Re: integrate

*To*: mathgroup at smc.vnet.net*Subject*: [mg72453] Re: integrate*From*: "David W. Cantrell" <DWCantrell at sigmaxi.net>*Date*: Thu, 28 Dec 2006 05:36:00 -0500 (EST)*References*: <emgg3i$2ca$1@smc.vnet.net>

"dimitris" <dimmechan at yahoo.com> wrote: For your third question: > Integrate[Sqrt[x^2 + 1]/Sqrt[1 + x^6], {x, 0, 10}] > (-(-1)^(1/6))*EllipticF[I*ArcSinh[10*(-1)^(1/3)], (-1)^(2/3)] > > N[%] > -0.10016641038463325 - 1.6857503548125956*I > > NIntegrate[Sqrt[x^2 + 1]/Sqrt[1 + x^6], {x, 0, 10}] > 2.056349237110889 > > --------> Any ideas to "help" Mathematica to give a correct answer? I have an answer, but getting there was not pleasant. I hope others will have better suggestions. Part 1: I wanted to know why the symbolic answer was wrong, so I looked at. In[7]:= Integrate[Sqrt[(1 + x^2)/(1 + x^6)], x] Out[7]= (-(-1)^(1/6))*Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[1 + (-1)^(2/3)*x^2]* Sqrt[1/(1 - x^2 + x^4)]*EllipticF[I*ArcSinh[(-1)^(1/3)*x], (-1)^(2/3)] The product of the algebraic factors involving x should be just 1, but I was not able to get Mathematica to give that: In[8]:= FullSimplify[Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[1 + (-1)^(2/3)*x^2]* Sqrt[1/(1 - x^2 + x^4)], x > 0] Out[8]= Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[(1 + (-1)^(2/3)*x^2)/(1 - x^2 + x^4)] Maybe I didn't try hard enough. Can someone suggest how I should get the above to simplify to 1, perhaps assuming x > 0? Anyway, I then simplified Out[7] myself to In[9]:= -(-1)^(1/6)*EllipticF[I*ArcSinh[(-1)^(1/3)*x], (-1)^(2/3)] and then plotted the real part of that from x = 0 to 10. A jump discontinuity is apparent, at roughly x = 2.45. If we ignore that discontinuity and try to evaluate the integral from x = 0 to 10 by the Fundamental Theorem, we get the same wrong answer that Mathematica got. So now I think I know what Mathematica did wrong. But I'm a bit surprised. Even when Mathematica returns antiderivatives having such discontinuities, it normally takes such discontinuities correctly into account when doing _definite_ integrals. But it didn't do so here, so there is indeed a bug to be exterminated. Part 2: I thought I would help Mathematica by taking the discontinuity, caused by a branch cut, into account by hand. But unfortunately it wasn't obvious exactly where the cut was. So I went to the Wolfram Functions site and looked under EllipticF. It was not helpful. The only relevant statement was "Branch cut locations: complicated." I then gave up on determining the exact location. But if I could have done that, it would have been nicer than what follows. (Anyone know exactly where the discontinuity is?) Part 3: I sought to transform the original integral into another one which Mathematica might handle better. My first thought was to use the substitution u = x^2, which led to In[16]:= 1/2*Integrate[Sqrt[(1 + u)/(u*(1 + u^3))], u] Out[16]= (-1)^(1/6)*Sqrt[1 - (-1)^(1/3)/u]*Sqrt[1 + (-1)^(2/3)/u]*u^(3/2)* Sqrt[1/(u - u^2 + u^3)]*EllipticF[I*ArcSinh[(-1)^(1/3)/Sqrt[u]],(-1)^(2/3)] Simplifying by hand and putting the result back in terms of x gave In[17]:= (-1)^(1/6)*EllipticF[I*ArcSinh[(-1)^(1/3)/x], (-1)^(2/3)]. Although different from In[9], this result still has a discontinuity. But thankfully, it's in a different place, roughly x = 0.4. That allows us to combine In[9] and In[17], using the former to integrate from x0 to 1 and the latter to integrate from 1 to x1. Thus, assuming x0 < 1 < x1, we can state that Integrate[Sqrt[(1 + x^2)/(1 + x^6)], {x, x0, x1}] is In[21]:= int = (-1)^(1/6)*(-2*EllipticF[I*ArcSinh[(-1)^(1/3)], (-1)^(2/3)] + EllipticF[I*ArcSinh[(-1)^(1/3)*x0], (-1)^(2/3)] + EllipticF[I*ArcSinh[(-1)^(1/3)/x1], (-1)^(2/3)]); In[22]:= int /. {x0 -> 0, x1 -> 10} Out[22]= (-1)^(1/6)*(EllipticF[I*ArcSinh[(1/10)*(-1)^(1/3)], (-1)^(2/3)] - 2*EllipticF[I*ArcSinh[(-1)^(1/3)], (-1)^(2/3)]) which is the precise symbolic result desired. In[23]:= N[%] Out[23]= 2.0563492371150103 + 3.3306690738754696*^-16*I David W. Cantrell

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