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Re: INDEFINITE INTEGRALS
*To*: mathgroup at smc.vnet.net
*Subject*: [mg18096] Re: INDEFINITE INTEGRALS
*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
*Date*: Thu, 17 Jun 1999 12:26:35 -0400
*Organization*: Universitaet Leipzig
*References*: <7jubb0$576@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Hi Jerry,
whats wron with ArcTan[x/Sqrt[a]]/Sqrt[a] it is the correct result.
The Log[-1+x]-Log[1+x] is simply a expanded version from the relation
ArcTan[x]==I Log[(1-I x)/(1+I x)]/x
(Abram. Stegun 4.4.28)
Regards
Jens
Blimbaum Jerry DLPC wrote:
>
> In order to see how well Mathematica does at integration, I tried
> several types of integrals where the values of the parameters changed the
> results. I compared the integrals for specific values of the parameters and
> then symbolically with assumptions. For example:
>
> Integrate[1/(1+x^2),x] vs. Integrate[1/(-1+x^2),x] and got the
> correct results of
>
> ArcTan[x] and 1/2 (Log[-1+x] -Log[1+x].>
> However, if i try the same thing except symbolically, namely
> Integrate[1/(a+x^2),x,Assumptions->{a>0}] and
> Integrate[1/(a+x^2),x,Assumptions->{a<0}] I got
> back ArcTan[x/Sqrt[a]]/Sqrt[a] for both.
> I tried variations on this theme using other functions as 1/(x
> Sqrt[a+x^2]), Sqrt[a x^2 + c], 1/Sqrt[c+a x^2], where I integrated them
> symbolically and then assumed a = -1 and +1. In all cases I got the same
> results as above.
>
> Am I doing this wrong or is something in Mathematica incomplete?
> The Mathematica book explicitly states that it checks for variation of the
> parameters, but these results dont seem to support that. (Mathematica
> 3.01)
>
> Thanks. Jerry Blimbaum NSWC Panama City, Florida
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