Re: Integral giving complex answer
- To: mathgroup at smc.vnet.net
- Subject: [mg59324] Re: Integral giving complex answer
- From: David Sagan <dcs16 at cornell.dot.edu>
- Date: Fri, 5 Aug 2005 01:23:02 -0400 (EDT)
- Organization: Cornell University
- References: <dcples$6mm$1@smc.vnet.net> <dcsbf7$ptn$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
I am using 5.0. Apparently, judging from the other posts, this got fixed
in 5.1
-- David
In article <dcsbf7$ptn$1 at smc.vnet.net>,
Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com> wrote:
> David Sagan wrote:
> > Hello:
> >
> > I am tring to do simple integrals but I am running into problems in that
> > Mathematica gives the answer using complex numbers. For example,
> > Integrate[1/(1 + 5x^2), x] gives a result in terms of logarithms of a
> > complex argument instead of the usual arctan formula. If I integrate
> > something like Integrate[1/(1 + a x^2), x] I get the answer in the form
> > I want using the arctan.
> >
> > My question is how to avoid getting the answer to Integrate[1/(1 +
> > 5x^2), x] in terms of complex logarithms. I know I could just integrate
> > 1/(1 + a x^2) and substitute a -> 5 later but in actuality I am dealing
> > with more complex integrals and it would be helpful if I did not have to
> > be making such substitutions.
> >
> > -- Thanks for any help, David Sagan
> >
> Hi David,
>
> What version/platform are you using? Here what I get with Mathematica
> 5.2 and 5.1.1 on Windows Xp (no complex logarithm, just arctan in both
> cases):
>
> In[1]:=
> $Version
>
> Out[1]=
> "5.2 for Microsoft Windows (June 20, 2005)"
>
> In[2]:=
> Integrate[1/(1 + 5*x^2), x]
>
> Out[2]=
> ArcTan[Sqrt[5]*x]/Sqrt[5]
>
> In[3]:=
> Integrate[1/(1 + a*x^2), x]
>
> Out[3]=
> ArcTan[Sqrt[a]*x]/Sqrt[a]
> In[1]:=
> $Version
>
> Out[1]=
> "5.1 for Microsoft Windows (January 27, 2005)"
>
> In[2]:=
> Integrate[1/(1 + 5*x^2), x]
>
> Out[2]=
> ArcTan[Sqrt[5]*x]/Sqrt[5]
>
> In[3]:=
> Integrate[1/(1 + a*x^2), x]
>
> Out[3]=
> ArcTan[Sqrt[a]*x]/Sqrt[a]
>
> Best regards,
> /J.M.