# Integral of the Square of the Elliptic Integral

Someone must know a good technique for

$$\int E^{2}(x)dx$$

Where $E$ is the complete elliptic integral of the second kind: $$E\left(k\right)=\int_{0}^{\frac{\pi}{2}}d\theta\sqrt{1-k^{2}\sin^{2}\left(\theta\right)}$$

This is essentially the working for another integral I posted (Simple Integral Involving the Square of the Elliptic Integral) though this is valuable in its own right.

• Is this particularly difficult?
– apg
Aug 28, 2014 at 22:51
• If you are just interested in a careful approximation of the integral, it is worth mentioning that $$E(x)^2 \geq\frac{\pi^2}{16}\left(\frac{3}{2}(2-x)+\sqrt{1-x}\right),$$ with a difference less than $0.0001$ for $x\in[-1/2,1/2]$. Aug 28, 2014 at 23:13
• Do you know where this comes from?
– apg
Aug 29, 2014 at 0:06
• The complete elliptic integral satisfies certain differential equations that leads to strong bounds. They are too long to fit in a comment, but I would be glad to give details as an "approximate answer" to your question, if you want. Aug 29, 2014 at 0:11
• Thanks for this. If anyone can do it exactly that'd be good, though for now I'll accept the top answer.
– apg
Aug 29, 2014 at 0:57

We have that the function: $$B(\lambda)=\sum_{n=0}^{+\infty}\left(\frac{1}{(2n-1)4^n}\binom{2n}{n}\right)^2\lambda^{2n} = \frac{1}{\pi}\int_{0}^{\pi}\sqrt{1+\lambda^2+2\lambda\cos(2\theta)}\,d\theta$$ satisfies the relation: $$B(\lambda) = 2\frac{1+\lambda}{\pi}\cdot E\left(\frac{4\lambda}{(1+\lambda)^2}\right)\tag{1}$$ and the differential equation: $$B = \left(\lambda+\frac{1}{\lambda}\right)\frac{dB}{d\lambda}+\left(1-\lambda^2\right)\frac{d^2 B}{d\lambda^2}. \tag{2}$$ Since all the coefficients of the Taylor series of $B(\lambda)$ are positive, $B(\lambda)$ is a convex function, as well as its derivatives. Now $(2)$ gives: $$\frac{B'}{B}(\lambda)\leq \frac{\lambda}{\lambda^2+1},\tag{3}$$ so: $$B(\lambda)\leq\sqrt{1+\lambda^2}.\tag{4}$$ Moreover, $$\frac{B'}{B}(\lambda)= \frac{\lambda}{(\lambda^2+1)+\frac{1}{1+\lambda\cdot\frac{B''}{B'}(\lambda)}},\tag{5}$$ hence: $$B(\lambda)\geq\sqrt{1+\frac{\lambda^2}{2}}.\tag{6}$$ Rearranging $(4)$ and $(6)$ through $(1)$ we get: $$\frac{\pi}{4}\,\sqrt{\frac{3}{2}\,(2-m)+\sqrt{1-m}}\leq E(m) \leq \frac{\pi}{2}\,\sqrt{1-\frac{m}{2}},\tag{7}$$ hence $E^2(m)$ can be very well approximated with continued-fraction-arguments that comes from $(2)$.
• Oh, sure. $$E(m)=\int_{0}^{\pi/2}\sqrt{1-m\sin^2\theta}\,d\theta.$$ Aug 29, 2014 at 1:08
If you're looking for good numerical approximations, then for example \eqalign{- 0.000038839155&+ ( 2.472698342+ ( - 0.1185303776+ (\cr & 0.599575302+ ( - 4.237027543+ ( 9.643963778+ (\cr - & 12.22813840+ ( 8.085033824- 2.189508102\,x ) x ) x ) x ) x ) x ) x ) x} is an optimal degree $8$ polynomial approximation on $[0,1]$, with maximum error approximately $0.000038949$.