       • 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>
• 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
>
>

--
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

```

• Prev by Date: Re: Why does Inverse[M] hesitate?
• Next by Date: Re: real telescopic sum becomes complex?