Schwarz-Christoffel mapping of the upper half-plane I am given the function $w(z)=\int_0^z \frac1{\sqrt{1-t^2} \sqrt{1-k^2t^2}}dz$ and shall show that this is mapping the upper half-plane onto a rectangle. We just discussed the Schwarz-Christoffel integral, and we can rewrite this as
$\int_0^z \frac1{\sqrt{t-1} \sqrt{t+1} \sqrt{kt-1} \sqrt{kt+1}}dt$, and since the exponents of the four factors are all $-1/2$, we have $\alpha_i-1=-1/2$ which tells us that we are dealing with four right angles.
But, isn't the map actually mapping the unit circle to the rectangle, and not the upper half-plane onto the rectangle? Where is my mistake?
Also, I want to show that the inverse function extends to a meromorphic function on $\mathbb C$. What is the trick here? I don't have any idea on it.
Best regards,
 A: As pointed out by J.M., this is an elliptic integral, whose inverse function is Jacobi's elliptic function $\text{sn}$.
As $z$ travels along the real line, the argument of $\sqrt{(1-t^2)(1-k^2t^2)}$ undergoes sudden $\pi/2$ changes as $z$ crosses the points $1,-1,1/k,-1/k$, provided $k$ is real. Hence the real line is mapped to the boundary of a rectangle of side lengths $\int_{-1}^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$ and $\int_1^{1/k}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$ (provided here that $|1/k|>1$). Hence no, it is not the unit disc which is mapped to a rectangle. However, the upper-half plane and unit disc are conformally equivalent and by a change of variables (eg. $z \mapsto (z-i)/(z+i)$) you can obtain a conformal map which does take the unit disc to the rectangle.
Proving that the inverse function extends to a meromorphic function on $\mathbb{C}$ can be done using the Lagrange inversion theorem.
A very detailed study of this integral can be found in Markushevish's centennial Theory of functions.
