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MathGroup Archive 1999

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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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