I am looking into elliptic integrals and I am slightly confused about something. The material I'm using for reference is mainly Wikipedia and Higher Transcendental Functions II (page 306 onward).

The integral I have to calculate is, for a 4th-degree polynomial $P$ :

$$I_1 = \int^1_0 \frac{dx}{\sqrt{P(x)}}$$

which in the book is called $I_1$. I have read up on the Legendre transformation, which allows one via rational transformations to "write" said $P$ as $(1 - \xi^2)(1-k^2\xi^2) = \eta^2$, whatever that means. My understanding of it is that this $k^2$ parameter is somehow linked to the starting polynomial $P$, so that aforementionned integral can be computed as the complete elliptic integral of the first kind :

$$K(k) = \int^1_0 \frac{d\xi}{\sqrt{(1 - \xi^2)(1-k^2\xi^2)}}$$

I was able to successfully compute $k(P)$, but before I go any further I want to be sure that I am understanding this correctly. Do I really have $I_1 = K(k(P))$, just like that ? I know that I can easily compute $K(k(P))$ using the arithmetic-geometric mean, so that's not a problem.

  • $\begingroup$ If the birational transformation used to map $P(x)$ into $(1-\xi^2)(1-k^2\xi^2)$ leaves the endpoints of $(0,1)$ unchanged or swapped, yes, you just have $I_1=K(k(P))$. Otherwise you might get an incomplete elliptic integral of the first kind. $\endgroup$ – Jack D'Aurizio Aug 27 '18 at 19:42
  • $\begingroup$ This is something else I'm a little confused about. In the book (page 318), the transformation involves $x$ in the expression of both $\xi$ and $\eta$. I don't think I really understand how the transformation actually works. I'm trying to think of it as a change of variables in an integral, where replacing $x$ by $y = f(x)$ would just change the endpoints to $f(0)$ and $f(1)$, but it doesn't seem to be that easy here. $\endgroup$ – Matrefeytontias Aug 27 '18 at 20:37

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