Elliptic Integral of the First Kind

Let the Modulus $k$ satisfy $0 < k^2 < 1$. (This may also be written in terms of the Parameter $m\equiv k^2$ or Modular Angle $\alpha \equiv \sin^{-1} k$.) The incomplete elliptic integral of the first kind is then defined as

$$F(\phi,k) = \int_0^\phi {d\theta\over\sqrt{1-k^2\sin^2\theta}}.$$ (1)

Let

$$t \equiv \sin\theta$$ (2)

$$dt = \cos\theta\,d\theta =\sqrt{1-t^2}\,d\theta$$ (3)

$$F(\phi,k) = \int_0^{\sin\phi} {1\over\sqrt{1-k^2t^2}}{dt\over \sqrt{1-t^2}}$$

$$= \int_0^{\sin\phi} {dt\over\sqrt{(1-k^2t^2)(1-t^2)}}.$$ (4)

Let

$$v \equiv \tan\theta$$ (5)

$$dv \equiv \sec^2\theta\,d\theta = (1+v^2)\,d\theta,$$ (6)

so the integral can also be written as

$$F(\phi,k) = \int_0^{\tan\phi}{1\over\sqrt{1-k^2{v^2\over 1+u^2}}} {du\over 1+v^2}$$

$$= \int_0^{\tan\phi}{dv\over\sqrt{1+v^2}\sqrt{(1+v^2)-k^2v^2}}$$ (7)

$$= \int_0^{\tan\phi}{dv\over\sqrt{(1+v^2)(1+k'v^2)}},$$ (8)

where $k'^2\equiv 1-k^2$ is the complementary Modulus.

The integral

$$I={1\over\sqrt{2}} \int_0^{\theta_0} {d\theta\over\sqrt{\cos\theta-\cos\theta_0}},$$ (9)

which arises in computing the period of a pendulum, is also an elliptic integral of the first kind. Use

$$\cos\theta = 1-2\sin^2({\textstyle{1\over 2}}\theta)$$ (10)

$$\sin({\textstyle{1\over 2}}\theta) = \sqrt{1-\cos\theta\over 2}$$ (11)

to write

$$\sqrt{\cos\theta-\cos\theta_0} = \sqrt{1-2\sin^2({\textstyle{1\over 2}}\theta)-\cos\theta_0}$$

$$= \sqrt{1-\cos\theta_0}\sqrt{1-{2\over 1-\cos\theta_0}\sin^2({\textstyle{1\over 2}}\theta)}$$

$$= \sqrt{2}\sin({\textstyle{1\over 2}}\theta_0)\sqrt{1-\csc^2({\textstyle{1\over 2}}\theta_0)\sin^2({\textstyle{1\over 2}}\theta)},$$ (12)

so

$$I={1\over 2}\int_0^{\theta_0} {d\theta\over\sin({\textstyle{... ...yle{1\over 2}}\theta_0)\sin^2({\textstyle{1\over 2}}\theta)}}.$$ (13)

Now let

$$\sin({\textstyle{1\over 2}}\theta)=\sin({\textstyle{1\over 2}}\theta_0)\sin\phi,$$ (14)

so the angle $\theta$ is transformed to

$$\phi=\sin^{-1}\left({\sin\theta\over\theta_0}\right),$$ (15)

which ranges from 0 to $\pi/2$ as $\theta$ varies from 0 to $\theta_0$. Taking the differential gives

$${\textstyle{1\over 2}}\cos({\textstyle{1\over 2}}\theta)\,d\theta=\sin({\textstyle{1\over 2}}\theta_0)\cos\phi\,d\phi,$$ (16)

or

$${\textstyle{1\over 2}}\sqrt{1-\sin^2({\textstyle{1\over 2}}\... ...\,d\theta=\sin({\textstyle{1\over 2}}\theta_0)\cos\phi\,d\phi.$$ (17)

Plugging this in gives

$$I = \int_0^{\pi/2} {1\over\sqrt{1-\sin^2({\textstyle{1\over 2}}\theta... ..._0)\cos\phi\,d\phi\over\sin({\textstyle{1\over 2}}\theta_0)\sqrt{1-\sin^2\phi}}$$

$$= \int_0^{\pi/2} {d\phi\over\sqrt{1-\sin^2({\textstyle{1\over 2}}\theta_0)\sin^2\phi}} = K(\sin({\textstyle{1\over 2}}\theta_0)),$$ (18)

so

$$I={1\over\sqrt{2}}\int_0^{\theta_0} {d\theta\over\sqrt{\cos\theta-\cos\theta_0}} = K(\sin({\textstyle{1\over 2}}\theta_0)).$$ (19)

Making the slightly different substitution $\phi=\theta/2$, so $d\theta=2\,d\phi$ leads to an equivalent, but more complicated expression involving an incomplete elliptic function of the first kind,

$$I = 2{1\over\sqrt{2}}{1\over\sqrt{2}}\csc({\textstyle{1\over 2}}\thet... ...\theta_0} {d\phi\over\sqrt{1-\csc^2({\textstyle{1\over 2}}\theta_0)\sin^2\phi}}$$

$$= \csc({\textstyle{1\over 2}}\theta_0)F({\textstyle{1\over 2}}\theta_0, \csc({\textstyle{1\over 2}}\theta_0)).$$ (20)

Therefore, we have proven the identity

$$\csc x F(x,\csc x)=K(\sin x).$$ (21)

The complete elliptic integral of the first kind, illustrated above as a function of $m=k^2$, is defined by

$$K(k) \equiv F({\textstyle{1\over 2}}\pi,k)$$ (22)

$$= \sum_{n=0}^\infty {(2n-1)!!\over (2n)!!} k^{2n} \int^{2\pi}_0 \sin^{2n}\theta \,d\theta$$ (23)

$$= {\textstyle{1\over 2}}\pi{\vartheta_3}^2(q)$$ (24)

$$= \sum_{n=0}^\infty {(2n-1)!!\over (2n)!!} k^{2n}{\pi\over 2}{(2n-1)!!\over (2n)!!}$$

$$= {\pi\over 2}\sum_{n=0}^\infty \left[{(2n-1)!!\over (2n)!!}\right]^2 k^{2n}$$ (25)

$$= {\textstyle{1\over 2}}\pi\, {}_2F_1({\textstyle{1\over 2}}, {\textstyle{1\over 2}}, 1; k^2)$$ (26)

$$= {\pi\over 2\sqrt{1-k^2}}P_{-1/2}\left({1+k^2\over 1-k^2}\right),$$ (27)

where

$$q=e^{-\pi K'(k)/K(k)}$$ (28)

is the Nome (for $\vert q\vert<1$), ${}_2F_1(a,b;c;x)$ is the Hypergeometric Function, and $P_n(x)$ is a Legendre Polynomial. $K(k)$ satisfies the Legendre Relation

$$E(k)K'(k)+E'(k)K(k)-K(k)K'(k)={\textstyle{1\over 2}}\pi,$$ (29)

where $E(k)$ and $K(k)$ are complete elliptic integrals of the first and Second Kinds, and $E'(k)$ and $K'(k)$ are the complementary integrals. The modulus $k$ is often suppressed for conciseness, so that $E(k)$ and $K(k)$ are often simply written $E$ and $K$, respectively.

The Derivative of $K(k)$ is

$${dK\over dk} \equiv \int_0^1 {dt \over \sqrt{(1-t^2)(1-k'^2t^2)}} = {E(k)\over k(1-k^2)} - {K(k)\over k}$$ (30)

$${d\over dk}\left({kk'^2 {dK\over dk}}\right)= kK,$$ (31)

so

$$E=k(1-k^2)\left({{dK\over dk}+{K\over k}}\right)=(1-k^2)\left({k {dK\over dk}+k}\right)$$ (32)

(Whittaker and Watson 1990, pp. 499 and 521).

See also Amplitude, Characteristic (Elliptic Integral), Elliptic Integral Singular Value, Gauss's Transformation, Landen's Transformation, Legendre Relation, Modular Angle, Modulus (Elliptic Integral), Parameter

References

Abramowitz, M. and Stegun, C. A. (Eds.). ``Elliptic Integrals.'' Ch. 17 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 587-607, 1972.

Spanier, J. and Oldham, K. B. ``The Complete Elliptic Integrals $K(p)$ and $E(p)$'' and ``The Incomplete Elliptic Integrals $F(p;\phi)$ and $E(p;\phi)$.'' Chs. 61-62 in An Atlas of Functions. Washington, DC: Hemisphere, pp. 609-633, 1987.

Whittaker, E. T. and Watson, G. N. A Course in Modern Analysis, 4th ed. Cambridge, England: Cambridge University Press, 1990.