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Introducing parameters in integrals

  1. Sep 30, 2013 #1
    1. The problem statement, all variables and given/known data
    Using the integral ∫dx/1+x^2 = pi/2 from 0 to infinity as a guide, introduce a parameter and then differentiate with respect to this parameter to evaluate the integral
    ∫dx/(x^2+a^2)^3 from 0 to infinity


    2. Relevant equations



    3. The attempt at a solution
    ∫(1/1+x^2) = tan^-1(x)
    introducing parameter x/a to replace x we get

    ∫(1/a)/(1+(x/a)^2) = tan^-1(x/a)

    ∫dx/(a^2+x^2) = 1/a*tan^-1(x/a)

    d/da(1/(x^2+a^2)) = -2a/(x^2+a^2)^2

    im kind of stuck here. i don't really understand how to do this question, if anyone can help me understand/ lead me to the correct path that would be appreciated.
     
  2. jcsd
  3. Sep 30, 2013 #2

    Zondrina

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    How about, let ##x = atan(θ)## then ##dx = asec^2(θ)dθ##.
     
  4. Sep 30, 2013 #3
    Sorry but I'm not sure where you're getting that from. My teacher used the method above to solve for a similar problem, but I don't fully understand it.
     
  5. Sep 30, 2013 #4

    Zondrina

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    It's called a trig substitution. It's designed to help you simplify the integral into something more manageable.

    There are three trig substitutions you should probably learn, and they come from recognizing patterns in the integral.

    The first paragraph of this article pretty much sums it all up: http://en.wikipedia.org/wiki/Trigonometric_substitution
     
  6. Sep 30, 2013 #5

    Office_Shredder

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    Zondrina's suggestion will help solve it, but it's not really the way the question asks you to do it. Consider
    [tex] \frac{d}{da} \int_{0}^{\infty} \frac{dx}{(x^2+a^2)} [/tex]

    Do you see how this will get you closer to the integral that you want to calculate?
     
  7. Sep 30, 2013 #6
    That's as far as I understand it unfortunately. I remember from a previous question we somehow manipulated the derivative of that to get this
    2a∫dx/(a^2+x^2)^2 = 1/a^2*tan^-1(x/a) +(1/a)*1/(1+(x^2/a^2))*(x/a^2)

    This is the the part I'm mostly having trouble understanding.

    I also tried Zondria's method and got to 1/a^5∫cos^4(θ)dθ and got stuck, but like you said I'm mostly interested on the other method of finding it right now (although I'll continue to try this one too to fully understand this).
     
    Last edited: Sep 30, 2013
  8. Sep 30, 2013 #7

    Office_Shredder

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    When you say "this is the part I'm mostly having trouble understanding", what part are you having trouble understanding exactly? Why that was helpful, or why you can take the derivative like that in the first place?
     
  9. Sep 30, 2013 #8
    I mean I know how to find the derivative of the integral, but after that I'm not sure what to do. The above is finding the integral of 2a/(a^2+x^2)^2 after finding the derivative of 1/(a^2+x^2). I don't know what the point is in finding the derivative though. That's what I'm confused on. In the original question I probably have to find the derivative twice, but after that I'm not sure what to do. I might be completely wrong about all this. Would be nice if someone pushed me in the right direction.
     
  10. Sep 30, 2013 #9

    Office_Shredder

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    Once you took the derivative, the power of x2 + a2 went from 1 to 2 in the denominator. You wish it was three though.... can you guess what you should do to increase the power once more?
     
  11. Sep 30, 2013 #10
    i already know that. find the derivative again but after that what do i do?

    so i have the second derivative = -2(x^2-3a^2)/(a^2+x^2)^3

    now what
     
  12. Oct 1, 2013 #11

    Office_Shredder

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    Whoops, forgot to mention that it might help to divide both sides by a before taking the next derivative.
     
  13. Oct 1, 2013 #12
    man, never mind I give up. I just don't understand this and I'll leave it at that.
     
  14. Oct 1, 2013 #13

    Office_Shredder

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    If you take this equation that you had
    and divide both sides by a before taking the derivative with respect to a, then the integral on the left hand side is a lot nicer.
     
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