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


Is this a bug in the docs?

  • $\begingroup$ Same here on 8.0.4/Win7-64 $\endgroup$ Jan 30, 2012 at 20:36
  • $\begingroup$ You can always submit any issues you find with the documentation to support@wolfram.com. $\endgroup$
    – Searke
    Jan 31, 2012 at 21:24
  • $\begingroup$ @Searke Done. :) $\endgroup$
    – JxB
    Jan 31, 2012 at 22:13

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$ Jan 30, 2012 at 20:58
  • $\begingroup$ @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". $\endgroup$
    – celtschk
    Jan 30, 2012 at 21:06
  • $\begingroup$ 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. $\endgroup$
    – Spawn1701D
    Jan 30, 2012 at 22:43

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.


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