How to Evaluate Complete Elliptic Integrals

Last Updated: December 21, 2018

Elliptic integrals are special functions that arise in many areas of mathematics and physics. In general, these functions cannot be written in terms of elementary functions. In this article, we evaluate the complete elliptic integrals of the first and second kinds in terms of power series.

It is recommended that you understand the Beta function and its related functions before proceeding.

Preliminaries

The complete elliptic integral of the first kind $K(k)$ arises when finding the period of a pendulum without the small-angle approximation. Note that some authors may choose to define it in terms of a modulus $m=k^{{2}}.$
- $K(k)=\int _{{0}}^{{1}}{\frac {{\mathrm {d}}t}{{\sqrt {(1-t^{{2}})(1-k^{{2}}t^{{2}})}}}}=\int _{{0}}^{{\pi /2}}{\frac {{\mathrm {d}}\phi }{{\sqrt {1-k^{{2}}\sin ^{{2}}\phi }}}}$
The complete elliptic integral of the second kind $E(k)$ arises when finding the arc length of an ellipse.
- $E(k)=\int _{{0}}^{{1}}{\sqrt {{\frac {1-k^{{2}}t^{{2}}}{1-t^{{2}}}}}}{\mathrm {d}}t=\int _{{0}}^{{\pi /2}}{\sqrt {1-k^{{2}}\sin ^{{2}}\phi }}{\mathrm {d}}\phi$

Steps

1
Set up the integral to be evaluated. We evaluate the complete elliptic integral of the first kind first; the second kind is not much different and uses the same techniques. We shall evaluate the trigonometric form, but note that Jacobi's form is a completely equivalent way of writing it.
- $K(k)=\int _{{0}}^{{\pi /2}}{\frac {{\mathrm {d}}\phi }{{\sqrt {1-k^{{2}}\sin ^{{2}}\phi }}}}$
2
Write the integral in terms of the binomial series.
- The binomial series is the Taylor expansion for the expression $(1+x)^{{\alpha }}$ for any real number $\alpha .$
  - $(1+x)^{{\alpha }}=\sum _{{m=0}}^{{\infty }}{\alpha \choose m}x^{{m}}$
- We can then write the integrand as such by identifying $x$ and $\alpha ,$ making sure to pull out any terms that are not dependent on $\phi .$
  - ${\begin{aligned}K(k)&=\int _{{0}}^{{\pi /2}}{\frac {{\mathrm {d}}\phi }{{\sqrt {1-k^{{2}}\sin ^{{2}}\phi }}}}\\&=\int _{{0}}^{{\pi /2}}\sum _{{m=0}}^{{\infty }}{-1/2 \choose m}(-1)^{{m}}k^{{2m}}\sin ^{{2m}}\phi {\mathrm {d}}\phi \\&=\sum _{{m=0}}^{{\infty }}{-1/2 \choose m}(-1)^{{m}}k^{{2m}}\int _{{0}}^{{\pi /2}}\sin ^{{2m}}\phi {\mathrm {d}}\phi \end{aligned}}$
- Notice that we are evaluating this integral term-by-term.
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3
Evaluate the integral using the Beta function.
- First, expand the binomial coefficients in terms of the Gamma function if necessary. Otherwise, leave it in terms of factorials. Remember that $m!=\Gamma (m+1).$
  - ${-1/2 \choose m}={\frac {\Gamma (1/2)}{m!\,\Gamma (1/2-m)}}$
- Second, recall the definition of the Beta function in terms of trigonometric functions.
  - ${\frac {\Gamma (\alpha )\Gamma (\beta )}{2\,\Gamma (\alpha +\beta )}}=\int _{{0}}^{{\pi /2}}\cos ^{{2\alpha -1}}\phi \sin ^{{2\beta -1}}\phi {\mathrm {d}}\phi$
- We identify $\alpha =1/2$ and $\beta =m+1/2.$
  - $K(k)=\sum _{{m=0}}^{{\infty }}{\frac {(-1)^{{m}}\Gamma (1/2)}{m!\,\Gamma (1/2-m)}}{\frac {\Gamma (1/2)\Gamma (1/2+m)}{2\,m!}}k^{{2m}}$
4
Use Euler's reflection identity and the fact that $\Gamma (1/2)={\sqrt {\pi }}$ .
- Euler's reflection identity is stated below.
  - $\Gamma (z)\Gamma (1-z)={\frac {\pi }{\sin(\pi z)}}$
- We can simplify our series using this formula if we let $z=1/2+m.$
  - $\Gamma (1/2+m)\Gamma (1/2-m)={\frac {\pi }{\sin(\pi (1/2+m))}}$
- We simplify further by making the observation that $(-1)^{{m}}\sin(\pi (1/2+m))=1$ for all $m.$
  - $K(k)={\frac {\pi }{2}}\sum _{{m=0}}^{{\infty }}{\frac {\Gamma ^{{2}}(1/2+m)}{\pi (m!)^{{2}}}}k^{{2m}}$
5
Use the double factorial identity.
- The double factorial identity can be related to the Gamma function in the following manner. See the tips for a derivation of this identity.
  - $(2m-1)!!={\frac {2^{{m}}\Gamma (1/2+m)}{{\sqrt {\pi }}}}$
- We can then simplify this series like so.
  - $K(k)={\frac {\pi }{2}}\sum _{{m=0}}^{{\infty }}\left[{\frac {(2m-1)!!}{2^{{m}}m!}}\right]^{{2}}k^{{2m}}$
- This series can also be written only with double factorials when making use of the identity $(2m)!!=2^{{m}}m!,$ which is sometimes encountered in the literature as well.
  - $K(k)={\frac {\pi }{2}}\sum _{{m=0}}^{{\infty }}\left[{\frac {(2m-1)!!}{(2m)!!}}\right]^{{2}}k^{{2m}}$
6
Expand the series.
- $K(k)={\frac {\pi }{2}}\left[1+\left({\frac {1}{2}}\right)^{{2}}k^{{2}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right)^{{2}}k^{{4}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right)^{{2}}k^{{6}}+\cdots \right]$
- The series has a few properties that immediately stand out. First, we can see that for small $k,$ the higher-order terms are suppressed, mainly due to the factorials. This is the justification for the small-angle approximation when analyzing a pendulum.
- Second, its region of convergence is $|k|<1.$ When $k=1,$ the integral diverges because the factorials cancel each other out in the large $m$ limit, although this divergence is very slow - $K(0.9999)\approx 5.645,$ for example.
- A physical example of when $k=1$ is when a pendulum is released from an angle of 180°, denoting an unstable equilibrium point. The period, being written in terms of this elliptic integral, then diverges, for the pendulum never falls down.
7
Verify the series for the complete elliptic integral of the second kind. Using the techniques presented in this article, the power series for this integral can also be found.
- $E(k)={\frac {\pi }{2}}\sum _{{m=0}}^{{\infty }}\left[{\frac {(2m-1)!!}{(2m)!!}}\right]^{{2}}{\frac {k^{{2m}}}{1-2m}}$
- $E(k)={\frac {\pi }{2}}\left[1-\left({\frac {1}{2}}\right)^{{2}}{\frac {k^{{2}}}{1}}-\left({\frac {1\cdot 3}{2\cdot 4}}\right)^{{2}}{\frac {k^{{4}}}{3}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right)^{{2}}{\frac {k^{{6}}}{5}}-\cdots \right]$
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Tips

In step 5, we used the double factorial identity that relates the double factorial to the gamma function $\Gamma (1/2+m).$
- This identity is not hard to derive, but we'll need Legendre's duplication formula. See this article for its derivation.
- ${\begin{aligned}{\frac {1}{\Gamma (1/2+m)}}&={\frac {2^{{2m}}}{{\sqrt {\pi }}}}{\frac {\Gamma (1+m)}{\Gamma (1+2m)}}\\\Gamma (1/2+m)&={\frac {{\sqrt {\pi }}}{2^{{2m}}}}{\frac {(2m)!}{m!}}\\&={\frac {{\sqrt {\pi }}}{2^{{m}}}}{\frac {(2m)(2m-1)\cdots (1)}{2^{{m}}(m)(m-1)\cdots (1)}}\end{aligned}}$
- Notice that $m!$ has $m$ terms in the product, which matches up with $m$ terms in $2^{{m}}.$ Therefore, all the terms of even parity cancel.
- ${\begin{aligned}\Gamma (1/2+m)&={\frac {{\sqrt {\pi }}}{2^{{m}}}}{\frac {(2m)(2m-1)(2m-2)(2m-3)\cdots (1)}{2(m)2(m-1)\cdots 2(1)}}\\&={\frac {{\sqrt {\pi }}}{2^{{m}}}}(2m-1)(2m-3)\cdots (1)\\&={\frac {{\sqrt {\pi }}}{2^{{m}}}}(2m-1)!!\end{aligned}}$
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How to Evaluate Complete Elliptic Integrals

Preliminaries

Steps

Expert Q&A

Tips

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