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Re: Integrate vs NIntegrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg89144] Re: [mg89129] Integrate vs NIntegrate
*From*: DrMajorBob <drmajorbob at att.net>
*Date*: Tue, 27 May 2008 07:13:31 -0400 (EDT)
*References*: <9489155.1211802216995.JavaMail.root@m08>
*Reply-to*: drmajorbob at longhorns.com
The fundamental theorem of calculus frequently does not apply when, as in
your example, Integrate is free to return a Complex-valued function.
(A better question is why you put in all those extraneous stars to inhibit
copy/pasting the code.)
Despite that, here's a slightly different approach where the FTC does
apply:
q = -1/2; a = 0; b = 3;
h[x_] = (1 + x^3)^q;
f[x_] = Integrate[h[t], {t, 0, x}]
If[x >= -1, (x^3)^(1/3) Hypergeometric2F1[1/3, 1/2, 4/3, -x^3],
Integrate[1/Sqrt[1 + t^3], {t, 0, x}, Assumptions -> x < -1]]
Res1 = N[f[b] - f[a]]
Res2 = NIntegrate[h[x], {x, a, b}]
1.65267
1.65267
Contrast that definition of f with yours:
Integrate[h[x], x]
(1/(3^(1/4) Sqrt[1 + x^3]))2 (-1)^(
1/6) Sqrt[-(-1)^(1/6) ((-1)^(2/3) + x)] Sqrt[
1 + (-1)^(1/3) x + (-1)^(2/3) x^2]
EllipticF[ArcSin[Sqrt[-(-1)^(5/6) (1 + x)]/3^(1/4)], (-1)^(1/3)]
If you plot both versions from a to b (use PlotRange->All to be sure),
you'll find that your version is complex on part of the range (between 2
and 3), but mine is real.
Bobby
On Mon, 26 May 2008 05:23:35 -0500, Armen Kocharyan
<armen.kocharyan at gmail.com> wrote:
> *q=-1/2; a=0; b=3;*
> *h[x_]=(1+x^3)^q;*
> *f[x_]=Integrate[h[x],x];*
> *Res1=N[f[b]-f[a]]*
> *Res2=NIntegrate[h[x],{x,a,b}]*
--
DrMajorBob at longhorns.com
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