Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2000
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2000

[Date Index] [Thread Index] [Author Index]

Search the Archive

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).



  • Prev by Date: Re: Header/Footer formatting question.
  • Next by Date: Re: Product with p!=j
  • Previous by thread: Integral of x/(1+x^4) problem (in Mathematica30)
  • Next by thread: Re: Integral of x/(1+x^4) problem (in Mathematica30)