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Re: A Problem with Simplify

  • To: mathgroup at smc.vnet.net
  • Subject: [mg87820] Re: A Problem with Simplify
  • From: Alexey Popkov <popkov at gmail.com>
  • Date: Fri, 18 Apr 2008 02:38:01 -0400 (EDT)
  • References: <ftkb7f$a9m$1@smc.vnet.net> <200804140943.FAA08090@smc.vnet.net>

Daniel Lichtblau:
> Sorry, I should have stated the situation more clearly.
>
> Yes, the antiderivatives you see have singularities at the parameter
> value in question (in all cases, when a is 1). Hence the results you
> show above. What I should have said is that one can get these
> singularities to cancel, if using the parametrized antiderivatives to
> obtain definite integrals a la Newton-Leibniz. Here is a standard simple
> example (modified slightly so as to have a->1 as the bad value).
>
> In[22]:= InputForm[i1 = Integrate[x^(-a),x]]
> Out[22]//InputForm= x^(1 - a)/(1 - a)
>
> Certainly it blows up as a->1.
>
> In[23]:= Limit[i1, a->1]
> Out[23]= -Infinity
>
> In[24]:= Limit[i1, a->1, Direction->1]
> Out[24]= Infinity
>
> But if we take a pair of values for x, say 2 and 3, to evaluate the
> definite integral, we in fact get the expected/desired logarithm result
> in the limit as a->1 (and independent of direction).
>
> In[27]:= InputForm[Limit[(i1/.x->3)-(i1/.x->2), a->1]]
> Out[27]//InputForm= Log[3/2]
>
> In[28]:= InputForm[Limit[(i1/.x->3)-(i1/.x->2), a->1, Direction->1]]
> Out[28]//InputForm= Log[3/2]
>
> To see that this really is reproducing the Fundamental Theorem of
> Calculus result (or at least doing a heckuva job to fool me), I'll show
> this instead as a definite integral from x0 to x.
>
> In[29]:= InputForm[Limit[i1 - (i1/.x->x0), a->1]]
> Out[29]//InputForm= Log[x] - Log[x0]
>
> I believe most of your examples will also behave fine when processed in
> this way. The Cos[a*z]/Sin[z] integrand might be an exception, but that
> indicates a limitation of Limit rather than a bug in Integrate.
> Depending on whether you give symbolic (z,z0) or exact numeric (3,2)
> input to Limit it either returns a complicated, but seemingly correct
> result, or else unevaluated.
>
>
> Daniel Lichtblau
> Wolfram Research

Thank you for the explanation. Now I understand. But I think that the
same way as there is an alternative to Solve (the function Reduce)
that carefully keeps track for all conditions for the all partial
answers, there should be also an alternative to Integrate that also
keeps track for conditions and gives the all partial answers! I am
grieved deeply that the alternative does not exists for this case!

In[1]:= Reduce[1/(a*x)==1,x]
Out[1]= a!=0&&x==1/a

This example is the true mathematics!


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