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MathGroup Archive 2007

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Re: strange behavior of Integrate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg75302] Re: strange behavior of Integrate
  • From: dimitris <dimmechan at yahoo.com>
  • Date: Tue, 24 Apr 2007 03:28:38 -0400 (EDT)
  • References: <200704220910.FAA20220@smc.vnet.net><f0hv8o$acb$1@smc.vnet.net>

>The difference in result is a subtle interaction between use of caching
>for certain intermediate results, and use of time constraints in some
>places (nobably simplifications). In effect a simplification attempt
>might time out the first attempt but succeed in later tries due to
>having more intermediate computations precomputed and cached.

Very clear explanation. I learn some new things. Hence, I really
appreciate your response.

>Integrate does make (very limited) use of FullSimplify.

I didn't know this. Thanks for pointing me out.

Using a code by Chris Chiasson I see that Integrate does use
FullSimplify
in this integral

Developer`ClearCache[]
Block[{$Output = {OpenWrite["C:\\msgStream.m"]}},
   TracePrint[(1/Pi)*Integrate[Log[x/(x^2 + 1)]*(1/(x^2 + 1)^m), {x,
0, Infinity}, Assumptions -> m >= 1],
     TraceInternal -> True]; Close /@ $Output];
Thread[Union[Cases[ReadList["C:\\msgStream.m",
HoldComplete[Expression]],
    symb_Symbol /; AtomQ[Unevaluated[symb]] &&
Context[Unevaluated[symb]] === "System`" :> HoldComplete[symb], {0,
Infinity},
    Heads -> True]], HoldComplete]

(*outout is ommited*)

MemberQ[FullSimplify, -1]
True

>Why is this integral important?

> > int = HoldForm[(1/Pi)*Integrate[Log[x/(x^2 + 1)]*(1/(x^2 + 1)^m), {x,=
0, Infinity}, Assumptions -> m >= 1]]

I study the book "Irresistible Integrals" by George Boros and Victor
Moll
(2nd Edition, Campridge University Press 2006).
I work various formulas by hand and Mathematica from this book and one
of worked examples is this integral (page 270).

(BTW, very good performance of Mathematica)

Is is a good reason or not?

Best Regards
Dimitris



Daniel Lichtblau
Wolfram Research


=CF/=C7 Daniel Lichtblau =DD=E3=F1=E1=F8=E5:
> dimitris wrote:
> > Hi fellas.
> >
> > In my travel (sic!) through definite integration I encountered a
> > strange
> > behavior (at least!) of Integrate. Of course may be this is something
> > well known but I haven't notice any relevant before. So I apologize
> > if I discuss an old issue.
> >
> > Anyway, here we go...
> >
> > $VersionNumber
> > 5.2
> >
> > Consider the integral
> >
> > In[1]:=
> > int = HoldForm[(1/Pi)*Integrate[Log[x/(x^2 + 1)]*(1/(x^2 + 1)^m), {x,
> > 0, Infinity}, Assumptions -> m >= 1]]
> >
> > Here is the definite integral by Mathematica
> >
> > In[2]:=
> > res1 = int // ReleaseHold
> > Infinity::indet: Indeterminate expression 0*Infinity encountered.
> >
> > Out[2]=
> > -((1/(4*Pi*Gamma[m]))*((-3 + 2*m)*Sqrt[Pi]*Gamma[-(3/2) +
> > m]*(PolyGamma[0, 1 - m] - PolyGamma[0, 3/2 - m]) +  Gamma[-(1/2) +
> > m]*(4^(1 + m)*m*Gamma[-2*m]*Gamma[m]*Gamma[1/2 + m] +
> >       Sqrt[Pi]*(EulerGamma + Log[4] + PolyGamma[0, -(1/2) + m]))))
> >
> > Observe first the Infinity::indet message.
> >
> > Despite the presence of the warning message the result is correct.
> > [...]
> > At this point, someone may believe that the strange behavior I was
> > talking about is this
> > warning message. Even though the presence of this message needs some
> > discussion,
> > (although they are known cases where Built in functions generate
> > warning messages
> > in spite of the results being correct) the next issue is by far more
> > interesting.
> >
> > Just evaluate AGAIN the integral
> >
> > In[10]:=
> > res2=int//ReleaseHold
> >
> > Infinity::indet: Indeterminate expression 0*Infinity encountered.
> > Out[10]=
> > (Gamma[-(1/2) + m]*(2*HarmonicNumber[1/2 - m] - HarmonicNumber[-(3/2)
> > + m] - Log[4] - 2*(EulerGamma + PolyGamma[0, m]) +  2*Pi*Tan[m*Pi]))/
> > (4*Sqrt[Pi]*Gamma[m])
> >
> > Integrate returns a different output for the same definite
> > integration!
> > The result is of course correct.
> > [...]
> > but I wonder how is this possible this!
> >
> > Note that
> >
> > In[15]:=
> > FullSimplify[res1]
> > Simplify[res2 == %]
> > [...]
> > whereas for example
> >
> > In[17]:=
> > FunctionExpand[res1]
> > [...]
> > It seams that the second time Integrate called FullSimplify!
> >
> > But I think Integrate uses Simplify and not FullSimplify.
> >
> > Any insight, explanations available?
> >
> > Dimitris
>
> The difference in result is a subtle interaction between use of caching
> for certain intermediate results, and use of time constraints in some
> places (nobably simplifications). In effect a simplification attempt
> might time out the first attempt but succeed in later tries due to
> having more intermediate computations precomputed and cached.
>
> Integrate does make (very limited) use of FullSimplify.
>
> Why is this integral important?
>
> Daniel Lichtblau
> Wolfram Research



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