Re: Integral of x/(1+x^4) problem (in Mathematica30)
- To: mathgroup at smc.vnet.net
- Subject: [mg21695] Re: [mg21674] Integral of x/(1+x^4) problem (in Mathematica30)
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Sat, 22 Jan 2000 02:52:58 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
You can get close to what you want simply by: In[1]:= FullSimplify[Integrate[x/(1+x^4), x]] Out[1]= 1 2 -(-) ArcCot[x ] 2 This is differs from the answer you wanted by Pi/2 and is also defined over the whole real axis. > From: Jan Krupa <krupa at alpha.sggw.waw.pl> > Organization: http://news.icm.edu.pl/ > Date: Fri, 21 Jan 2000 04:00:33 -0500 (EST) > To: mathgroup at smc.vnet.net > Subject: [mg21695] [mg21674] Integral of x/(1+x^4) problem (in Mathematica30) > > 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). > >