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Differentiating an integral

  1. Feb 26, 2008 #1


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    I'm confused about differentiating an improper integral. Consider the function

    [tex]F(r)=\int_0^\infty J_0(rx)\,dx=\frac{1}{r}\int_0^\infty J_0(m)\,dm=\frac{1}{r}[/tex]

    where I've solved the integral by making the substitution [itex]m=rx[/itex] (I think this is OK). Now I would like to find [itex]\frac{\partial F}{\partial r}[/itex]. From the solution I know that this is [itex]-\frac{1}{r^2}[/itex], but I would like to do it another way, by differentiating inside the integral. I thought it was allowable to write

    [tex]\frac{\partial F(r)}{\partial r}=\int_0^\infty \frac{\partial}{\partial r}J_0(rx)\,dx=\int_0^\infty -x\,J_1(rx)\,dx[/tex]

    but this integral doesn't converge. Where have I gone wrong? Thanks!
  2. jcsd
  3. Feb 26, 2008 #2
    Are you sure it doesn’t converge? You should be able to switch the order of differentiation and integration provided that the limits of integration do not depend on the dependent variable of the derivative.
  4. Feb 26, 2008 #3


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    Mathematica says it doesn't converge. If all of my operations are valid, however, then it does converge and equals [itex]-\frac{1}{r^2}[/itex]. This would mean that

    [tex]\int^\infty_0 m\,J_1(m)\,dm=1[/tex]

    I'm wary about this equation, though. I haven't seen it in any tables of integrals.
  5. Feb 26, 2008 #4
    I wonder if mathematic uses any convergence tests. I'm never 100% confident in answers produced by computer algebra systems. Perhaps consider re-asking the question on either the mathematica news group or the symbolic math newsgroup.
  6. Feb 27, 2008 #5
    Improper integrals of Bessel functions are definitely a weak point of Mathematica. Only recently I encountered a very similar problem. (It was convergent after all.)

    Do you have any additional parameters in your integrand which might cause mathematica's convergence tests to fail? If so, try setting them to some fixed value. (In my case, Assuming[] nd With[] was not enough.)

    You can also use NIntegrate to check if it converges. It might produce a warnign that it is not, but as long as it gives a reasonable numerical answer it's a good indication that it converges:smile:
  7. Feb 28, 2008 #6
  8. Feb 28, 2008 #7


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    Thanks, very interesting. I also enjoyed the paper (found via the second link) on speaking math to a computer.
  9. Feb 29, 2008 #8
    The Bessel functions satisy the recurrence relations

    [tex]J_{n-1}+J_{n+1}=\frac{2\,n}{x}\,J_n\quad \text{and} \quad J_{n-1}-J_{n+1}=2\,J_n'[/tex]

    Adding these, you get

    [tex]x\,J_{n-1}=n\,J_n+x\,J_n'\overset{n=2}\Rightarrow x\,J_1=2\,J_2+x\,J_2'[/tex]

    while integrating from [itex]0[/itex] to [itex]\infty[/itex]

    [tex]\int_0^\infty x\,J_1\,d\,x=2\int_0^\infty J_2\,d\,x+\int_0^\infty x\,J_2'\,d\,x[/tex]

    The first one equals 2 since [itex]\int_0^\infty J_2\,d\,x=1[/itex] and the second one can be evaluated by integrating by parts, i.e.

    [tex]\int_0^\infty x\,J_2'\,d\,x=x\,J_2(x)\Big|_0^\infty-\int_0^\infty J_2\,d\,x=-1[/tex]


    [tex]\int_0^\infty x\,J_1(x)\,d\,x=1[/tex]
  10. Mar 2, 2008 #9


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    Very appreciated, thanks. Wow, now I can show that

    [tex]\int_0^\infty x\,J_0(x)\,d\,x=0[/tex]
    [tex]\int_0^\infty x\,J_2(x)\,d\,x=2[/tex]

    Believe it or not, this is contributing to a journal article in cell biology that I'm working on. When tissue cells attach to a surface, they exert stress on the substrate that can be modeled with elasticity theory. Bessel functions arise naturally from assuming that the stress is applied over a circular area.
    Last edited: Mar 2, 2008
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