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Yukawa Potential Integral

  1. Mar 26, 2013 #1
    While preparing for an exam I came across an integral of the form

    [tex]\int_0^\infty dx\;e^{-\alpha x}\sin{q x}[/tex]

    with [itex]q,\alpha>0[/itex].

    My question will be regarding my solution to the integral which I present as follows:

    I expand the sine function as a Taylor series and differentiate with respect to alpha to yield

    [tex]\begin{align}e^{-\alpha x}\sin{q x} &= \sum_{n=0}^\infty (-1)^n\dfrac{q^{2n+1}}{(2n+1)!}x^{2n+1}e^{-\alpha x} \\
    &= \sum_n (-1)^{n+1}\dfrac{q^{2n+1}}{(2n+1)!}\dfrac{d^{2n+1}}{d\alpha^{2n+1}}e^{-\alpha x}\end{align}[/tex]

    After integrating with respect to x and differentiating with respect to alpha I arrive at

    [tex]\int_0^\infty dx\;e^{-\alpha x}\sin{q x}=\dfrac{q}{\alpha^2}\sum_n \left(i\dfrac{q}{\alpha}\right)^{2n}.[/tex]

    Here comes the troubling part. For [itex]q/\alpha<1[/itex] this geometric series converges nicely to

    [tex]\dfrac{q}{q^2+\alpha^2}.[/tex]

    However, Mathematica tells me that the integral, unlike my geometric series above, will still converge for [itex]q/\alpha\geq 1[/itex].

    I guess my question is a) Where have I gone wrong in my solution such that it is only valid for the case [itex]q/\alpha<1[/itex]? b) Is there a more straightforward way of performing this integral?

    Thanks in advance for any insight you all may offer.

    I realize this is more of a math question, but it came up while performing the Fourier transform of the Yukawa potential and I thought that the physics community here would be well acquainted with this integral.

    [Edit]: I want to make clear that this is not a homework problem. I was simply curious if I could perform the integral by hand.
     
  2. jcsd
  3. Mar 26, 2013 #2
    As a simpler route, I think you can just write ##\sin(qx) = [e^{iqx} - e^{-iqx}]/(2i)##, which turns the integral into a sum of two easy integrals of exponentials.
     
  4. Mar 26, 2013 #3
    Oh wow... I was originally trying to do just that except that I was thinking I would use the residue theorem... which makes no sense for this problem. Wow that is easy. Thanks, The_Duck.

    However, I am still curious as to where my overly-complicated solution goes wrong. Any thoughts?
     
  5. Mar 26, 2013 #4

    vanhees71

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    That's a trick any physicist has learn at a point. You have a series, converging for some values, and then after summing you do an analytic continuation. Most prominent examples is renormalization theory in quantum field theory: You evaluate an integral in [itex]d[/itex] dimensions and analytically continuate to the physical value [itex]d=4[/itex] after subtracting divergent terms of the type [itex]1/(d-4)[/itex] (dimensional regularization) and so on.

    You example is a nice example. Of course the geometric series converges only for [itex]|z|<1[/itex]. For these values you have
    [tex]\sum_{k=0}^{\infty} z^k=\frac{1}{1-z}.[/tex]
    This shows, why the series diverges for [itex]|z|>1[/itex]: The analytic function defined by the series only for [itex]|z|<1[/itex] has a pole at [itex]z=1[/itex], and a well-known theorem from complex function theory tells you that the Taylor series around a point [itex]z_0[/itex] has the radius of divergence determined by the largest disk not containing any singularities of the corresponding function. Here [itex]z_0=0[/itex] and the closest (and in this case only) singularity is at [itex]z=1[/itex].

    The function as an analytic (meromorphic) function is uniquely defined on the entire complex plane except at the pole [itex]z=1[/itex]. Thus the function is valid on this much larger domain than the convergence region of the original power series tells you.

    The integral is obviously convergent for any real [itex]q[/itex] for [itex]\text{Re} \; \alpha>0[/itex].
     
  6. Mar 27, 2013 #5
    Thank you vanhees71 for your response. I believe I understand the main idea of what you are saying and will do some playing around with this idea.
     
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