Elliptic integral

An elliptic integral is an integral of a rational function of $x$ and $\sqrt{P(x)}$, where $P$ is a cubic or quartic polynomial with distinct roots. The standard first, second and third kinds describe the essential normal forms. Choosing the full endpoint yields complete versions such as the complete elliptic integral of the first kind.

For real modulus $0\le k\le 1$, the displayed integral is the incomplete elliptic integral of the second kind, with amplitude $\varphi$. The modulus convention matters: some sources instead use the parameter $m=k^2$. The arc length of an ellipse reduces to this integral, because a trigonometric parametrization gives a square root of $1-k^2{\sin }^2\theta$. At $k=0$ it reduces to $\varphi$; at $k=1$ the real integrand is $|\cos \theta |$.

For real $|k|<1$, define $E(\varphi ,k)={\int }_0^{\varphi }\sqrt{1-k^2{\sin }^{2}x}dx$ and $F(\varphi ,k)={\int }_0^{\varphi }(1-k^2{\sin }^{2}x{)}^{-1/2}dx$. Smoothness of the integrands permits differentiation under the integral sign. Subtracting the two integrands gives $E-F=-k^2{\int }_0^{\varphi }{\sin }^{2}x/\sqrt{1-k^2{\sin }^{2}x}dx$, which is $k{\partial }_{k}E$. This proves the displayed identity for $k\ne 0$. At $k=0$, both sides extend continuously with value zero. The variable $k$ is the modulus; derivatives with respect to the parameter $m=k^2$ instead obey ${\partial }_{m}E=(E-F)/(2m)$, with removable value $-\frac{1}{2}{\int }_0^{\varphi }{\sin }^{2}xdx$ at zero.

For $0\le k\le 1$, define $E(k)={\int }_0^{\pi /2}\sqrt{1-k^2{\sin }^2\theta }d\theta$. An ellipse with semi-major and semi-minor axes $a\ge b>0$ and eccentricity of a conic $e=\sqrt{1-b^2/a^2}$ has perimeter $4aE(e)$. Differentiating its parametrization gives the speed $a\sqrt{1-e^2{\sin }^2\theta }$, and fourfold symmetry reduces the full arc length to a quadrant. The binomial series gives $E(k)=(\pi /2)(1-k^2/4-3k^4/64+O(k^6))$ near zero.

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 $k^{\prime }=\sqrt{1-k^2}$,