Re: Integral of x/(1+x^4) problem (in Mathematica30)
- To: mathgroup at smc.vnet.net
- Subject: [mg21702] Re: Integral of x/(1+x^4) problem (in Mathematica30)
- From: "Kevin J. McCann" <kevin.mccann at jhuapl.edu>
- Date: Sat, 22 Jan 2000 02:53:13 -0500 (EST)
- Organization: Johns Hopkins University Applied Physics Lab, Laurel, MD, USA
- References: <86aamp$8ho@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
If you plot
f[x_]=-(1/2) ArcTan[1/x^2]
and
g[x_]=0.5*ArcTan[x^2)]
you will see that they differ by Pi/4. Also
Limit[f[x],{x->0}] = -Pi/4
The two answers differ by an integration constant, and are thus
equally good answers. Depends on you criterion for "goodness".
I tend to agree with you on this, but mathematically it makes no
difference.
Kevin
--
Kevin J. McCann
Johns Hopkins University APL
Jan Krupa <krupa at alpha.sggw.waw.pl> wrote in message
news:86aamp$8ho at smc.vnet.net...
> It is well known that antiderivative of x/(1+x^4) is
> 0.5*arctan(x^2) over the interval (-oo,oo).
>
> I tried it in mathematica3.0:
>
> Integrate[x/(1+x^4), x] gives
>
> -(1/2) ArcTan(1/x^2) which is correct but
> only for (-oo,0) or (0,oo).
>
> Of course one can define function F(x)= -(1/2) ArcTan(1/x^2) when x is
> not
> equal 0 and F(x)=0 when x=0.
>
> F(x)=0.5*ArcTan(x^2)+Pi/4 (because we have ArcTan(1/a)+ArcTan(a)=Pi/2
> for x not eq 0. )
>
> How one can force Mathematica to achieve rather the better result
> 0.5*ArcTan(x^2) than -0.5*ArcTan(1/x^2) ?
>
> Jan
>
> P.S. MuPAD also gives -0.5*ArcTan(1/x^2).
>
>