Problem with EllipticE documentation

4

The complete elliptic integral of the second kind, EllipticE, is defined as,

Integrate[Sqrt[1-m Sin[t]^2],{t,0,z}]

According to the version 8 docs, the first of the "Possible Issues" is supposed to evaluate Integrate[Sqrt[1-m Sin[t]^2],{t,0,z}] as

If[(m Sin[z]^2 \[NotElement] Reals || 
    Re[m Sin[z]^2] <= 1) && (Csc[z]^2/m \[NotElement] Reals || 
    Re[Csc[z]^2/m] <= 0 || Re[Csc[z]^2/m] >= 1) && 
    2 + m Cos[2 z] != m, 
  EllipticE[z, m], 
  Integrate[Sqrt[1 - m Sin[t]^2], {t, 0, z}, Assumptions -> 
     2 + m Cos[2 z] == m || (((2 - m + m Cos[2 z]) Csc[z]^2)/m \[Element] Reals && -2 < 
   Re[((2 - m + m Cos[2 z]) Csc[z]^2)/m] < 0) || (Re[m Sin[z]^2] >
    1 && m Sin[z]^2 \[Element] Reals)]]

whereas I get simply

EllipticE[z,m]

Is this a bug in the docs?

JxB

Posted 2012-01-30T20:18:32.480

Reputation: 4 891

Same here on 8.0.4/Win7-64 – Sjoerd C. de Vries – 2012-01-30T20:36:19.910

You can always submit any issues you find with the documentation to support@wolfram.com. – Searke – 2012-01-31T21:24:28.600

@Searke Done. :) – JxB – 2012-01-31T22:13:31.263

Answers

3

I just tried it with both Mathematica 7 and 8, and Mathematica 7 gives the result from the documentation, while Mathematica 8 indeed gives just EllipticE[z, m].

Therefore I conclude Wolfram modified Integrate but forgot to update this piece of documentation.

celtschk

Posted 2012-01-30T20:18:32.480

Reputation: 18 543

Are you sure it's an improvement? The doc page says "The defining integral converges only under additional conditions". That sounds like it is describing a known property of the integral. I suppose that shouldn't change from version to version. – Sjoerd C. de Vries – 2012-01-30T20:58:39.723

@SjoerdC.deVries: I admit that I didn't read the documentation, but relied on the quote above. Since it was referred to as "Issue" I assumed ― wrongly, as it seems ― that it was a limitation of Integrate. OTOH it's not the only place where Mathematica makes implicit assumtions on integration: Integrate[x^n,x] gives x^(1 + n)/(1 + n), without any special case for n=-1. I now changed "improved" to "modified". – celtschk – 2012-01-30T21:06:22.087

You can see by checking than Elliptic[10,Pi] does not evaluate but Elliptic[10,0] does. If you see the conditions the first one do not meet them. So I assume the conditions are implemented inside the symbolic function. – Spawn1701D – 2012-01-30T22:43:59.527

1

As I have also commented above, this is merely an omission in updating the help file for the EllipticE. All the information about the integral is hidden behind the symbolic transcendental function EllipticE.

Spawn1701D

Posted 2012-01-30T20:18:32.480

Reputation: 1 771