Re: Re: Re: Integrate vs NIntegrate

*To*: mathgroup at smc.vnet.net*Subject*: [mg89206] Re: [mg89182] Re: [mg89147] Re: [mg89129] Integrate vs NIntegrate*From*: DrMajorBob <drmajorbob at att.net>*Date*: Thu, 29 May 2008 07:04:38 -0400 (EDT)*References*: <9489155.1211802216995.JavaMail.root@m08> <op.ubr1yjh52c6ksp@bobbys-imac> <2b8d8f40805261936u2ec81454r1b6ba16be0897662@mail.gmail.com> <200805271114.HAA01380@smc.vnet.net> <29744585.1211975587467.JavaMail.root@m08>*Reply-to*: drmajorbob at longhorns.com

> Not a proof, of course, but I would bet a fair amount on it [Integrate] > being right. I believe someone recently said we should ignore predictions from those who haven't staked actual money on it. But anyway, is Integrate correct in BOTH these calculations? q = -1/2; a = 0; b = 3; h[x_] = (1 + x^3)^q; f[x_] = Integrate[h[x], x]; f[b] - f[a] // N -2.55387 - 2.42865 I f[x_] = Integrate[h[t], {t, 0, x}]; f[b] - f[a] // N 1.65267 Along with the OP, I'm betting the first result from Integrate is wrong, in view of NIntegrate[h[x], {x, a, b}] 1.65267 If NIntegrate is wrong instead, that hardly improves matters. Bobby On Wed, 28 May 2008 03:46:27 -0500, Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > I would say that Integrate is correct. Evidence: > > In[1]:= q = -2^(-1); > > In[2]:= f = Integrate[(1 + x^3)^q, x]; > > In[3]:= g = Chop[N[D[Normal[f + O[x]^12], x]]] > > Out[3]= -0.3125000000000013*x^9 + 0.37500000000000105*x^6 - > 0.5000000000001258*x^3 + > 0.9999999999999998 > > and, on the other hand: > > In[4]:= N[Normal[(1 + x^3)^q + O[x]^10]] > Out[4]= -0.3125*x^9 + 0.375*x^6 - 0.5*x^3 + 1. > > > Not a proof, of course, but I would bet a fair amount on it being right. > > Andrzej Kozlowski > > > On 27 May 2008, at 20:14, DrMajorBob wrote: > >> Because they're completely different. One is an antiderivative in >> Complex >> variables; the other is a definite integral from 0 to x (so you know >> that >> it is zero at zero, among other things). Antiderivatives tend to have >> branch cuts, poles, etc., and Mathematica's chosen branch is complex >> in >> part of your region. >> >> Or... it's entirely possible Mathematica's answer for >> Integrate[h[x], x] >> is simply WRONG. >> >> Let's check whether it has the correct derivative: >> >> q = -1/2; a = 0; b = 3; >> h[x_] = (1 + x^3)^q; >> f[x_] = Integrate[h[x], x]; >> FullSimplify[f'[x] == h[x], a < x < b] >> >> Sqrt[-(1 + x) (-1 + (-1)^(1/3) x)] == ( >> 2 (-1)^(1/12) Sqrt[-(-1)^(1/6) ((-1)^(2/3) + x) (1 + x^3)])/( >> Sqrt[1 - I/Sqrt[3] + (-1 - I/Sqrt[3]) x] Sqrt[ >> 3 + I Sqrt[3] + I (3 I + Sqrt[3]) x]) >> >> As you see, FullSimplify doesn't think Integrate got a correct >> result. Or >> at least, it couldn't verify it. >> >> Here's an attempt to provide FullSimplify some adult supervision: >> >> simple= FullSimplify[ #, a<x<b]&; >> test= f'[x]== h[x]//simple >> >> Sqrt[(-(1 + x))*(-1 + (-1)^(1/3)*x)] == >> (2*(-1)^(1/12)*Sqrt[(-(-1)^(1/6))*((-1)^(2/3) + x)*(1 + x^3)])/ >> (Sqrt[1 - >> I/Sqrt[3] + (-1 - I/Sqrt[3])*x]*Sqrt[3 + I*Sqrt[3] + I*(3*I + >> Sqrt[3])*x]) >> >> test2 = simple[test /. {(-1)^(1/3) -> -1, (-1)^(2/3) -> >> ComplexExpand[Exp[I*(Pi/3)]], (-1)^(1/6) -> ComplexExpand[Exp[I*(Pi/ >> 6)]], >> (-1)^(1/12) -> ComplexExpand[Exp[I*(Pi/12)]]}] >> >> 1 + x == (((1/2 + I/2)*(3 - I*Sqrt[3])*Sqrt[(-(2*I + (I + >> Sqrt[3])*x))*(1 >> + x^3)])/Sqrt[3 - I*Sqrt[3] + (-3 - I*Sqrt[3])*x])*Sqrt[3 + I*Sqrt[3]= >> + I*(3*I + Sqrt[3])*x] >> >> test3 = test2 /. (z_)^(r_) :> ComplexExpand[z]^r >> >> 1 + x == (((1/2 + I/2)*(3 - I*Sqrt[3])*Sqrt[(-Sqrt[3])*x - Sqrt[3= ]*x^4 >> + I*(-2 - x - 2*x^3 - x^4)])/Sqrt[3 - 3*x + I*(-Sqrt[3] - >> Sqrt[3]*x)])*Sqrt[3 - 3*x + I*(Sqrt[3] + Sqrt[3]*x)] >> >> simple[(#1^2 & ) /@ test3] >> >> 2*I + (3*I + Sqrt[3])*x == 0 >> >> I don't think Integrate has the right answer. >> >> Bobby >> >> On Mon, 26 May 2008 21:36:31 -0500, Armen Kocharyan >> <armen.kocharyan at gmail.com> wrote: >> >>> Dear Bob, >>> >>> Sorry for the asterisks, it wasn't intentional. I just cut and paste= >>> from my >>> notebook. >>> >>> I understand. But why would Mathematica give different results for >>> the >>> following? >>> >>> Integrate[h[t], {t, 0, x}] and Integrate[h[x], x] >>> >>> >>> Regards, >>> Armen >>> >>> 2008/5/27 DrMajorBob <drmajorbob at att.net>: >>> >>>> The fundamental theorem of calculus frequently does not apply >>>> when, as >>>> in >>>> your example, Integrate is free to return a Complex-valued function=

**References**:**Re: Integrate vs NIntegrate***From:*DrMajorBob <drmajorbob@att.net>