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	↑	→ #294 Infinite recursion during integration

#293 integrate 1/(1+x^2)

Edit detail for #293 integrate 1/(1+x^2) revision 2 of 2

	1 2
Editor: flyboy788 Time: 2009/06/19 12:31:37 GMT-7
Note:

added:

From flyboy788 Fri Jun 19 12:31:37 -0700 2009
From: flyboy788
Date: Fri, 19 Jun 2009 12:31:37 -0700
Subject: 
Message-ID: <20090619123137-0700@axiom-wiki.newsynthesis.org>

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

fricas

(1) -> integrate(1/(1+x^4),x=%minusInfinity..%plusInfinity)

$\label{eq1}\frac{\pi \ {\sqrt{2}}}{2}$

(1)

Type: Union(f1: OrderedCompletion?(Expression(Integer)),...)

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

fricas

integrate(1/(1+x^4),x)

$\label{eq2}\frac{{{\sqrt{2}}\ {\log \left({{x \ {\sqrt{2}}}+{{x}^{2}}+ 1}\right)}}-{{\sqrt{2}}\ {\log \left({-{x \ {\sqrt{2}}}+{{x}^{2}}+ 1}\right)}}+{2 \ {\sqrt{2}}\ {\arctan \left({\frac{{\sqrt{2}}+{2 \ x}}{\sqrt{2}}}\right)}}-{2 \ {\sqrt{2}}\ {\arctan \left({\frac{{\sqrt{2}}-{2 \ x}}{\sqrt{2}}}\right)}}}{8}$

(2)

Type: Union(Expression(Integer),...)

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,

fricas

integrate((x^4+2*a*x^2+1)^-1, x=0..%plusInfinity, "noPole")

$\label{eq3}0$

(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:

fricas

D(integrate((x^4+2*a*x^2+1)^-1, x=0..b, "noPole"), b)

$\label{eq4}\frac{1}{{{b}^{4}}+{2 \ a \ {{b}^{2}}}+ 1}$

(4)

Type: Expression(Integer)

another similar evaluation --kratt6, Thu, 10 Jan 2008 01:28:46 -0800 reply

The following also seems to be wrong:

fricas

ex := integrate(1/(a+x**4),x=0..%plusInfinity, "noPole")

   There are no library operations named ** 
      Use HyperDoc Browse or issue
                                 )what op **
      to learn if there is any operation containing " ** " in its name.

   Cannot find a definition or applicable library operation named ** 
      with argument type(s) 
                                 Variable(x)
                               PositiveInteger

      Perhaps you should use "@" to indicate the required return type, 
      or "$" to specify which version of the function you need.

compare with

fricas

integrate(1/(1+x**4),x=0..%plusInfinity, "noPole")

   There are no library operations named ** 
      Use HyperDoc Browse or issue
                                 )what op **
      to learn if there is any operation containing " ** " in its name.

   Cannot find a definition or applicable library operation named ** 
      with argument type(s) 
                                 Variable(x)
                               PositiveInteger

      Perhaps you should use "@" to indicate the required return type, 
      or "$" to specify which version of the function you need.

... --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)

... --flyboy788, Fri, 19 Jun 2009 12:31:37 -0700 reply

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