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

**References**:**Question about an integral***From:*richard speers <rspeers@skidmore.edu>