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Integration by parts homework problem

  1. May 4, 2010 #1
    How do I integrate the following:

    [itex]\int_0^\infty r e^{-ar} \sin{(Kr)} dr[/itex]

    i tried writing [itex] r e^{-ar} = -\frac{d}{da} e^{-ar}[/itex] and using integration by parts but i couldn't get anywhere. any ideas?
  2. jcsd
  3. May 4, 2010 #2
    Re: Integration

    Would i.b.p not work if you let dv = re^(-ar)dr? You would have to use another iteration of i.b.p.... I'll check it out when I've got some paper in front of me, but give that a try.
  4. May 4, 2010 #3
    Re: Integration

    but then we would have to use integration by parts to evaluate v from dv. So it would end up being a nested integration by parts within an integration by parts? that seems awfully complicated, no?
  5. May 4, 2010 #4
    Re: Integration

    You can use integration by part, but it is easier to write

    sin(K r) = Im[exp(i k r)]

    Then you have the integral of

    r exp(-b r)

    wih b = a - i k

    and you need to take the imaginary part of the answer.

    And, as you already noted, you can compute the integral without the factor r in the integrand and differentiate w.r.t. b to bring a factor of r down.
  6. May 5, 2010 #5
    Re: Integration

    why do i only need to consider the imaginary part of the answer?
  7. May 5, 2010 #6
    Re: Integration

    Because that will yield the sin(kr) factor in the integrand. Note that you do that at the very end after you have differentiated w.r.t. the constant in the exponential to bring down the factor r.
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