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Topic:
Goursat pseudoelliptics and the Wolfram Integrator
Replies:
5
Last Post:
Dec 18, 2012 12:15 PM




Re: Goursat pseudoelliptics and the Wolfram Integrator
Posted:
Dec 15, 2012 3:43 PM


"Nasser M. Abbasi" schrieb: > > On 12/14/2012 3:47 PM, clicliclic@freenet.de wrote: > > > > I was curious how Mathematica would handle the pseudoelliptic integrals > > recently referred to on this newsgroup. So I plugged two examples into > > the Wolfram Integrator (which presumably represents Mathematica 8): > > > > <http://integrals.wolfram.com/index.jsp> > > > > Example 1 (cubic radicand): > > > > Integrate[(k*x^2  1)/((a*k*x + b)*(b*x + a) > > *Sqrt[x*(1  x)*(1  k*x)]), x] > > > > ... eeeek! The Integrator replies in terms of incomplete elliptic F, > > incomplete elliptic Pi, and the imaginary unit. But the antiderivative > > just is: > > > > 2/(Sqrt[a*b]*Sqrt[(a + b)*(a*k + b)]) > > *ArcTan[Sqrt[a*b]*Sqrt[x*(1  x)*(1  k*x)] > > /(Sqrt[(a + b)*(a*k + b)]*x)] > > > > I tried this on version 9 of Mathematica. I get the same as with > the above web site you mention. > > But when I plot the result for some values of a,b,k and compare with > what you have above, the plots do not look the same? Not even a scaled > shift, but they look different (near origin). Away from origin, they > seem to become closer to each others. (i.e. for very large x) > Here is the results to see: > > http://12000.org/my_notes/misc_items/121512/integrals.html
To my eyes, the plotted results seem to agree up to a piecewise constant  so the Mathematica result should be ok since differentiation on Derive confirms the ArcTan antiderivative without problem.
> > I tried it on Maple16. Was not able to plot the result in > Maple, I get an empty plot. But Maple also gives result in > terms of incomplete elliptic functions also. > > > Example 2 (quartic radicand): > > > > Integrate[(k*x^2  1)/((a*k*x + b)*(b*x + a) > > *Sqrt[(1  x^2)*(1  k^2*x^2)]), x] > > > > ... "Mathematica could not find a formula for your integral. Most likely > > this means that no formula exists." Waouw! Here the elementary > > antiderivative is: > > > > 2/(Sqrt[(a + b)*(a*k + b)]*Sqrt[(a  b)*(a*k  b)]) > > *ArcTanh[Sqrt[(a + b)*(a*k + b)]*Sqrt[(1  x^2)*(1  k^2*x^2)] > > /(Sqrt[(a  b)*(a*k  b)]*(1  x)*(1  k*x))] > > > > I tried this also, yes, Mathematica 9 does not do it. > > Maple16 does solve it. It gives answers in terms of incomplete > elliptic as well. result is in the above link as well.
So Maple wins! The treatment of elliptic integrands in Mathematica is really ridiculous.
> > > The theory behind these integrals is given in: Edouard Goursat, Note sur > > quelques intégrales pseudoelliptiques, Bulletin de la Société > > Mathématique de France 15 (1887), 106120, online at: > > > > <http://www.numdam.org/item?id=BSMF_1887__15__106_1> > > > > This was written 125 years ago  apparently too recent for the "Risch" > > integrator of Mathematica 8. I expect that FriCAS can do the second > > integral too. How do Maple and Sympy behave? > > > > Do not use sympy, may be someone who has that can try them. >
Thanks for the results! And I hadn't known MMA9 was out.
Martin.



