1
$\begingroup$

I want to plot the integral $$I(\phi) = \int_0^{\phi} \frac{\mathrm{d} \theta}{\sqrt{1 +\sin(\theta)^2}}$$

In Mathematica notation, it is a case of an elliptic integral of the first kind with $m=-1$,

$F(\phi| m) = \int_0^{\theta} \frac{\mathrm{d} \theta}{\sqrt{1-m \sin(\theta)^2}}$, so I have $I(\phi) = F(\phi| -1)$.

In Maple notation, it is a case of an elliptic integral of the first kind with $k = i$,

$F(\sin(\phi), k) = \int_0^{\theta} \frac{\mathrm{d} \theta}{\sqrt{1-k^2 \sin(\theta)^2}}$, so I have $I(\phi) = F(\sin(\phi), i)$.

Upon plotting $I(\phi)$ and $F(\phi| -1)$ in Mathematica I find perfect agreement.

Upon plotting $I(\phi)$ (in red) and $F(\sin(\phi), i)$ (in green) in Maple I get the result below:

Maple output

How can such a difference arise?

EDIT:

Code for Mathematica:

F1 = Integrate[1/Sqrt[1 + Sin[p]^2], {p, 0, phi}]
F2 = EllipticF[phi, -1]

Code for Maple:

F1:= int(1/sqrt(1+sin(theta)^2), theta=0..phi)
F2:= EllipticF(sin(phi), I)
Improve this question
$\endgroup$
  • 1
    $\begingroup$ People here generally like users to post code as Mathematica code instead of images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. $\endgroup$ – user9660 Apr 17 '16 at 17:24
  • 1
    $\begingroup$ (1) It seems the question may be properly be about the behavior of Maple, which is off-topic on this site. (2) The green graph looks like Re@EllipticF[ArcSin[Sin@\[Theta]], I] -- as Louis implies, no one can be sure about the problem with your code, if you don't share it. $\endgroup$ – Michael E2 Apr 17 '16 at 17:41
  • $\begingroup$ I edited to give my code. Thanks for your help! $\endgroup$ – Nigel1 Apr 17 '16 at 17:53
  • 2
    $\begingroup$ I don't have/know Maple, but from the Maple doc. here, it seems that Maple's EllipticF[z, k] is equivalent to EllipticF[ArcSin[z], k^2] in Mathematica, in which case the problem is due to the periodicity of sine and the branch cuts of arc sine. $\endgroup$ – Michael E2 Apr 17 '16 at 18:26
  • 2
    $\begingroup$ The main problem, as Michael notes, is that Maple and Mathematica are using different argument conventions. In particular, Maple's EllipticF() is in fact directly equivalent to Mathematica's InverseJacobiSN[]. EllipticF[] uses a different argument convention, and is built so as not to have unneeded branch cuts. $\endgroup$ – J. M.'s ennui♦ Apr 20 '16 at 0:39
4
$\begingroup$

In Mathematica notation

$Assumptions = ϕ ∈ Reals

F[ϕ_, m_] := 
Integrate[1/Sqrt[1 - m Sin[θ]^2], {θ, 0, ϕ}]

Plot[F[ϕ, -1], {ϕ, - Pi,  π}, PlotStyle -> Red]

enter image description here

In Maple notation

F[\[Phi]_, k_] := 
Integrate[1/Sqrt[1 - k^2 Sin[\[Theta]]^2], {\[Theta], 0, \[Phi]}]

Plot[F[\[Phi], I], {\[Phi], -2 Pi, 2 \[Pi]}]

enter image description here

In maple

enter image description here

Improve this answer
$\endgroup$
  • $\begingroup$ And when you plot $F(\sin(\phi),i)$ in Maple what do you get? I get the strange plot in the question. $\endgroup$ – Nigel1 Apr 17 '16 at 17:35
  • 1
    $\begingroup$ @Nigel1 same result, I edit my answer $\endgroup$ – vito Apr 17 '16 at 17:55
  • $\begingroup$ Thanks - but when you compare it to "built in" Maple EllipticK function with code: EllipticF(sin(phi), I) do you reproduce my plot in question or the Mathematica result? $\endgroup$ – Nigel1 Apr 17 '16 at 18:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.