Elliptic integral of the first kind ($F(z,k)$)

For a parameter $k$, one form of the incomplete elliptic integral of the first kind isIts value depends on the choices of square-root branches and, under analytic continuation around the branch points, on the integration path.

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 $B={\int }_0^{1}dt/\sqrt{1-t^4}$. 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 $F_1^{\prime }(w)=1/\sqrt{w(1-w^2)}$, is $2i/(1-z^4)$. Thus the two square maps differ by an affine change of coordinate.

For $0<k<1$, the complete elliptic integral of the first kind is

It is the complete first-kind member of the elliptic integral family.

Here $m$ is the parameter, equal to the squared modulus. Near $m=1$ the endpoint integrand is locally $(1-m+s^2{)}^{-1/2}$. Matching its integral to the outer secant integral gives $L$. The next correction is of order $(1-m)\log (1/(1-m))$. Keeping the parameter/modulus convention explicit prevents a factor-of-two error in the complementary small quantity.

For the complementary modulus $k^{\prime }=\sqrt{1-k^2}$,