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Help with numerical integration

  1. Mar 28, 2005 #1
    Hello there, I've not been here in a while, but I'm stuck doing this integration and wondered if some of you kind people would help :smile:

    [tex]\int_0^\infty \frac{1} {(1+x)\sqrt{x}} dx[/tex]

    (appologies for the lack of spacing in there...)

    anyways, I know that when x tends to infinity, the integral can be approximated to,


    [tex]\int_0^\infty \frac{1} {(x)\sqrt{x}} dx[/tex]

    but I cant seem to find this identity in any of my tables anywhere...

    The reason I need it is because I'm in the processes of writing some c code to analytically calculate this with a specified degree of acuracy, (am going to use the trapezium method of integration I think) so it would be nice to know if the answers I get out of it are any good or not!

    thanks for you time :smile:
     
  2. jcsd
  3. Mar 28, 2005 #2
    use rules of exponents.

    1/(x(sqrtx)) = x^(-3/2)
     
  4. Mar 28, 2005 #3
    awesome, thanks DeadWolfe :)
     
  5. Mar 28, 2005 #4

    dextercioby

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    That integral (the first) can be very simply computed analytically to yield the result [itex] \pi [/itex]...Heck,u can even define [itex] \pi [/itex] by it

    [tex] \pi=:\int_{0}^{+\infty} \frac{dx}{(1+x)\sqrt{x}} [/tex]

    HINT:Make the obvious substitution
    [tex] \sqrt{x}=t [/tex]

    Daniel.
     
  6. Mar 28, 2005 #5
    wow, even more helpful, thanks alot :smile:

    also, would I be correct in saying that the limit of the integrand as x-->infinity is pi, in that case?
    or have I got completely mudled up? :confused:
     
  7. Mar 28, 2005 #6

    dextercioby

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    Define

    [tex] P(x)=:\int_{0}^{x} \frac{dt}{(1+t)\sqrt{t}} [/tex]

    Show that

    [tex] P(x)=2\arctan x [/tex]

    Then it's easy to say

    [tex] \lim_{x\rightarrow +\infty} P(x)=\pi [/tex]

    Not the integrand!!The integrands's (inferior) limit to [itex] 0 [/itex] is [itex] +\infty [/itex],while its limit to [itex] +\infty [/itex] is [itex] 0 [/itex] (:wink:)

    Daniel.
     
  8. Mar 28, 2005 #7
    thanks very much for your help dextercioby, it'll be most useful!

    now all I've got to do is figure out how to do this with C :wink:

    also, where you have,

    [tex] P(x)=:\int_{0}^{x} \frac{dt}{(1+t)\sqrt{t}} [/tex]

    that'll yield the same result as for

    [tex] F(x)=:\int_{0}^{+\infty} \frac{dx}{(1+x)\sqrt{x}} [/tex]

    with the substitution

    [tex] \sqrt{x}=t [/tex]

    right?
    :smile:
     
    Last edited: Mar 28, 2005
  9. Mar 28, 2005 #8

    dextercioby

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    Nope.The second (the one with F(x)) is a number,while the first is a function...So in the second case,the notation is incorrect...

    Daniel.
     
  10. Mar 28, 2005 #9
    ah I see, ok, so the first one, with x = +infinity, will equal pi, the same as the bottom one would (had I wrote it correctly).
     
  11. Mar 28, 2005 #10

    dextercioby

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    Yes.U got it :smile:

    Daniel.
     
  12. Mar 28, 2005 #11
    finally :tongue: thanks very much for your help :smile:
     
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