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). > >