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I would like to know whether there is any solution available on the inversion of elliptic integrals of the third kind (incomplete)?

That means that given $\Pi(n,u,m) = f(x)$, I would like to obtain $u$ as a function of $x$. Here, $m = (sin(\alpha))^2$ is the parameter and $n$ is the characteristic.

I know how to invert elliptic integrals of the first kind (incomplete), but I haven't been able to find anything useful in order to invert a general elliptic integral of the third kind (incomplete, non-circular case).

Thanks for any help!

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2 Answers 2

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I don't know if you are still interested, but here you go:

  1. Write $f(x)=\Pi(n,u,m)$ as a series $f(x)=S(u)$ in $u$ around $u_0=0$.
  2. Inverse/reverse that series $S(u)$ to the reverse series $T(v)$ with the property that $T(S(u))=u$. Then insert $f(x)$ into $T(v)$ to get a series representation $u=T(f(x))$ of $u$.

This can be done with help of a computer algebra system like Mathematica or Maple. A numerical approach is given here: T. Fukushima: Numerical Inversion of General Incomplete Elliptic Integral

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Yes, but here is only the idea since using the Riemann Theta function is complicated. A rational expression of the function inverts an Abelian integral as introduced here:

$$\int \frac {dx}{\sqrt{\sum\limits_{n=0}^d a_nx^n}}\tag1$$

and Ellipic Pi $\Pi(a,t,b)$ has this integral representation in the Wolfram language:

$$\Pi(a,\sin^{-1}(x),b)=\int\frac{dx}{\sqrt{1-x^2}(1-ax^2)\sqrt{1-bx^2}}=\int\frac{dx}{\sqrt{1-(2a+b+1)x^2+(a^2+2a+b)x^4-(a^2b+a^2+2ab)x^6+a^2bx^8}}$$

which fits the general form in $(1)$. Therefore, it is invertible in terms of the Riemann Theta function.

Inversion of hyperelliptic integrals of arbitrary genus with application to particle motion in general relativity

shows how to use the function for the more advanced user.

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