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