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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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