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MathGroup Archive 2004

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Re: Zero testing

  • To: mathgroup at smc.vnet.net
  • Subject: [mg53159] Re: Zero testing
  • From: Bill Rowe <readnewsciv at earthlink.net>
  • Date: Mon, 27 Dec 2004 06:41:29 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

On 12/25/04 at 4:01 AM, ab_def at prontomail.com (Maxim) wrote:

>I didn't claim that Integrate[1/((x - Pi/4)^2), x] was
>ComplexInfinity: my post says "Out[3] is also equal to ComplexInfinity
>(the denominator is zero) and therefore wrong as well", and you're
>talking about In[3]; I hope you can see the difference between the
>two. The denominator of Out[3] does equal zero, therefore the whole
>expression is identical to ComplexInfinity, which is why it is an
>incorrect result for Integrate[1/((x - Pi/4)^2), x].

Since

In[1]:=
a = Pi/4; b =  I*Log[-(-1)^(3/4)]; 
Limit[(a - b)/x, x -> 0]
Integrate[1/((x - a)*(x - b)), x]
Integrate[1/(x - b), {x, 0, 1}]
Out[2]=
DirectedInfinity[Pi - 4*I*Log[-(-1)^(3/4)]]
Out[3]=
-((4*(-2*I*ArcTan[Log[-(-1)^(3/4)]/x] -  2*Log[Pi - 4*x] + 
  Log[x^2 + Log[-(-1)^(3/4)]^2]))/
   (2*Pi - 8*I*Log[ -(-1)^(3/4)]))
Out[4]=
(1/2)*((-I)*Pi - 2*I*ArcTan[Log[-(-1)^(3/4)]] - 
   Log[Log[-(-1)^(3/4)]^2] + 
   Log[1 + Log[-(-1)^(3/4)]^ 2])
In[5]:=
FullSimplify[Out[3]]
Out[5]=
ComplexInfinity

That is applying FullSimplify to your Out[3] results in ComplexInfinity. So, it would seem the only way to assert Out[3] is "wrong" would be if Mathematica should be expected to automatically apply FullSimplify to the results of Integrate[1/((x - a)*(x - b)), x].

>Certainly I do not suggest that FullSimplify be applied at every step
>of the evaluation or anything of the sort. 

If you aren't suggesting FullSimplify to be applied to the result, how do you expect Mathematica to reduce an arbitrary expression to zero and reconize a zero in the denominator?
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