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#293 integrate 1/(1+x^2)
  
last edited 14 years ago by flyboy788

Edit detail for #293 integrate 1/(1+x^2) revision 1 of 2
1 2
Editor: nhanlm0601
Time: 2009/06/19 12:31:37 GMT-7
Note: 
changed:
-
\begin{axiom}
integrate(1/(1+x^4),x=%minusInfinity..%plusInfinity)
\end{axiom}

is obviously wrong. UPDATE: *this* problem is fixed in FriCAS.

\begin{axiom}
integrate(1/(1+x^4),x)
\end{axiom}

is an antiderivative only for $x\not\in\{-\frac{\sqrt2}2,\frac{\sqrt2}2\}$. The correct antiderivative, i.e., which is defined on all of $\mathbb R$ can be obtained from the above expression by replacing $\atan\frac1z$ with $\frac\pi2-\atan z$. UPDATE: *this* problem is also fixed in FriCAS, although the result could be simpler.  For example, mathematica 5.2 gives::

  In[10]:= Integrate[1/(1+x^4), x]

  Out[10]= (-2 ArcTan[1 - Sqrt[2] x] + 2 ArcTan[1 + Sqrt[2] x] - 
 
                               2                         2
         Log[-1 + Sqrt[2] x - x ] + Log[1 + Sqrt[2] x + x ]) / (4 Sqrt[2])


However,

\begin{axiom}
integrate((x^4+2*a*x^2+1)^-1, x=0..%plusInfinity, "noPole")
\end{axiom}

is wrong, also in FriCAS.  However, the solution really depends on the value of a. Mathematica 5.2 gives::

  In[2]:= Integrate[(x^4+2*a*x^2+1)^-1, {x,0,Infinity}]

                                     2
  Out[2]= If[Im[Sqrt[-a - Sqrt[-1 + a ]]] > 0 && 
  
                              2
      Im[Sqrt[-a + Sqrt[-1 + a ]]] > 0, 
 
      -I                             2                        2
     (-- Pi) / (Sqrt[-a - Sqrt[-1 + a ]] Sqrt[-a + Sqrt[-1 + a ]] 
      2
 
                              2                          2
        (Sqrt[-a - Sqrt[-1 + a ]] + Sqrt[-a + Sqrt[-1 + a ]])), 
 
                      1
     Integrate[---------------, {x, 0, Infinity}, 
                        2    4
               1 + 2 a x  + x
 
      Assumptions -> 
 
                               2                                    2
       Im[Sqrt[-a - Sqrt[-1 + a ]]] <= 0 || Im[Sqrt[-a + Sqrt[-1 + a ]]] <= 0]
 
      ]

The definite version seems to be correct:
\begin{axiom}
D(integrate((x^4+2*a*x^2+1)^-1, x=0..b, "noPole"), b)
\end{axiom}

From kratt6 Thu Jan 10 01:28:46 -0800 2008
From: kratt6
Date: Thu, 10 Jan 2008 01:28:46 -0800
Subject: another similar evaluation
Message-ID: <20080110012846-0800@axiom-wiki.newsynthesis.org>

The following also seems to be wrong:
\begin{axiom}
ex := integrate(1/(a+x**4),x=0..%plusInfinity, "noPole")
eval(ex, a=1)
\end{axiom}
compare with
\begin{axiom}
integrate(1/(1+x**4),x=0..%plusInfinity, "noPole")
\end{axiom}

From dsadsadas Sun Sep 14 18:14:52 -0700 2008
From: dsadsadas
Date: Sun, 14 Sep 2008 18:14:52 -0700
Subject: 
Message-ID: <20080914181452-0700@axiom-wiki.newsynthesis.org>

Name: '#293 integrate (1/(1+x^4),x = %minusInfinity..%plusInfinity)' => '#293 integrate (1/(1+x^4)' 


From japp Tue Oct 14 06:02:32 -0700 2008
From: japp
Date: Tue, 14 Oct 2008 06:02:32 -0700
Subject: From Fricas
Message-ID: <20081014060232-0700@axiom-wiki.newsynthesis.org>

http://fricas.svn.sourceforge.net/viewvc/fricas/trunk/src/algebra/gaussian.spad.pamphlet?r1=257&r2=358&view=patch


From gdr Fri Oct 17 17:13:47 -0700 2008
From: gdr
Date: Fri, 17 Oct 2008 17:13:47 -0700
Subject: Patch applied to OpenAxiom
Message-ID: <20081017171347-0700@axiom-wiki.newsynthesis.org>

I'm not sure it solves the issue though.

From kratt6 Wed Nov 12 09:16:26 -0800 2008
From: kratt6
Date: Wed, 12 Nov 2008 09:16:26 -0800
Subject: 
Message-ID: <20081112091626-0800@axiom-wiki.newsynthesis.org>

Name: '#293 integrate 1/(1+x)^4' => '#293 integrate 1/(1+x^4)' 


From kratt6 Tue Jan 27 21:05:46 -0800 2009
From: kratt6
Date: Tue, 27 Jan 2009 21:05:46 -0800
Subject: 
Message-ID: <20090127210546-0800@axiom-wiki.newsynthesis.org>

Name: '#293 integrate 1/(4x^2-1)' => '#293 integrate 1/(1+x^4)' 


From faboabdo Wed Feb 25 10:53:23 -0800 2009
From: faboabdo
Date: Wed, 25 Feb 2009 10:53:23 -0800
Subject: 
Message-ID: <20090225105323-0800@axiom-wiki.newsynthesis.org>

Name: '#293 integrate 1/(1+x^4)' => '#293 integrate 1x/(1+x^4)' 


From nhanlm0601 Tue May 26 17:12:46 -0700 2009
From: nhanlm0601
Date: Tue, 26 May 2009 17:12:46 -0700
Subject: 
Message-ID: <20090526171246-0700@axiom-wiki.newsynthesis.org>

Name: '#293 integrate 1x/(1+x^4)' => '#293 integrate 1/(1+x^4)'

Submitted by : (unknown) at: 2007-11-17T22:22:23-08:00 (16 years ago)
Name :
Axiom Version :
Category : Severity : Status :
Optional subject :  
Optional comment :

axiom
integrate(1/(1+x^4),x=%minusInfinity..%plusInfinity)
LatexWiki Image(1)
Type: Union(f1: OrderedCompletion? Expression Integer,...)

is obviously wrong. UPDATE: this problem is fixed in FriCAS?.

axiom
integrate(1/(1+x^4),x)
LatexWiki Image(2)
Type: Union(Expression Integer,...)

is an antiderivative only for LatexWiki Image. The correct antiderivative, i.e., which is defined on all of LatexWiki Image can be obtained from the above expression by replacing LatexWiki Image with LatexWiki Image. UPDATE: this problem is also fixed in FriCAS?, although the result could be simpler. For example, mathematica 5.2 gives: 

  In[10]:= Integrate[1/(1+x^4), x]

  Out[10]= (-2 ArcTan[1 - Sqrt[2] x] + 2 ArcTan[1 + Sqrt[2] x] - 

                               2                         2
         Log[-1 + Sqrt[2] x - x ] + Log[1 + Sqrt[2] x + x ]) / (4 Sqrt[2])

However,

axiom
integrate((x^4+2*a*x^2+1)^-1, x=0..%plusInfinity, "noPole")
LatexWiki Image(3)
Type: Union(f1: OrderedCompletion? Expression Integer,...)

is wrong, also in FriCAS?. However, the solution really depends on the value of a. Mathematica 5.2 gives: 

  In[2]:= Integrate[(x^4+2*a*x^2+1)^-1, {x,0,Infinity}]

                                     2
  Out[2]= If[Im[Sqrt[-a - Sqrt[-1 + a ]]] > 0 && 

                              2
      Im[Sqrt[-a + Sqrt[-1 + a ]]] > 0, 

      -I                             2                        2
     (-- Pi) / (Sqrt[-a - Sqrt[-1 + a ]] Sqrt[-a + Sqrt[-1 + a ]] 
      2

                              2                          2
        (Sqrt[-a - Sqrt[-1 + a ]] + Sqrt[-a + Sqrt[-1 + a ]])), 

                      1
     Integrate[---------------, {x, 0, Infinity}, 
                        2    4
               1 + 2 a x  + x

      Assumptions -> 

                               2                                    2
       Im[Sqrt[-a - Sqrt[-1 + a ]]] <= 0 || Im[Sqrt[-a + Sqrt[-1 + a ]]] <= 0]

      ]

The definite version seems to be correct:

axiom
D(integrate((x^4+2*a*x^2+1)^-1, x=0..b, "noPole"), b)
LatexWiki Image(4)
Type: Expression Integer

another similar evaluation --kratt6, Thu, 10 Jan 2008 01:28:46 -0800 reply
The following also seems to be wrong:
axiom
ex := integrate(1/(a+x**4),x=0..%plusInfinity, "noPole")
LatexWiki Image(5)
Type: Union(f1: OrderedCompletion? Expression Integer,...)
axiom
eval(ex, a=1)
LatexWiki Image(6)
Type: Expression Integer

compare with

axiom
integrate(1/(1+x**4),x=0..%plusInfinity, "noPole")
LatexWiki Image(7)
Type: Union(f1: OrderedCompletion? Expression Integer,...)

... --dsadsadas, Sun, 14 Sep 2008 18:14:52 -0700 reply
Name: #293 integrate (1/(1+x^4),x = %minusInfinity..%plusInfinity) => #293 integrate (1/(1+x^4)

From Fricas --japp, Tue, 14 Oct 2008 06:02:32 -0700 reply
http://fricas.svn.sourceforge.net/viewvc/fricas/trunk/src/algebra/gaussian.spad.pamphlet?r1=257&r2=358&view=patch

Patch applied to OpenAxiom? --gdr, Fri, 17 Oct 2008 17:13:47 -0700 reply
I'm not sure it solves the issue though.

... --kratt6, Wed, 12 Nov 2008 09:16:26 -0800 reply
Name: #293 integrate 1/(1+x)^4 => #293 integrate 1/(1+x^4)

... --kratt6, Tue, 27 Jan 2009 21:05:46 -0800 reply
Name: #293 integrate 1/(4x^2-1) => #293 integrate 1/(1+x^4)

... --faboabdo, Wed, 25 Feb 2009 10:53:23 -0800 reply
Name: #293 integrate 1/(1+x^4) => #293 integrate 1x/(1+x^4)

... --nhanlm0601, Tue, 26 May 2009 17:12:46 -0700 reply
Name: #293 integrate 1x/(1+x^4) => #293 integrate 1/(1+x^4)