Calculating the Laplace transform of a Bessel function

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  • #1
fluidistic
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Homework Statement


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.


Homework Equations


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


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.
 

Answers and Replies

  • #2
STEMucator
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I don't believe that integral is possible by elementary means... I think you can only approximate it.
 
  • #3
fluidistic
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I don't believe that integral is possible by elementary means... I think you can only approximate it.
Ok thanks. This probably mean there's a trick I'm missing to solve the problem.
 
  • #4
STEMucator
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I might be mistaken here, but try expressing it as a series.
 
  • #5
fluidistic
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I might be mistaken here, but try expressing it as a series.
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.
 
  • #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.
 
  • #7
fluidistic
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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.
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.
 
  • #8
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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.
 
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  • #9
fluidistic
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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.
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.
 
  • #10
I like Serena
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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.
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.
 
  • #11
fluidistic
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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.
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.
 
  • #12
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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.
 
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  • #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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