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Re: Question about an integral
*To*: mathgroup at smc.vnet.net
*Subject*: [mg54413] Re: [mg54381] Question about an integral
*From*: Murray Eisenberg <murray at math.umass.edu>
*Date*: Sun, 20 Feb 2005 00:08:36 -0500 (EST)
*Organization*: Mathematics & Statistics, Univ. of Mass./Amherst
*References*: <200502190732.CAA06170@smc.vnet.net>
*Reply-to*: murray at math.umass.edu
*Sender*: owner-wri-mathgroup at wolfram.com
$Version
5.1 for Microsoft Windows (October 25, 2004)
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)]]
Integrate[1/x^2, {x, 2, A}, Assumptions -> {A > 0}]
If[A > 2, 1/2 - 1/A, Integrate[1/x^2, {x, 2, A}, Assumptions -> A <= 2]]
Integrate[1/x^2, {x, 2, A}, Assumptions -> {0 < A <= 2}]
1/2 - 1/A
You didn't say to which previous version of Mathematica you refer, but
in 5.1, obviously Mathematica is being cautious -- and, I think (without
my carefully checking all cases), correct!
richard speers 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.
>
>
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
Lederle Graduate Research Tower phone 413 549-1020 (H)
University of Massachusetts 413 545-2859 (W)
710 North Pleasant Street fax 413 545-1801
Amherst, MA 01003-9305
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