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Tried using $x=a\sin\left(\theta\right)$ $$\rightarrow \intop_{0}^{\pi/2}\sqrt{\frac{a^{2}-\left(a\sin\left(\theta\right)\right)^{2}}{1-\left(a\sin\left(\theta\right)\right)^{2}}}a\cos{\left(\theta\right)d\theta}$$ $$\iff \intop_{0}^{\pi/2}\frac{a^{2}\cos^{2}\left(\theta\right)}{\sqrt{1-\left(a\sin\left(\theta\right)\right)^{2}}}{d\theta}$$

Which looks similar to the complete elliptical of first or second kind, is there a way to make the conversion? Thanks

pikachu
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  • It is an incomplete elliptic integral of the second kind: $aE(a,1/a)$. – Gary Jul 25 '21 at 06:35
  • thanks @Gary, I am after a two term approximation for the original integral and thought I might able to convert it and then approximate, is there a way to obtain an approximation without converting? – pikachu Jul 25 '21 at 06:51
  • For what range of $a$ would you like to approximate? – Gary Jul 25 '21 at 06:58
  • Of course, forget to add: fro $a\rightarrow0$ and for $a\rightarrow1$ – pikachu Jul 25 '21 at 07:06
  • @Gary. I don't know where I have a mistake but I have $a E\left(\sin ^{-1}(a)|\frac{1}{a^2}\right)$. – Claude Leibovici Jul 25 '21 at 07:11
  • @ClaudeLeibovici I used a different defintion. I should have written $aE(a;1/a)$, see https://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_second_kind – Gary Jul 25 '21 at 07:46
  • @Gary. This problem of notation with elliptic integrals !! Thanks for clarifying. – Claude Leibovici Jul 25 '21 at 07:59
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    The Taylor expansion about $a=0$ is easy: $$ \int_0^{\pi /2} {\frac{{a^2 \cos ^2 t}}{{\sqrt {1 - a^2 \sin ^2 t} }}dt} = \frac{\pi }{4}a^2 \left( {1 + \frac{1}{8}a^2 + \frac{3}{{64}}a^4 + \frac{{25}}{{1024}}a^6 + \frac{{245}}{{16384}}a^8 + \cdots } \right). $$ – Gary Jul 25 '21 at 08:17
  • I'm also trying to approximate $\intop_{a}^{1}\sqrt{\frac{x^{2}-a^{2}}{1-x^{2}}}dx$ and having trouble because of the integration points, I get: $$ \intop_{\pi/2}^{\sin^{-1}\left(1/a\right)}\frac{a^{2}\cos^{2}\left(\theta\right)}{\sqrt{1-\left(a\sin\left(\theta\right)\right)^{2}}}{d\theta}$$ – pikachu Jul 25 '21 at 09:40
  • When $a\rightarrow0$ from above, the first term in the Taylor expansion $\cos^{2}\left(x\right)$ explodes – pikachu Jul 25 '21 at 10:50
  • We have $$ \int_a^1 {\sqrt {\frac{{x^2 - a^2 }}{{1 - x^2 }}} dx} = \sqrt {1 - a^2 } \left( {1 - \frac{{a^2 }}{{16}} + \cdots } \right) - \tanh ^{ - 1} \sqrt {1 - a^2 } \left( {\frac{{a^2 }}{2} + \frac{{a^4 }}{{16}} + \cdots } \right) $$ as $a\to 0+$. – Gary Jul 26 '21 at 07:08
  • @Gary is this also by expanding around 1 and integrating term by term? I get a different series – pikachu Jul 26 '21 at 11:54
  • Note that this is near $a=0$ and not $a=1$. Also I realised that the factor $\sqrt{1-a^2}$ in the front can itself be expanded using the binomial series. – Gary Jul 26 '21 at 11:55
  • Yes, around 0 I get: $\sqrt{\frac{-x^2}{x^2 - 1}} -a^2 \sqrt{\frac{-x^2/(x^2 - 1)}{2x^2 }} +...$ – pikachu Jul 26 '21 at 12:05
  • I expanded the integrand around $a=0$ and integrated term-by-term with respect to $x$ from $a$ to $1$. After simplification, I arrived at the formula above. – Gary Jul 26 '21 at 12:21

2 Answers2

1

You can convert this integral in terms of elliptic functions. You need to make an assumption about the parameter $a$:

If $0<a<1$ we can continue with your approach:

First, note

$$ J=\intop_{0}^{\pi/2} \frac{\sin^{2}\left(\theta\right)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta} = \frac{K(k)-E(k)}{k^2}$$

To show this recall the definition of the complete elliptic integrals of the first and second kind, respectively:

$$K(k) = \intop_{0}^{\pi/2} \frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta}$$

$$E(k) = \intop_{0}^{\pi/2} \sqrt{1-k^2\sin^2\left(\theta\right)}{d\theta}$$

Hence

$$\frac{1}{k^2}\left[K(k)-E(k)\right] = \frac{1}{k^2} \intop_{0}^{\pi/2} \left[\frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}} - \sqrt{1-k^2\sin^2\left(\theta\right)}\right]{d\theta} =\frac{1}{k^2} \intop_{0}^{\pi/2} \left[\frac{1}{\sqrt{1-k^2\sin^2\left(\theta\right)}} - \frac{1-k^2\sin^2(\theta)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}\right]{d\theta} = \intop_{0}^{\pi/2} \frac{\sin^{2}\left(\theta\right)}{\sqrt{1-k^2\sin^2\left(\theta\right)}}{d\theta} $$

Then

$$ I = \intop_{0}^{\pi/2}\frac{a^{2}\cos^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} = a^2\intop_{0}^{\pi/2}\frac{1-\sin^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} = a^2\intop_{0}^{\pi/2}\frac{1}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} - a^2\intop_{0}^{\pi/2}\frac{\sin^{2}\left(\theta\right)}{\sqrt{1-a^2\sin^2\left(\theta\right)}}{d\theta} = a^2K(a) -K(a)+E(a) =E(a)-a'^2K(a)$$

where $a' = \sqrt{1-a^2}$ is the complementary modulus

Hence

$$\boxed{\intop_{0}^{a}\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}dx= E(a)-a'^2K(a)}$$ Note that Wolfram use a slight different notation for Elliptic integrals.

There are some nice approximations for the complete Elliptic integrals.

$$ K(k) \approx \frac{\pi}{2} \left(\frac{16-5k^2}{16-9k^2}\right) \quad 0\leq k\leq 0.67$$

$$ E(k) \approx \frac{\pi}{2} \left(\frac{16-7k^2}{16-3k^2}\right) \quad 0\leq k \leq 0.71$$

and you can find others when $k$ is close to $1$.

You can also find series expansions for both functions.

Bertrand87
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0

Approximating the elliptic integral seems to be difficult.

What you could do is to expand the integrand around $a=1$ $$\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}=1+\frac{a-1}{1-x^2}-\frac{(a-1)^2 x^2}{2 \left(x^2-1\right)^2}-\frac{(a-1)^3 x^2}{2 \left(x^2-1\right)^3}-\frac{(a-1)^4 \left(x^2 \left(x^2+4\right)\right)}{8 \left(x^2-1\right)^4}+O\left((a-1)^5\right)$$ and integrate termwise to have for the integral (without simplifications)

$$a+(a-1) \tanh ^{-1}(a)+\frac{(a-1) \left(\left(a^2-1\right) \tanh ^{-1}(a)+a\right)}{4 (a+1)}-$$ $$\frac{1}{16} (a-1)^3 \left(\tanh ^{-1}(a)-\frac{a \left(a^2+1\right)}{\left(a^2-1\right)^2}\right)+\frac{(a-1) \left(-9 a^5+40 a^3+9 \left(a^2-1\right)^3 \tanh ^{-1}(a)+9 a\right)}{384 (a+1)^3}$$

Gary
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Claude Leibovici
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  • This is about $a=1$ and not $a=0$, right? – Gary Jul 25 '21 at 07:48
  • @Gary. Yes, it is around $a=1$ – Claude Leibovici Jul 25 '21 at 08:00
  • @ClaudeLeibovici is this kind of analysis also valid for around a=0? Expanding around a=0 gives $\sqrt{\frac{x^{2}}{(x^{2}-1)}}-\frac{a^{2}\sqrt{\frac{x^{2}}{(x^{2}-1)}}}{2x^{2}}+...$ – pikachu Jul 25 '21 at 08:22
  • @pikachu The expansion about $a=0$ is easy, see my comment above. You expand $$ \frac{1}{{\sqrt {1 - a^2 \sin ^2 \theta } }} = \sum\limits_{n = 0}^\infty {( - 1)^n \binom{- 1/2}{n}a^{2n}\sin ^{2n} \theta } $$ and integrate term-by-term using the beta function. – Gary Jul 25 '21 at 08:25