From a translated version of this (page 111):

One can easily recognise that in this case, the lemniscatic integral $$u' = \int^t_0\frac{dt}{\sqrt{4t(1-t^2)}}$$ represents the interior of each of the two half-planes in which the plane is divided by the real axis conformally on the interior of a square with sides $\int^1_0\frac{dt}{\sqrt{4t(1-t^2)}}$.

I guess that it's saying that the integral conformally maps half-planes to a square with sides $\int^1_0\frac{dt}{\sqrt{4t(1-t^2)}}$, but I can't see the reason why.


1 Answer 1


Changing notation slightly (using $s$ as the dummy variable of integration, and dropping the prime on $u$, which in the original appears to denote a choice of lower limit of integration), the derivative of the mapping $$ u(t) = \int_{0}^{t} \frac{ds}{\sqrt{4s(1 - s^{2})}} $$ is $$ \frac{du}{dt} = \frac{1}{\sqrt{4t(1 - t^{2})}} = \frac{1}{\sqrt{4t(1 - t)(1 + t)}}. \tag{1} $$ The derivative is non-vanishing in the open upper half-plane, so $u$ is conformal in the open upper half-plane. Further, the derivative has singularities at $t = -1$, $0$, and $1$, at each of which the asymptotic behavior is $(t - t_{0})^{-1/2}$. Consequently, $u$ itself has a square root-like branch point at each of $t = -1$, $0$, and $1$. Particularly, $u$ maps straight angles to right angles at each branch point.

The singularities of the derivative split the real axis into four intervals: $$ I_{1} = (-\infty, -1),\quad I_{2} = (-1, 0),\quad I_{3} = (0, 1),\quad I_{4} = (1, \infty). $$ The sign of the radicand $4t(1 - t)(1 + t)$ changes each time $t$ crosses a branch point, so the square root is alternately real and imaginary. That is, the images of these intervals under $u$ are line segments, horizontal when $du/dt$ is real and vertical when $du/dt$ is imaginary.

Because each real interval $(a, b)$ maps to a line segment, the length of the image $u(a, b)$ is $$ \left|\int_{a}^{b} \frac{ds}{\sqrt{4s(1 - s^{2})}}\right|. $$ Particularly, $I_{3}$ maps to a segment of length $$ \ell = \int_{0}^{1} \frac{ds}{\sqrt{4s(1 - s^{2})}}. $$ The substitution $s = 1/r$ gives $ds = -dr/r^{2}$, so by change of variables, the image of $I_{4}$ has length $$ \left|\int_{1}^{\infty} \frac{ds}{\sqrt{4s(1 - s^{2})}}\right| = \left|\int_{1}^{0} \frac{-dr}{r^{2}\sqrt{(4/r)(1 - r^{-2})}}\right| = \left|\int_{0}^{1} \frac{dr}{\sqrt{4r(r^{2} - 1)}}\right| = \ell. $$ Similarly, the images of $I_{1}$ and $I_{2}$ have length $\ell$.

Putting this all together, $u$ maps the boundary of the upper half plane to a square of length $\ell$, and the map is conformal in the open half-plane.


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