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Re: INDEFINITE INTEGRALS
*To*: mathgroup at smc.vnet.net
*Subject*: [mg18069] Re: [mg18020] INDEFINITE INTEGRALS
*From*: "Andrzej Kozlowski" <andrzej at tuins.ac.jp>
*Date*: Tue, 15 Jun 1999 01:43:29 -0400
*Delivery-date*: Tue Jun 15 02:15:42 1999
*Sender*: owner-wri-mathgroup at wolfram.com
The reason that Mathematica gives these answers is that they are quite
correct, just they are not given in the form you expected. One can check
this:
In[1]:=
Integrate[1/(a+x^2),x]
Out[1]=
x
ArcTan[-------]
Sqrt[a]
---------------
Sqrt[a]
In[2]:=
TrigToExp[%]
Out[2]=
I x I x
I (Log[1 - -------] - Log[1 + -------])
Sqrt[a] Sqrt[a]
---------------------------------------
2 Sqrt[a]
You can see now that if a is negative, say -b^2 than Sqrt[a] is I*b and you
get the answer you expected. You can induce Mathematica to give you that
answer if from the start you use -b^2 in place of a
In[3]:=
Integrate[1/(-b^2+x^2),x]
Out[3]=
x
ArcTanh[-]
b
-(----------)
b
In[4]:=
TrigToExp[%]
Out[4]=
x x
-Log[1 - -] + Log[1 + -]
b b
-(------------------------)
2 b
--
Andrzej Kozlowski
Toyama International University
JAPAN
http://sigma.tuins.ac.jp
http://eri2.tuins.ac.jp
----------
>From: Blimbaum Jerry DLPC <BlimbaumJE at ncsc.navy.mil>
To: mathgroup at smc.vnet.net
>To: mathgroup at smc.vnet.net
>Subject: [mg18069] [mg18020] INDEFINITE INTEGRALS
>Date: Sun, Jun 13, 1999, 4:52 AM
>
> 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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