5
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.