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Please help with numerical integration

  1. Aug 5, 2005 #1
    Im supposed to solve
    integral 10 to +infinity ((sin(1/x)/(1+x^3))dx with error precision of e=0.5*10^-4. Can someone please give me detailed explenation of solving this. (Supposedly by Simpson but i get lost in the way.

    P.S. sorry for bad spelling and lack of proper formula notions.
     
  2. jcsd
  3. Aug 5, 2005 #2
    I process from integral i get
    1/3[ln((x+1)/sqrt(x^2-x+1))+sqrt(3)*arctg(2*sqrt(3)*x/3-sqrt(3)/3)] from M to +infinity <=1/4*10^-2

    than i get

    -ln((M+1)/(sqrt(M^2-M+1))) +sqrt(3)*(pi/2-arctg(2*sqrt(3)*M/3-sqrt(3)/3))<=7.5*10^-3

    and i cant find any exact solution to solve that
     
  4. Aug 5, 2005 #3

    lurflurf

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    break the interval up
    10=x0<x1<x2<...<xn-1<xn=infinity
    for
    int(10,x1)
    int(x1,x2)
    .
    .
    .
    int(xn-2,xn-1)
    use simpsons rule
    for
    int(xn-1,infinity)
    chose xn-1 large
    estimate the integral on each subinterval accurate enough so that total error is within limit
     
  5. Aug 5, 2005 #4
    Thanks, but i am totaly lost when i need to find int(xn-1,infinity). I try to make it lower than error estimate but i get either logaritmic inverse-trigonometric inequality as above or another insolvable integral.
     
  6. Aug 5, 2005 #5

    lurflurf

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    Don't over think it
    [tex]|{\int_x^\infty \frac{\sin(\frac{1}{x})}{1+x^3}}|<\int_x^\infty \frac{1}{x^3}dx=\frac{1}{2x^2}[/tex]
     
    Last edited: Aug 5, 2005
  7. Aug 5, 2005 #6
    Thank you very, very much. You are a life savior.
     
  8. Aug 5, 2005 #7
    Does someone who has Maple or Mathlab can give me aproximate solution of the integral, posiblly of int(10,45), and int(10,142). Error comarision goes to absurd. I get 3.3*10^-4 on pocket calculator for int(10,35). Simpson with above given boundries gives more than 10 times larger values.
     
  9. Aug 5, 2005 #8

    saltydog

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    [tex]\int_{10}^{45} \frac{Sin(1/x)}{1+x^3}dx\approx 0.000329176[/tex]

    [tex]\int_{10}^{142} \frac{Sin(1/x)}{1+x^3}dx\approx 0.000332717[/tex]

    As per Mathematica's NIntegrate.
     
    Last edited: Aug 5, 2005
  10. Aug 6, 2005 #9

    lurflurf

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    I don't know if you want to stay with the original integral or not. In any case a substitution will make this much easier.
    u=1/x
    [tex]\int_{10}^\infty \frac{\sin(\frac{1}{x})}{1+x^2}dx=\int_0^\frac{1}{10}\frac{u\sin(u)}{1+u^3}du[/tex]
    then since x(sin(.1)/.1)<sin(x)<x for 0<x<.1
    [tex]\frac{\sin(.1)}{.3}\log(1.001)=10\sin(1/10)\int_0^\frac{1}{10}\frac{u^2}{1+u^3}du<\int_0^\frac{1}{10}\frac{u\sin(u)}{1+u^3}du<\int_0^\frac{1}{10}\frac{u^2}{1+u^3}du=\frac{1}{3}\log(1.001)[/tex]
    The average of these yeild a good approximation.
    simpsons rule will meet the error tolerance
    I~(.1-0)/6(f(0)+4f(.05)+f(.1))
    where f(x)=x sin(x)/(1+x^3)
     
    Last edited: Aug 6, 2005
  11. Aug 6, 2005 #10
    Thank. This afternoon i finaly finished the monster with int(10,150) by Simpson. Got aprox 0.000324. I cant realy believe its finished.
     
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