[Date Index]
[Thread Index]
[Author Index]
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*: Fri, 21 Jan 2000 04:00:33 -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).
Prev by Date:
**Re: Header/Footer formatting question.**
Next by Date:
**Re: Plotting A verticle line?**
Previous by thread:
**Re: emmathfnt**
Next by thread:
**Integral of x/(1+x^4) problem (in Mathematica30)**
| |