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MathGroup Archive 2006

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Re: (revision) NIntegrate that upper limit is infinite

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71963] Re: (revision) NIntegrate that upper limit is infinite
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Wed, 6 Dec 2006 06:03:43 -0500 (EST)
  • References: <el12c7$5j5$1@smc.vnet.net>

I think Maxim and Daniel cover completely the subject, so
you are sure that your integral diverges.

Here I provide another short of evidence.

G[a_] := (-(a - Cos[0.5])^(-3/2)/(3*10^2))*Sin[0.5]^2 + (1/(1 -
0.1*0.1))*
    ((0.005*(1 + Cos[0.5]) + 0.001/1000)*(1 +
Cos[0.5])^(1/2)*(ArcTan[(a - Cos[0.5])^(1/2)/(1 + Cos[0.5])^(1/2)] -
Pi/2) +
     (0.005*(1 - Cos[0.5]) - 0.001/1000)*(1 -
Cos[0.5])^(1/2)*ArcTanh[(1 - Cos[0.5])^(1/2)/(a - Cos[0.5])^(1/2)])

P[a_] := N[LegendreP[-2^(-1) + I, a]]

First see how the setting of MaxRecursion affects the estimated value
for the integral.
It seems that there is not a value that the integral converges (even
with setting so small the required PrecisionGoal)

Block[{Message}, ({#,Re[NIntegrate[G[a]*P[a], {a, 1, Infinity},
SingularityDepth -> 1000, MaxRecursion -> #1,
      PrecisionGoal -> 4]]} & ) /@ Range[6, 34,
2]]//TableForm[#,TableHeadings->{None,{"maxrecursion","estimated
value"}},TableAlignments->Center]&

Also take a look of the following table:

(TableForm[#1, TableHeadings -> {None, {"upper limit", "estimated
value"}}, TableAlignments -> Center] & )[
  ({#1, Chop[NIntegrate[G[a]*P[a], {a, 1, #1}, SingularityDepth ->
1000, PrecisionGoal -> 4, MaxRecursion -> 24]]} & ) /@
   {1000, 10000, 20000, 50000, 100000, 200000, 300000, 500000, 700000,
1000000, 2000000, 2500000, 3000000}]

Best Regards
Dimitris


Evanescence wrote:
> Dear all:
> I am very sorry for my article contain a lot of mistake before.
> So I check it and state my questions again.
> I am really sorry that cause a lot of confuse for everybody.
> My questions is as follows:
> First I definite a function that is:
> G[a_] := ((-(a -Cos[0.5])^(-3/2))/(3*(10)^2))*(Sin[0.5])^(2) +
> (1/(1-0.1*0.1))*((0.005*(1 + Cos[0.5]) + 0.001/1000)*(1 +
> Cos[0.5])^(1/2)*(ArcTan[(a - Cos[0.5])^(1/2)/(1 + Cos[0.5])^(1/2)] -
> Pi/2) + (0.005*(1 - Cos[0.5]) - 0.001/1000)*(1 -
> Cos[0.5])^(1/2)*(ArcTanh[(1 - Cos[0.5])^(1/2)/(a - Cos[0.5])^(1/2)]))
>
> another function is:
> P[a_]:=Re[N[LegendreP[(-1/2)+I,a]]]       where I=(-1)^(1/2)
>
> then
> NIntegrate[G[a]*P[a], {a, 1, Infinity}]
>
> but get error message is:
> NIntegrate's singularity handling has failed at point
> {a}={2.7013362243395366*(10^150)}for the specified
> precision goal.
> Try using larger values for any of $MaxExtraPrecision
> or the options WorkingPrecision, or SingularityDepth and MaxRecursion.
>
> NIntegrate::"inum":
> "Integrand G[a]*P[a] is not numerical at {a}=
> {2.7013362243395366`*(10^150)}.
>
> Please solve my confuse ,thank you very much!
>
> By the way
> G[Infinity]*P[Infinity]=0 ¡A G[1] is singular point¡AP[1]=1
> and I try Plot the graph for G[a]¡AP[a]¡Aand G[a]*P[a] ¡Afrom the
> graph of  G[a]*P[a] 
> it should convergence for a-->Infinity.


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