= Lemniscatic integral
{title2=$F(z)=\int_0^z(1-w^4)^{-1/2}\,dw$}
Choose the <holomorphic square root> equal to one at zero on the <unit disc>. The resulting <elliptic integral> maps that disc conformally onto the <square> with vertices $\pm B,\pm iB$, where $B=\int_0^1dt/\sqrt{1-t^4}$. For the inverse <Cayley transform between the half-plane and disk> $h(z)=i(1+z)/(1-z)$, the square of the <derivative> of $F_1\circ h$, with $F_1'(w)=1/\sqrt{w(1-w^2)}$, is $2i/(1-z^4)$. Thus the two <square> maps differ by an affine change of coordinate.
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