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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.

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  • $\begingroup$ mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html $\endgroup$ Oct 19, 2017 at 17:36
  • $\begingroup$ 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 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 $m.$ $\endgroup$ Oct 19, 2017 at 17:44
  • $\begingroup$ @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. $\endgroup$
    – abc123
    Oct 19, 2017 at 17:47
  • $\begingroup$ @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. $\endgroup$
    – abc123
    Oct 19, 2017 at 17:48

1 Answer 1

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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

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

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  • $\begingroup$ 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! $\endgroup$
    – abc123
    Oct 25, 2017 at 1:50
  • $\begingroup$ 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? $\endgroup$
    – abc123
    Nov 5, 2017 at 17:59
  • $\begingroup$ The imaginary modulus transformation is only intended for purely imaginary moduli (or equivalently, negative parameter values). $\endgroup$ Nov 5, 2017 at 20:33
  • $\begingroup$ 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. $\endgroup$
    – abc123
    Nov 7, 2017 at 22:53
  • $\begingroup$ @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. $\endgroup$ Nov 7, 2017 at 22:57

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