will allow the integral to converge.
However, upon evaluation of the definite integral using different programs like Wolfram-Alpha, I get answers which are imaginary, which could not be the case since the integral is on the real line. Wolfram-Alpha gives $-2.64996 - 4.58986 i$. I assume it is because the roots that are evaluated using these programs are not the real roots.
Is there a way for me to transform the integral for the singularities to disappear? I attempted shifting the curve so that the singularities lie at $x = -1/2$ and $x = 1/2$. However, I realize that the problem with integration over the endpoints still exist. I planned on using quadrature algorithms such as Gaussian and Clenshaw-Curtis to carry out the numerical integration. Any help would be appreciated.
Split the computation in the middle at $1/2$ and use partial integration to remove the singularity. $$ \int_0^{1/2}(z-z^2)^{-2/3}dz = \Bigl[3z^{1/3}(1-z)^{-2/3}\Bigr]_0^{1/2}+2\int_0^{1/2}z^{1/3}(1-z)^{-5/3}dz $$ The first term has a finite value, and the second has a continuous and bounded integrand, so that standard numerical methods should give a reliable result.
You should get real results from WA if you make the radicant positive, that is, replace $(z^2-z)^{-2/3}$ with $-(z-z^2)^{-2/3}$. (The graph is wrong in that light, values should be negative.) Remember that CAS are aware of complex numbers and will always use the main branch of the root, $w^a$ evaluates to $\exp(a\,{\rm Ln}(w))=\exp(a\ln|w|+i\,a\arg(w))$. In this case this leads to a constant factor $\frac{1+i\sqrt3}2$ relative to the real result, that is the real result is twice the real part of the WA value.