4

I am trying to integrate the integral $\displaystyle \int \dfrac{1}{\sqrt{1 - \beta \cos(\theta)}}d\theta$, where $0 \leq \beta \leq 1$.

I believe that this is the incomplete elliptic integral of the first kind. I have found several books on special functions that extensively covered the complete elliptic integral, but not so much for the incomplete elliptic integral.

I did recently stumble on the following formula online:

$\displaystyle \int \dfrac{1}{\sqrt{1 - \beta \cos(\theta)}}d\theta = \dfrac{2\sqrt{\dfrac{\beta \cos(\theta) -1}{\beta-1}}F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta - 1}\bigg)}{\sqrt{1 - \beta \cos(\theta)}}$

where $F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta - 1}\bigg)$ is the incomplete elliptic integral of the first kind.

My questions are:

(1) Is this formula correct, and where can I find more resources on this to further verify my integral?

(2) When $\beta = \dfrac{1}{2}$ then $F\bigg(\dfrac{\theta}{2}, \dfrac{2\beta}{\beta -1}\bigg)$, then the $\dfrac{2\beta}{\beta -1}$ is a negative value. Is that possible?

Any suggestion is much appreciated.

J. M. ain't a mathematician
  • 76,540
abc123
  • 51
  • 4
  • http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html – Jack D'Aurizio Oct 19 '17 at 17:36
  • Depends on how you define $F(.,.).$ Wolfram Alpha confirms your function (with the Wolfram definition), but Maple does not, because the arguments of $F(.,.)$ have different meanings, see http://functions.wolfram.com/EllipticIntegrals/EllipticF/02/0001/ and compare this to the link given by @jack-daurizio. The second argument is sometimes the modulus $k$ and sometimes the parameter $m.$ – gammatester Oct 19 '17 at 17:44
  • @JackD'Aurizio thank you, I did came across that site awhile back. To me it seems to define what an incomplete elliptical integral is, but not so much on how to actually integrate it. I'm hoping to integrate it as it will help answer a minimum question that I'm working on. – abc123 Oct 19 '17 at 17:47
  • @gammatester thank you for your suggestion. Wolfram did give me that nice formula, but I'm curious about how that formula came about as the answer (to me) is not that obvious. I will definitely go back and look at both yours and jack-daurizio suggestions. – abc123 Oct 19 '17 at 17:48

1 Answers1

3

Using

$$\cos\theta=1-2\sin^2\frac{\theta}{2}$$

we have

$$\int\frac1{\sqrt{1 - \beta \cos(\theta)}}\mathrm d\theta=\frac1{\sqrt{1-\beta}}\int\frac1{\sqrt{1 + \frac{2\beta}{1-\beta} \sin^2\left(\frac{\theta}{2}\right)}}\mathrm d\theta$$

from which the form of the incomplete elliptic integral of the first kind should already be apparent:

$$\frac1{\sqrt{1-\beta}}\int\frac1{\sqrt{1 + \frac{2\beta}{1-\beta} \sin^2\left(\frac{\theta}{2}\right)}}\mathrm d\theta=\frac2{\sqrt{1-\beta}}F\left(\frac{\theta}{2}\middle|\frac{2\beta}{\beta-1}\right)$$

(Note my use of the parameter instead of the modulus.)

Since you mention that $0\le\beta\le1$, we're not quite out of the woods yet, because in most practical computing environments, the modulus or parameter should be within $(0,1)$. To that end, apply the imaginary-modulus transformation to finally yield

$$\frac2{\sqrt{1-\beta}}F\left(\frac{\theta}{2}\middle|\frac{2\beta}{\beta-1}\right)=\frac2{\sqrt{1+\beta}}F\left(\arcsin\left(\sqrt{\frac{1+\beta}{1-\beta\cos\theta}}\sin\frac{\theta}{2}\right)\middle|\frac{2\beta}{1+\beta}\right)$$

J. M. ain't a mathematician
  • 76,540
  • yes, since $0 \leq \beta \leq 1$ was the concerning part for me as, if we use say $\beta = \dfrac{1}{2}$ then our value is negative which will violate the criteria for elliptic integral, thus this was the most troubling issue that I was trying to understand how it is possible for it be the incomplete elliptic integral. I am not familiar with imaginary-modulus transformation, but thanks for the pointer, I will definitely check it out. thanks! – abc123 Oct 25 '17 at 01:50
  • Can the imaginary modulus transformation be apply in any setting? Does it have to be in a complex space in order for this transformation to be useful, or can it be apply to any space? – abc123 Nov 05 '17 at 17:59
  • The imaginary modulus transformation is only intended for purely imaginary moduli (or equivalently, negative parameter values). – J. M. ain't a mathematician Nov 05 '17 at 20:33
  • Do you know of any references/resources on the imaginary modulus transformations that I can use? I've been looking but haven't found any good book beside the handbook for engineers. I know you have mention the website in your previous post, but I preferred books. thanks. – abc123 Nov 07 '17 at 22:53
  • @abc123, "I know you have mention the website in your previous post" - the DLMF is also a book, so I don't understand the pickiness over this. – J. M. ain't a mathematician Nov 07 '17 at 22:57
  • I apologize if my comment came off as nitpicking your response, that was not my intention. I should have phrased my question better: I have found a reference on the transformation in a book called handbook for engineer, but it only list the transformation. And I am interested in finding more info on the how and why of this transformation. Since special functions is a broad subject, I figure you're the best person to ask for suggestion on other books that perhaps go into more details than just stating what the transformation is. – abc123 Nov 08 '17 at 14:41
  • @abc, the Byrd/Friedman book is indeed a good reference. I don't have access to my books at the moment, you could try looking for what you need in Cayley or Greenhill. – J. M. ain't a mathematician Nov 08 '17 at 15:31