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MathGroup Archive 2005

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Re: Question about an integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54455] Re: [mg54381] Question about an integral
  • From: DrBob <drbob at bigfoot.com>
  • Date: Sun, 20 Feb 2005 00:11:13 -0500 (EST)
  • References: <200502190732.CAA06170@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Integrate[1/x^2, {x, 2, A}]
(-2 + A)*If[Re[A] >= 0 ||
     Im[A] != 0, 1/(2*A),
    Integrate[1/(2 + (-2 + A)*x)^
       2, {x, 0, 1},
     Assumptions ->
       !(Re[A] >= 0 || Im[A] !=
         0)]]

Simplify[%, A > 0]
1/2 - 1/A

That's the wrong answer if A < 0, at least, so it's not odd if Mathematica
doesn't return it for unknown A.

Bobby

On Sat, 19 Feb 2005 02:32:53 -0500 (EST), richard speers <rspeers at skidmore.edu> wrote:

> The previous version of Mathematica integrated  1/x^2 from x=2 to x=A
> very nicely, giving 1/2-1/A.  The new
> version won't do the integral.  Please help.
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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