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Need the Integral of

  1. May 14, 2004 #1
    Could someone show me how the integral of f(v) equals 1, where f(v) is

    (4/(Sqrt Pi)) x (m/2kT)^(3/2) x (v^2) x (e^(-mv^2/2kT))

    I got through the integral and it didn't equal 1, but it's supposed to, so could someone help me out.
  2. jcsd
  3. May 14, 2004 #2


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    Could you show us what you have done so far?

    And what is the region of integration?
  4. May 14, 2004 #3
    Ooops, sorry. The integration is from zero to infinity. I'll type in what I've done so far if someone can tell me how to get the integral and various other mathematical characters in the post, it's much easier that way. But so far my methods have included substitution, integration by parts, and taking the limit of the function.
  5. May 15, 2004 #4


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    Look here: https://www.physicsforums.com/misc/howtolatex.pdf

    Or go to this thread and click on various examples and the code will pop up in a small window so you see how it is done: https://www.physicsforums.com/showthread.php?t=8997
  6. May 15, 2004 #5
    you should be able to get the indefinate integral using integration by parts. use v^2 = du and e^(-mv^2/2kt) = v. Than in the new integral you create, you can substitute e^u and it should work out fine (in other words, do integration by parts and than use substitution). Whenever you have a parametre for an integral as an infinite, you have to take the limit of the integral as it approaches that infinite. For example, in this case you would have to than show that (lim x--> infinite of F(x)) - F(0) = 1 (thats what you do when you have infinite parameters for integration). Try it out and see what you get.
  7. May 18, 2004 #6

    Yeah, those were the methods that I used to integrate it, integration by parts, substitution, and taking the limit. I talked to my Calculus 3 teacher yesterday about it, and she said it was something like a complex integral where you can't just solve it using those methods...or something like that.
  8. May 18, 2004 #7
    when you do integration by parts you will obtain a new integral that you have to solve. This [unless im mistaken] can be solved using substitution. Altho you may have to do integration by parts again (i suspect you would)
    Last edited: May 18, 2004
  9. May 18, 2004 #8


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    Yeah .. it seems that way.

    I tried to use IBP & substitution.

    I assume k, m, T, & [tex]\pi[/tex] are constants.

    Then I get:


    I'm not sure how to do in an integral the likes of [itex]\int\frac{v^2}{e^v^2}[/itex], or [itex]\intv^2e^-v^2[/itex] for that matter.

    With substitution of [itex]u=v^2[/itex], I get:

    Last edited: May 20, 2004
  10. May 18, 2004 #9

    Maybe i'm just more tired than i tihnk, but with that substitution i believe you should get:



  11. May 18, 2004 #10
    You are tired.

  12. May 18, 2004 #11
    Huh? Makes no sense..
  13. May 19, 2004 #12

    oh...yeah...i didn't change the varible of integration, i left it [tex] dv [/tex] even though i typed [tex] du [/tex]. So yeah, disregard that post then.

    edit: actually he never put the [tex] dv [/tex] up there, so i'm assuming that was my error.
  14. May 19, 2004 #13
    yay! problem solved..
  15. May 19, 2004 #14

    Just the roblem of my mistakes when sleep-deprived. Oh well...i suppose i could go actually try to solve it now...here i go
  16. May 19, 2004 #15
    Dun try now... get sleep and then try. :)
  17. May 19, 2004 #16
    actually i think it can't be solved analytically...integration by parts fails, given that [tex] \frac{1}{e^x^2} [/tex] cannot be integrated. So...i can't seem to come up with any standard single variable calculus techniques that would help...so yeah, sleep first, boggle later.
  18. May 19, 2004 #17
    The thing is, the integral of the whole expression is supposed to divide out to 1. The integral is supposed to end up being the inverse of the constant you pull out when you integrate, giving you a 1/1 or 1 answer. This suggests that standard integration techniques won't work, because the when you integrate, you have to end up with the Sqrt( Pi ) / 4 in the answer to cancel out the 4/Sqrt ( Pi ) in the constant.
  19. May 19, 2004 #18


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    But like you said in your post, if you require concepts from Calc III, it's beyond what I can do.
  20. May 19, 2004 #19

    Tom Mattson

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    This integral can be done the same way that the integral of exp(-x2) can be done. First, write the integral of x2exp(-x2) from zero to infinity. Then write the integral of y2exp(-y2) from zero to infinity (they're both exactly the same as your integral). Now multiply the integrands together double integrate over x and y. When you convert to polar coordinates, you will get an integral that can be done by parts. Just don't forget to take the square root at the end.


    dx dy=rdr dθ
  21. Oct 21, 2007 #20

    Sorry to hash up this very old thread, but I need help with this exact problem and I don't understand the solution given.

    I'm trying to find the integral of x2exp(-x2 / a2) from -infinity to infinity.

    I've done what's recommended here, but I end up with a double integral of 1/4 sin2(2θ) r5 exp[-r2] drdθ from zero to infinity and zero to infinity. How do I solve this?
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