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Integral of square of Bessel function

  1. Jul 30, 2009 #1
    Hi there,

    I am starting with the Bessel functions and have some problems with it. I am getting stuck with this equation. I could not find this kind of integral in the handbooks.

    1. [tex]\int_0^aJ_0^2(bx)dx[/tex]


    Besides of this, I have other equations in similar form but I think this integral is the key to solve others:

    2. [tex]\int_0^\infty J_0^2(bx)e^{-\frac{x}{c}}dx[/tex]

    3. [tex]\int_0^\infty J_0^2(bx)e^{-\frac{x^2}{c}}dx[/tex]

    3. [tex]\int_0^\infty J_0^2(bx)\frac{x}{c}dx[/tex]


    Please help me. It is highly appriciated.
     
  2. jcsd
  3. Jul 31, 2009 #2
    I asked Maple, and got something in terms of the Struve H function.

    [tex]\int _{0}^{a}\! \left( {{\rm J}_0\left(bx\right)} \right) ^{2}{dx}=-
    a \left( -2\,{{\rm J}_0\left(ba\right)}+\pi \,
    {{\rm J}_0\left(ba\right)}{\rm H}_1 \left(ba \right) -\pi \,
    {{\rm J}_1\left(ba\right)}{\rm H}_0 \left(ba \right)
    \right) /2
    [/tex]

    added: This is wrong. I for got the square. This is only [itex]\int _{0}^{a}\! {{\rm J}_0\left(bx\right)} {dx}[/itex]
     
    Last edited: Jul 31, 2009
  4. Jul 31, 2009 #3
    Hi g_edgar,

    Thank you for your reply. I tried with Maple too and I got this:

    [tex]a*hypergeom([1/2, 1/2], [1, 1, 3/2], -a^2*b^2)[/tex]


    The equation you got must be the result of this integral: [tex]\int _{0}^{a}\! {{\rm J}_0\left(bx\right)}{dx}[/tex]


    I have to search for the generalized hypergeometric function. I have a little knowledge on this.
     
  5. Jul 31, 2009 #4

    jasonRF

    User Avatar
    Science Advisor
    Gold Member

    I looked these up in the book "integrals of bessel functions" by Luke, McGraw-Hill 1962.

    [tex]
    \int_0^1 dt \ J_0^2(b t) & = & 2\ J_1(b) \sum_{k=0}^{\infty}
    \frac{(-1)^k}{2k+1} J_{2k+1}(b).
    [/tex]



    [tex]
    \int_0^\infty dt e^{-pt} J_0^2(bt) & = & \frac{k {\mathbf{K}}(k)}{\pi b}
    [/tex]

    for

    [tex]
    Re(p)>0
    [/tex]

    where
    [tex]
    k^2 & = & \frac{4 b^2}{p^2 + 4 b^2}
    [/tex]
    and
    [tex]
    {\mathbf{K}}(k)}
    [/tex]
    is the complete elliptic integral of the first kind.




    [tex]
    \int_0^\infty dt e^{-p^2t^2} J_0^2(bt) & = & \frac{\Gamma(\frac{1}{2})}{2p}
    \ _3F_3 (\frac{1}{2},1,\frac{1}{2}; 1,1,1 | - \frac{b^2}{p^2} ),
    [/tex]
    [tex]
    for Re(p^2)>0
    [/tex]


    Are you sure this converges? Given the asymptotic expansion of [tex]J_0[/tex] I'm skeptical.
     
  6. Jul 31, 2009 #5
    Thanks jasonRF for the results. I have that book too. Could you tell me in which parts and pages you found that?


    I made a mistake with the last one. It should be:
    [tex]\int_0^a J_0^2(bx)\frac{x}{c}dx [/tex]
     
    Last edited: Jul 31, 2009
  7. Apr 12, 2011 #6
    Hi there,

    I found this integral at Gradshteyn:

    [tex]\int_{0}^{1} x\, J_{\nu}(\alpha\,x)J_{\nu}(\beta\,x)\,dx = \frac{\beta J_{\nu-1}(\beta)J_{\nu}(\alpha) - \alpha J_{\nu-1}J_{\nu}(\beta)}{\alpha^2 - \beta^2}.[/tex]

    Then, taking the limit [tex]\beta\rightarrow\alpha[/tex] you can find

    [tex]\int_{0}^{1}x\,J_{\nu}^2(\alpha\,x) dx = -\frac{1}{2\alpha}\left[J_{\nu-1}(\alpha)J_{\nu}(\alpha) + \alpha J_{\nu-1}^{\prime}(\alpha)J_{\nu}(\alpha) - \alpha J_{\nu-1}(\alpha)J_{\nu}^{\prime}(\alpha)\right].[/tex]

    I hope it is useful.
     
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