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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: [mg71876] Re: NIntegrate that upper limit is infinite
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Sat, 2 Dec 2006 05:10:43 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <ekp4q1$2h9$1@smc.vnet.net>

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)])
-------^ [1] 
--------------------------------------------------------------------------------------------^ 
[2]
[1] SetDelayed, that is :=, for function definition is better.
[2] Missing parenthesis

So, the correct expression 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)
-------^ [3] -----------------^ [4]
[3] SetDelayed, that is :=, for function definition is better.
[4] Sqrt[-1] is written capital I in Mathematica.

Therefore, the correct expression is

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

> then
> NIntegrate[G[a]*P[a],{a,1,infinite}]
----------------------------^^^^^^^^ [5]
[5] Positive infinity is written Infinity in Mathematica.

Hence, the correct expression is

NIntegrate[G[a]*P[a], {a, 1, Infinity}]

> but get the error message 
----------^^^
Which one did you get? There are dozens of possible messages...

Regards,
Jean-Marc


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