Integral of x/(1+x^4) problem (in Mathematica30)
- To: mathgroup at smc.vnet.net
- Subject: [mg21674] Integral of x/(1+x^4) problem (in Mathematica30)
- From: Jan Krupa <krupa at alpha.sggw.waw.pl>
- Date: Sat, 22 Jan 2000 02:52:28 -0500 (EST)
- Organization: http://news.icm.edu.pl/
- Sender: owner-wri-mathgroup at wolfram.com
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).