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Integral resulting in Bessel function

  1. Jan 15, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx = \frac{\pi \sqrt{a}}{2} J_{1} \left( 2 \sqrt{a} \right)[/tex] where [tex]J_{1}[/tex] is the Bessel function of the first kind of order 1.

    2. Relevant equations

    3. The attempt at a solution

    Some calculations I did already

    [tex]\int_{0}^{\infty} \sin \left(x\right) \sin \left(\frac{a}{x}\right) \ dx= \int_{0}^{\infty} \sum_{k=0}^{\infty }(-1)^{k}\frac{x^{2k+1}}{2k+1!} \cdot \sum_{l=0}^{\infty }(-1)^{l}\frac{a^{2l+1}x^{-2l-1}}{2l+1!} \ dx[/tex]

    [tex]=?????? \int_{0}^{\infty} \sum_{l=0}^{\infty } \sum_{k=0}^{\infty }(-1)^{k+l}\frac{x^{2(k-l)}}{(2k+1)!(2l+1)!} a^{2l+1} \ dx[/tex]


    [tex]\frac{\pi \sqrt{a}}{2}J_{1}(2\sqrt{a})=\frac{\pi \sqrt{a}}{2} \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{2l+1}l!(1+l)!} 2^{l+\frac{1}{2}}a^{l+\frac{1}{2}}[/tex]
    [tex]=\pi \sum_{l=0}^{\infty}\frac{(-1)^l}{2^{l+\frac{3}{2}}l!(1+l)!} a^{l+1}[/tex]

    I put ??? because I think this step is not allowed because of the singularity of [tex]\sin \left(\frac{a}{x}\right)[/tex] at x=0. Can someone confirm if this equality is true?
  2. jcsd
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