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Show integral is equal to Bessel function

  1. Sep 13, 2011 #1
    Hi guys,

    I'm pretty sure the following is true but I'm stuck proving it:
    \frac{1}{2\pi}\int_{-1}^1 \left(\frac{e^{\sqrt{1-y^2}}}{\sqrt{1-y^2}}+\frac{e^{-\sqrt{1-y^2}}}{\sqrt{1-y^2}}\right) e^{iyx} dy&=\frac{1}{2\pi i}\mathop\oint\limits_{|t|=1} \exp\left\{\sqrt{x^2-1}(t-1/t)/2\right\}t^{-1}dt\\

    I tried the substitutions:


    but not getting it. Also, I think only the first substitution converts the domain of integration into the required circle around the origin but that 's not too clear to me as well. Anyone can make a suggestion what to try next? Not sure if I should have posted this in the homework section.

    Edit: Ok, I made a mistake in the Bessel integral notation. It's z/2 and not iz/2 so I changed it above and will re-do my calculations. May change things for me.

    Last edited: Sep 13, 2011
  2. jcsd
  3. Sep 29, 2011 #2
    You can expand the exponential term on the rhs as a product of two series:


    and so it becomes
    [itex]\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{(\sqrt{x^2-1})^{n+k}}{2^{n+k}n!k!}(-1)^k\frac{1}{2\pi i}\oint\frac{t^{n-k}}{t}dt[/itex]

    The contour integral is now 1 if [itex]n=k[/itex] and 0 otherwise, giving the series
    which is the series definition for [itex]J_{0}\left(\sqrt{x^2-1}\right)[/itex].
  4. Sep 29, 2011 #3
    Ok, but that only shows how the RHS is equal to the Bessel function. I was wanting to figure out how to show the LHS is equal to the RHS.
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