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.