Riemann surfaces: computing $f(z) = \int_0^z \frac{dx}{\sqrt{P(x)}}$

Question

I am trying to validate that the Schwartz-Christoffel mapping does indeed take the upper-half plane $\mathbb{H} = \{ z: \mathrm{Re}(z) > 0\}$ to a polygon.

This involves integrals of functions of the type and computing at every point.

$$ f(z) = \int_0^z \frac{dx}{\sqrt{(x-a)(x-b)}}$$

where $a < b \in \mathbb{R}$.

I would like the case $a = 1,b=-1$. I know this integral can be computed exactly, but not if the denominator $P(z)$ were a cubic or quartic like $x^4 - 1$.

Here is a numpy script I wrote... sorry if it's wrong.

import numpy as np
impott matplotlib.pyplot as plt

f = lambda z: np.cumsum(z, axis=1) + 1j*np.cumsum(z, axis=0)    
n = 200
dt = 1.0/n
y = (np.ones((n,2*n)))/n
y = f(y)-1
w = y[(y.imag > 0.001)] # semicircle of radius 1

g = lambda t: np.sqrt(1 - t**2)**-0.5 # function to be integrated

plt.plot((g(w)).real, g(w).imag, 'b.')

plt.axis("Equal")
plt.show()

Here's the picture I came up with... defintely not a polygon. Maybe I have to change the original mesh?

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I don't want to compute the antiderivative exactly, since I would like it to work for more complex examples like $\int_0^z \frac{dx}{\sqrt{(x-1)(x-2)(x-3)}}$. Instead I proposed two strategy:

• build a mesh representing values of $\frac{1}{\sqrt{P(z)}}$
• transform the mesh into the values of $\int_0^z$ at each point
• if the roots of $P(z)$ are real and we stick to $\mathrm{Re}(z)> 0$ the square-root is still single-valued at least

Answer 1

Note that the integral is $$ \int^z \frac{dx}{\sqrt{x-a}\sqrt{x-b}}, $$ and in general for complex variables $$ \sqrt{x-a}\sqrt{x-b} \not\equiv \sqrt{(x-a)(x-b)}. $$

I'm not sure what you were trying to do with your code, so I can't help you there. I don't see where in your code you do the integration, it seems to plot some other function entirely: the issue is not with the mesh. When I did your example in Mathematica, $a=-b=1$, and I got this:

-((I \[Pi])/2) + 2 ArcSinh[Sqrt[-1 + z]/Sqrt[2]] /. z -> x + I y;
ParametricPlot[{Re@%, Im@%}, {x, -8, 8}, {y, 0, 8}, PlotRange -> {{-1, 4}, {-2, 2}}]