Questions tagged [elliptic-integrals]

Questions on elliptic integrals, integrals that involve the square root of a cubic or quartic polynomial.

An elliptic integral is most generally defined as $$\int R\left(t,\sqrt{P(t)}\right)\,dx$$ where $R$ is a rational function and $P$ is a cubic or quartic polynomial with no repeated roots. They arise in many fields of mathematics and physics.

Every elliptic integral may be expressed in terms of three standard forms (arguments follow Mathematica/mpmath conventions):

  • The first kind: $$F(\varphi,m)=\int_0^\varphi\frac1{\sqrt{1-m\sin^2t}}\,dt$$
  • The second kind: $$E(\varphi,m)=\int_0^\varphi\sqrt{1-m\sin^2t}\,dt$$
  • The third kind: $$\Pi(n,\varphi,m)=\int_0^\varphi\frac1{(1-n\sin^2t)\sqrt{1-m\sin^2t}}\,dt$$

These incomplete integrals become complete when $\varphi=\frac\pi2$; their notations become $K(m),E(m)$ and $\Pi(n,m)$ respectively.

The inverse of $F(\varphi,m)$ for a fixed $m$ leads to the Jacobian .

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Asymptotic expression for the complete Elliptic integral of the first kind

On the Wikipedia page Elliptic integral it states, that the complete elliptic integral of the first kind has asymptotic expression $$K(k) =…
HarryT
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An identity for the complete elliptic integral of the first kind

On the Wolfram webpage, one can find the following identity for the complete elliptic integal of the first kind $K(z)$: $$K(z) = \frac{2}{1+\sqrt{1-z}} K\Big( \Big( \frac{1-\sqrt{1-z}}{1+\sqrt{1-z}} \Big)^2 \Big)$$ See…
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Elliptic Type Integral

I am trying to solve an integral of the type $\int\limits_{0}^{+\infty} x^a(x^2+b)^{-c} dx$, Where $a$, $b$ and $c$ are positive parameters. I am not sure whether this integral is doable. But I assume that if it is, the result should be a function…
Candidate
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Complete Elliptical Integral for the parameters greater than 1 or less than -1

I am trying to compute the Complete elliptical integral of second kind kind in Mathematica with Parameter m=-19.7 .Following is the response from Mathematica. Input:EllipticE[-19.71] Output:4.81841 I want to know if this is correct or rather how…
Tarun
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