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Calculating the Laplace transform of a Bessel function

  1. Nov 10, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hi guys! I'm basically stuck at "starting" (ouch!) on the following problem:
    Using the integral representation of the Bessel function [itex]J_0 (x)=\frac{1}{\pi} \int _0 ^\pi \cos ( x\sin \theta ) d \theta[/itex], find its Laplace transform.


    2. Relevant equations
    [itex]\mathbb{L} [f(x)]=\int _0 ^{\infty} e^{-sx} f(x)dx[/itex].


    3. The attempt at a solution
    So I simply applied the formula above and could not solve the integral in theta.
    Namely [itex]\mathbb{L} [J_0 (x)]=\frac{1}{\pi} \int _0^{\infty} e^{-sx} \int _0 ^ \pi \cos (x \sin \theta ))d\theta dx[/itex].
    My only idea is to evaluate the theta integral first, treating x as a constant. So I've something of the form [itex]\int \cos (k \sin \theta ))d\theta[/itex] to calculate. I am not sure this is the way to go. And if it is, I don't have any idea on how to evaluate the integral.
    Any idea is welcome.
     
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  3. Nov 10, 2012 #2

    Zondrina

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    I don't believe that integral is possible by elementary means... I think you can only approximate it.
     
  4. Nov 10, 2012 #3

    fluidistic

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    Ok thanks. This probably mean there's a trick I'm missing to solve the problem.
     
  5. Nov 10, 2012 #4

    Zondrina

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    I might be mistaken here, but try expressing it as a series.
     
  6. Nov 10, 2012 #5

    fluidistic

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    Hmm I don't think that's the way to go. Otherwise they wouldn't have stated "Using the integral representation..."
    Thanks for the help by the way.
     
  7. Nov 10, 2012 #6

    I like Serena

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    Bessel functions cannot simply be integrated.
    Otherwise all the mathematical toolkits around wouldn't specifically include them.

    A typical first step would be to reverse the order of integration.
    And to replace ##\cos u = \frac 1 2(e^{iu} + e^{-iu})##.
    Then you can integrate with respect to x.
     
  8. Nov 10, 2012 #7

    fluidistic

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    Ok thanks for the help. But I don't see how this simplify things. If I understand you well, you suggest to rewrite the Laplace transform as [itex]\frac{1}{\pi} \int _0^\pi \int _0 ^\infty e^{-sx} \cdot \frac{1}{2} (e^{i x \sin \theta } - e^{-i x \sin \theta }) dx d\theta[/itex]. Then solve the x integral first. But the introduction of the complex exponential does not look nice to me. I'm sure I'm missing something.
     
  9. Nov 11, 2012 #8
    If you write it nice it does. I'll do the first one:

    [tex]\int_0^{\pi} \int_0^{\infty}e^{-sx}e^{ix\sin(\theta)}dxd\theta[/tex]
    [tex]\int_0^{\pi} \int_0^{\infty}e^{x(i\sin(\theta)-s)}dxd\theta[/tex]
    [tex]\int_0^{\pi} \frac{1}{i\sin(\theta)-s} e^{x(i\sin(\theta)-s)}\biggr|_0^{\infty} d\theta[/tex]

    Now, decide what restrictions must be placed on [itex]s[/itex] to make that inner expression well-behaved, then proceed.
     
    Last edited: Nov 11, 2012
  10. Nov 11, 2012 #9

    fluidistic

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    Thanks for the help. But I still don't see it.
    What troubles me is the imaginary part in the exponential.
    One restriction over s is that [itex]s\neq i \sin \theta[/itex]. I don't really see what could be the other restriction. I guess it has to see with the convergence of the integral but that imaginary unit boggles me.

    Edit: for example I know that if [itex]s=i \sin (\theta ) +ax[/itex] with a>0 then the integral would be a Gaussian and would converge but that's a very special case.
     
  11. Nov 11, 2012 #10

    I like Serena

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    Compare the exponential to the polar representation of an imaginary number (and split it like that).
    $$z=r e^{i \phi}$$
    The imaginary part of the exponent identifies the angle of the imaginary number.
    The real part identifies the length.

    Note that a length of the form ##e^{-x s}## tends to zero if x tends to infinity.
    But only if s is a positive (real) number.

    In other words, just fill in ##\infty## respectively 0 for x.
     
  12. Nov 11, 2012 #11

    fluidistic

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    I see. Despite knowing the representation of a complex number under this form, I did not notice it here. :frown:
    [itex]\mathbb{L} [J_0 (x)]=\frac{1}{2\pi}\int _0 ^\pi \underbrace{ \int _0^\infty e^{-sx}(e^{-ix \sin \theta } +e^{ix \sin \theta })dx } _I d\theta [/itex].
    Where [itex]I=\left ( -\frac{1}{s+i\sin \theta } \right ) e^{-xs}e^{ix\sin \theta } \big | _{x=0} ^{x=\infty } + \left ( \frac{1}{i \sin ( \theta ) -s } \right ) e^{-xs} e^{-ix \sin \theta } \big | _{x=0}^{x=\infty }[/itex].
    So for s>0, [itex]I=\frac{1}{s+i \sin \theta } + \frac{1}{s-i \sin \theta }=\frac{2s}{s^2+ \sin ^2 (\theta )}[/itex].
    Thus [itex]\mathbb{L} [J_0 (x)]= \frac{1}{\pi s} \int _0 ^\infty \frac{d \theta }{1+ \left ( \frac{\sin \theta }{s}\right ) ^2}[/itex]. Hmm once again I'm stuck on an integral. I don't see any "u substitution" that would work. It seems like it's related to the arctangent function.
    I might have messed up some algebra somewhere, I'm going to recheck.
     
  13. Nov 11, 2012 #12
    That last integral should go from 0 to pi. So we need to evaluate:

    [tex]\frac{1}{\pi} \int_0^{\pi} \frac{s}{s^2+\sin^2(t)}dt[/tex]

    Whenever you have a rational expression of sin or cos, can make the substitution [itex]z=\tan(t/2)[/itex]. Then:

    [tex]\sin(t)=\frac{2z}{1+z^2}[/tex]

    [tex]dt=\frac{2}{1+z^2}dz[/tex]

    Now it's a messy rational expression in z. Might be a good candidate for the Residue Theorem. Keep in mind partial fractions is an indirect application of the Residue Theorem. That is, the coefficients of the partial fraction decomposition are the residues.
     
    Last edited: Nov 11, 2012
  14. Nov 11, 2012 #13

    I like Serena

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    Alternatively, you can look up difficult integrals in a table (or use a mathematical program).
    The following table on wiki is applicable: http://en.wikipedia.org/wiki/List_o...e_integrals_involving_trigonometric_functions

    The relevant equation is:
    $$\int_0^{2\pi} \frac{dx}{a+b\cos x}=\frac{2\pi}{\sqrt{a^2-b^2}}$$
    To use this, you'll have to convert ##\sin^2 t## to ##\cos 2t## and you're good to go.

    If nothing else, it helps to verify long calculations (that tend to contain a couple of mistakes for some reason).
     
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