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MathGroup Archive 2006

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Re: NIntegrate that upper limit is infinite

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71894] Re: NIntegrate that upper limit is infinite
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Sat, 2 Dec 2006 05:11:45 -0500 (EST)
  • References: <ekp4q1$2h9$1@smc.vnet.net>

Your code contain many mistakes.

Let assume that

G[a_?NumericQ] := ((-(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_?NumericQ] := Re[N[LegendreP[(-1/2) + I, a]]]

Plot[G[a], {a, 1, 10}]
Plot[P[a], {a, 1, 20}]
Plot[P[a]G[a], {a, 1, 20}]

Then

Block[{Message}, NIntegrate[G[a]*P[a], {a, 1, Infinity},
SingularityDepth -> 1000]]
-0.00259109


Regards
Dimitris


Evanescence wrote:
> Dear all:
> My question 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,infinite}]
> but get the error message such that I do not get the correct answer
> so,please solve my confuse,thank you!
> 
> Evanescence 2006/12/1


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