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MathGroup Archive 1999

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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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