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MathGroup Archive 2007

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Re: GenerateConditions->False gives fine result!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73976] Re: GenerateConditions->False gives fine result!
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Sat, 3 Mar 2007 23:54:55 -0500 (EST)
  • References: <es9238$3ig$1@smc.vnet.net>

Hi and thanks a lot regarding answering to my messages/questions
apperared recently
about Integrate.

> This is because the integral does in fact diverge in the Riemann sense. T=
aking the simpler example with {t->3, x->2}:

Now I did plots of the integrand I understand where was my mistake.
Thanks again!

In fact what confused me (which should have confused!) was the fact
that

In[78]:=
(1/(2*Pi*I))*Integrate[q/(E^(2*I*q)*Sqrt[q^2 + 9]), {q, -Infinity,
Infinity}, GenerateConditions -> False]
N[%]
Out[78]=
-((3*BesselK[1, 6])/Pi)
Out[79]=
-0.001283348797177626

In[82]:=
Chop[(1/(2*Pi*I))*NIntegrate[q*(Exp[(-I)*q*x]/Sqrt[q^2 + t^2]) /. {t -
> 3, x -> 2}, {q, -Infinity, Infinity},
    Method -> Oscillatory]]
SequenceLimit::seqlim: The general form of the sequence could not be \
determined, and the result may be incorrect.
Out[82]=
-0.0012833487971660698

I=2Ee. the two results are almost the same; this lead me to ignore the
messages from SequenceLimit!
Very amateurish mistake!

Anyway...Integrate rules! I need more reading!

Kind Regards
Dimitris


=CF/=C7 Bhuvanesh =DD=E3=F1=E1=F8=E5:
> This is because the integral does in fact diverge in the Riemann sense. T=
aking the simpler example with {t->3, x->2}:
>
> In[1]:= integrand = q/(E^(2*I*q)*Sqrt[q^2 + 9]);
>
> In[2]:= Limit[integrand, q->Infinity]
>
> Out[2]= (1 + I) Interval[{-1, 1}]
>
> GenerateConditions->False, in addition to checking convergence and lookin=
g for singularities, also does Hadamard-type integrals. Here's another exam=
ple:
>
> In[1]:= Integrate[1/(x^(1 + I/2)*(1 + x)), {x, 0, 1}]
>
>                                     1
> Integrate::idiv: Integral of ---------------- does not converge on {0, 1}.
>                               1 + I/2
>                              x        (1 + x)
>
>                          1
> Out[1]= Integrate[----------------, {x, 0, 1}]
>                    1 + I/2
>                   x        (1 + x)
>
> In[2]:= Integrate[1/(x^(1 + I/2)*(1 + x)), {x, 0, 1}, GenerateCondition=
s->False] //InputForm
>
> Out[2]//InputForm= (-PolyGamma[0, -I/4] + PolyGamma[0, 1/2 - I/4])/2
>
> Bhuvanesh,
> Wolfram Research.



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