Re: (revision) NIntegrate that upper limit is infinite

*To*: mathgroup at smc.vnet.net*Subject*: [mg71943] Re: (revision) NIntegrate that upper limit is infinite*From*: ab_def at prontomail.com*Date*: Tue, 5 Dec 2006 06:04:41 -0500 (EST)*References*: <el12c7$5j5$1@smc.vnet.net>

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. Evaluate Series[G[a], {a, Infinity, 1}]: the leading term will be around -3.*^-18 and it's not clear whether it's numerical noise or whether the limit is really finite. So let's work with exact (or at least arbitrary-precision) numbers: ga = Unevaluated@ G[a] /. (DownValues@ G /. x_Real :> Rationalize[x]) pa = LegendreP[-1/2 + I, a] Then Series[ga, {a, Infinity, 1}] // FullSimplify tells us that ga is O(1/sqrt(a)) and Series[pa, {a, Infinity, 1}] // FullSimplify[ComplexExpand[#], a > 0]& N[%, 20] // Chop shows that pa is asymptotic to (C1*Sin[Log[a]] + C2*Cos[Log[a]])/Sqrt[a]. Note that here again it can better to use N only in the end. Now make the change of variables a = E^t, estimate the higher order terms, and it will be obvious that the integral diverges. Here is an illustration: Plot[(ga*pa /. a -> E^t)*E^t /. t -> SetPrecision[tt, 60], {tt, 0, 100}] But don't rely on plots too much; trying to explore the singularity at a = 1 this way probably wouldn't be such a great idea. Maxim Rytin m.r at inbox.ru