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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,

J. M. ain't a mathematician
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Marie. P.
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    Note: $w(z)$ is one of the canonical examples of an elliptic integral, and the inverse function of $w(z)$ is in fact one of Jacobi's elliptic functions. There should be a discussion of this particular Schwarz-Christoffel mapping in McKean and Moll's book. – J. M. ain't a mathematician Dec 02 '11 at 00:14
  • See also the discussion here. – J. M. ain't a mathematician Dec 05 '11 at 02:23
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    Hi @Guesswhoitis., I studied the section in McKean's book just now, and I understand how the real line, traces out the perimeter of a rectangle. However, how do we know that the UHP is necessarily mapped to the interior of the rectangle? This doesn't seem obvious to me, for some reason. Is it just simply testing a point from the UHP, e.g., integrating the above integral from 0 to +i, and try to show that +i actually gets mapped to the interior of the rectangle? Thanks, – User001 Jul 18 '15 at 01:23

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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.

Bruno Joyal
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