# An Unconventional Elliptic Integral?

I came across the following integral recently: $$I = \int_a^b d \lambda \sqrt{(\lambda^2-a^2)(b^2 - \lambda^2)}$$ The author of the paper claims that this integral can be transformed into an elliptic integral, giving the answer: $$I = b [(a^2 + b^2) E(k) - 2a^2 K(k)] \quad k^2 = \frac{b^2 - a^2}{b^2}$$ I tried to manipulate $I$ into one of the standard elliptic integrals, but I couldn't quite get it right. The most promising form I have obtained is via the substitution: $$\lambda = \sqrt{\frac{a^2 + b^2}{2}} \sin \theta$$ Which gives the integral: $$I = \sqrt{\frac{a^2 +b^2}{2}} \int d \theta \, \cos \theta \sqrt{\left(\frac{a^2-b^2} 2 \right)^2 - \left(\frac{a^2+b^2} 2\right)^2 \cos^4 \theta }$$ Am I on the right track? Is there some nice trick to evaluate this integral?

## 1 Answer

You have a typo in your expression, your expression is missing an overall factor $\frac13$.
You can verify that by comparing the integral and expression you have at $a \to 0$ and $b = 1$.
At that limit, $I \to \frac13$ while your expression $$b ((a^2 + b^2) E(k) - 2a^2 K(k))\quad\to\quad 1((0^2+1^2)\times 1 - 2\times 0 ) = 1$$

Let $c^2 = b^2 - a^2$ and $\displaystyle\;k = \frac{c}{b}$. Change variable to

$$u = \frac1c \sqrt{b^2 - \lambda^2} \quad\iff\quad \lambda = \sqrt{b^2 - c^2u^2} = b\sqrt{1-k^2u^2}$$

Notice $$\lambda^2 - a^2 = (b^2 - a^2) - (b^2 - \lambda^2) = c^2(1-u^2) \quad\text{ and }\quad d\lambda = -\frac{bk^2u du}{\sqrt{1-k^2u^2}}$$ The integral at hand equals to

$$I = \int_1^0 c^2u\sqrt{1-u^2}\left(-\frac{bk^2u du}{\sqrt{1-k^2u^2}}\right) = b^3k^4\int_0^1 \frac{u^2(1-u^2)}{\sqrt{(1-u^2)(1-k^2u^2)}}du$$ To evaluate this integral, one can use the fact (trick?)

$$\frac{d}{du}\left[u\sqrt{(1-u^2)(1-k^2u^2)}\right] = \frac{1 - 2u^2 + k^2u^2 - 3k^2u^2(1-u^2)}{\sqrt{(1-u^2)(1-k^2u^2)}}$$ and $u\sqrt{(1-u^2)(1-k^2u^2)}$ vanishes at $u = 0$ and $1$. With this, one can transform the integral to

$$I = \frac{b^3k^2}{3}\int_0^1\frac{1 - 2u^2 + k^2u^2}{\sqrt{(1-u^2)(1-k^2u^2)}} du$$

Notice $$1 - 2u^2 + k^2u^2 = 1 + (k^2-2)\frac{1 - (1-k^2u^2)}{k^2} = \frac{1}{k^2}\left((2-k^2)(1-k^2u^2) - (2-2k^2)\right)$$ We can simplify the integral to

\begin{align} I = &\frac{b^3}{3}\left[(2-k^2)\int_0^1\sqrt{\frac{1-k^2u^2}{1-u^2}}du - (2-2k^2)\int_0^1\frac{1}{\sqrt{(1-u^2)(1-k^2u^2)}}du\right]\\ = & \frac{b^3}{3}\left[(2-k^2)E(k) - (2-2k^2)K(k)\right]\\ = & \frac{b}{3}\left[(b^2+a^2)E(k) - 2a^2K(k)\right] \end{align} Up to a factor $\frac13$, this is the expression you have.

• Very nice solution, for sure ! – Claude Leibovici May 18 '17 at 6:05