Evaluate integral of Bessel function.

In summary: I mean Happy Christmas!In summary, the conversation revolved around trying to evaluate the indefinite integral of J_2(x) using various identities involving Bessel functions. The closest solution involved using the identities (1), (3), and (4), but the integral could not be further simplified. It was suggested to try calculating a definite integral instead and to write out J_0(x) in its series form for a possible solution. The conversation also touched on the possibility of finding a simpler representation of the integral using higher integer orders of J_0(x) and J_1(x).
  • #1
yungman
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I am trying to evaluate[tex]\int J_{2}(x)dx[/tex]

I have been trying to use all the identities involving Bessel function to no prevail. The ones I used are:

[tex]\frac{d}{dx}[x^{-p}J_{p}(x)]=-x^{-p}J_{p+1}(x)[/tex] (1)

[tex]\frac{d}{dx}[x^{p}J_{p}(x)]=-x^{p}J_{p-1}(x)[/tex] (2)

[tex]J_{p-1}(x)]+J_{p+1}(x)=\frac{2p}{x}J_{p}(x)[/tex] (3)

[tex]J'_{0}(x)=-J_{1}(x)[/tex] (4)


So far this is the closest:

[tex]\int J_{2}(x)dx=\int x[x^{-1}J_{2}(x)]dx[/tex] Using (1) with p=1 [tex]=-\int x \frac{d}{dx}[x^{-1}J_{1}](x)]dx[/tex]

[tex]u=x,du=dx, v=x^{-1}J_{1}(x)[/tex] using intergate by parts:

[tex]\int J_{2}(x)dx=-J_{1}+\int x^{-1}J_{1}(x)dx[/tex] Using (3) p=1 [tex]\int J_{2}(x)dx=-J_{1}(x)+\frac{1}{2}\int J_{0}(x)dx+\frac{1}{2}\int J_{2}(x)dx[/tex]

[tex]\Rightarrow \frac{1}{2}\int J_{2}(x)dx=-J_{1}(x)+\frac{1}{2}\int J_{0}(x)dx \Rightarrow \int J_{2}(x)dx=-2J_{1}(x)+\int J_{0}(x)dx[/tex]



I have not been able to go any further, can anyone help or at least give me a hint. Is there a general approach of solving these problems of integration. I worked on integrating of Bessel function of order 3, order 1. The methods are very different.


Thanks
Alan
 
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  • #2
Anyone please!
 
  • #3
yungman said:
I am trying to evaluate[tex]\int J_{2}(x)dx[/tex]

I have been trying to use all the identities involving Bessel function to no prevail. The ones I used are:

[tex]\frac{d}{dx}[x^{-p}J_{p}(x)]=-x^{-p}J_{p+1}(x)[/tex] (1)

[tex]\frac{d}{dx}[x^{p}J_{p}(x)]=-x^{p}J_{p-1}(x)[/tex] (2)

[tex]J_{p-1}(x)]+J_{p+1}(x)=\frac{2p}{x}J_{p}(x)[/tex] (3)

[tex]J'_{0}(x)=-J_{1}(x)[/tex] (4)


So far this is the closest:

[tex]\int J_{2}(x)dx=\int x[x^{-1}J_{2}(x)]dx[/tex] Using (1) with p=1 [tex]=-\int x \frac{d}{dx}[x^{-1}J_{1}](x)]dx[/tex]

[tex]u=x,du=dx, v=x^{-1}J_{1}(x)[/tex] using intergate by parts:

[tex]\int J_{2}(x)dx=-J_{1}+\int x^{-1}J_{1}(x)dx[/tex] Using (3) p=1 [tex]\int J_{2}(x)dx=-J_{1}(x)+\frac{1}{2}\int J_{0}(x)dx+\frac{1}{2}\int J_{2}(x)dx[/tex]

[tex]\Rightarrow \frac{1}{2}\int J_{2}(x)dx=-J_{1}(x)+\frac{1}{2}\int J_{0}(x)dx \Rightarrow \int J_{2}(x)dx=-2J_{1}(x)+\int J_{0}(x)dx[/tex]



I have not been able to go any further, can anyone help or at least give me a hint. Is there a general approach of solving these problems of integration. I worked on integrating of Bessel function of order 3, order 1. The methods are very different.


Thanks
Alan

I'm curious that you are trying to calculate an indefinite integral for [itex] J_2(x) [/itex]. Since Bessel functions are so widely used in applications, could you recast the problem as a definite integral?

It seems that you've achieved some simplification in the last step. Couldn't you write [itex] J_0(x) [/itex] out in its series form and integrate it term by term? You'd have to worry about convergence of the resulting infinite series, but that might give you a way to get a usable numerical answer if that's what you need.
 
  • #4
AEM said:
I'm curious that you are trying to calculate an indefinite integral for [itex] J_2(x) [/itex]. Since Bessel functions are so widely used in applications, could you recast the problem as a definite integral?

It seems that you've achieved some simplification in the last step. Couldn't you write [itex] J_0(x) [/itex] out in its series form and integrate it term by term? You'd have to worry about convergence of the resulting infinite series, but that might give you a way to get a usable numerical answer if that's what you need.

Thanks for the reply, this is the section of "Identities of Bessel Functions" where they switch the order around as shown in (1) (2) (3) etc. They start out as indefinite integral, but in real application, it would be a definite integral, you are correct on that. This is just as exercise.

Yes, writing out [tex]J_{0}[/tex] would be the last resort, I am just thinking is there any answer using simple J function like those shown.

Thanks
 
  • #5
yungman said:
Yes, writing out [tex]J_{0}[/tex] would be the last resort, I am just thinking is there any answer using simple J function like those shown.

I just calculated the integral of [itex] J_0(x) [/itex] for fun and it's not bad. I haven't looked at the Bessel identities in years (many years), but I'll take a look at them for a change of pace. Will let you know if I come up with something.
 
  • #6
AEM said:
I just calculated the integral of [itex] J_0(x) [/itex] for fun and it's not bad. I haven't looked at the Bessel identities in years (many years), but I'll take a look at them for a change of pace. Will let you know if I come up with something.


I know I can integrate it as a series as shown:

[tex]\int J_{0}(x)=\int \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!\Gamma(n+1)2^{2n}}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!\Gamma(n+1)2^{2n}}\int x^{2n} dx=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!\Gamma(n+1)2^{2n}}\frac{x^{2n+1}}{(2n+1)}[/tex]

[tex]\int J_{0}(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n!)^{2}2^{2n}}\frac{x^{2n+1}}{(2n+1)}[/tex]

I thought there might be some fancy identity that I miss! Could it be that's it??

Thanks
Alan
 
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  • #7
Anyone willing to give me a Christman preasent?!
 
  • #8
yungman said:
Anyone willing to give me a Christman preasent?!

Here's a tiny one: [itex] \Gamma(n+1) = n! [/itex] for n integer.
 
  • #9
AEM said:
Here's a tiny one: [itex] \Gamma(n+1) = n! [/itex] for n integer.

This is still in series form, should their be a representation in [tex]f(J_{p})[/tex] form instead of series form. I added your suggestion to post #6 already. I did spent more time this morning, I tried all 8 of the identities, I have not manage to find anything that can bring it to a simplier form.

I have a suspicion that in this kind of integrals all the higher integer order is a function of [tex] J_{0}(x)[/tex] and [tex]J_{1}(x)[/tex] where [tex]J_{1}(x)=-J'_{0}(x)[/tex]

Thanks, Merry Christmas
Alan
 
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  • #10
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html" can be of use. E.g., you immediately see that the derivative of the Bessel function can be expressed in terms of Bessel functions of one order higher and lower.
 
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  • #11
Count Iblis said:
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html" can be of use. E.g., you immediately see that the derivative of the Bessel function can be expressed in terms of Bessel functions of one order higher and lower.

I am not familiar with contour integral and I don't think I should go too deep into this for the PDE. I am stuck on this section for about a month and this is only a very small part of the PDE! I am hopping to find a bessel function identity for this but I am starting to think there is not easy answer.
Thanks
 
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  • #13
Well, here's what Maple produces when asked to integrate [itex] J_2(x) [/itex]

x*BesselJ(2,x)*LommelS1(-1,1,x)-x*BesselJ(1,x)*LommelS1(0,2,x)

According to Maple,

[itex] LommelS1( \mu, \nu, z) [/itex] and [itex] LommelS2( \mu, \nu, z) [/itex]

solve the following differential equation:

[itex] z^2 y'' + z y' + (z^2 - \nu^2) y = z^(\mu +1) [/itex]

So I think you're probably right there is no simple identity for your integral.
 
  • #14
AEM said:
Well, here's what Maple produces when asked to integrate [itex] J_2(x) [/itex]

x*BesselJ(2,x)*LommelS1(-1,1,x)-x*BesselJ(1,x)*LommelS1(0,2,x)

According to Maple,

[itex] LommelS1( \mu, \nu, z) [/itex] and [itex] LommelS2( \mu, \nu, z) [/itex]

solve the following differential equation:

[itex] z^2 y'' + z y' + (z^2 - \nu^2) y = z^(\mu +1) [/itex]

So I think you're probably right there is no simple identity for your integral.

Thanks for your time. What is maple?

Do you know of a site that calculate answer for Bessel function if I type in the question?

Thanks
 
  • #15
yungman said:
Thanks for your time. What is maple?

Do you know of a site that calculate answer for Bessel function if I type in the question?

Thanks

First, this will give you info on the relationship between Lommel functions and Bessel functions:

http://en.wikipedia.org/wiki/Lommel_function

As you probably noticed Lommel's D.E. is a nonhomogeneous Bessel equation.

Second, Maple is a symbolic manipulation program, similar to Mathematica. It does all kinds of good stuff and is a very valuable resource. The full blown version is rather pricey, but you may be able to get a student version if you are affiliated with a college, or university as a student for a lot less. Mathematica is also pricey and has a student version, I believe. The limitations on the student versions aren't too severe, things like limitations on the size of matrices that you can enter, etc.

Third, I'm not aware of a website that will calculate Bessel functions for you, although I'm sure if you Google Bessel functions you'll come up with something.

Merry Christmas!
 
  • #17
Thanks guys for all the help.

I am not a student so I can't get any discount. I'll check out Wolfram and see what happen. Beside I am almost done with the section on Bessel series expansion...Good riddens! I can't wait to move onto Lagendre and the rest of the PDE. I spent way too much time on this already. I wasted over a week trying to learn Bessel function of the second kind just to realize I don't need to learn it for the series expansion require for the PDE! It is because I study on my own, that is why a lot of simple basic questions that I should ask the professor in school end up have to come here to ask.

Merry Christmas

Alan
 

Related to Evaluate integral of Bessel function.

1. What is a Bessel function?

A Bessel function is a mathematical function that arises in many areas of physics and engineering, particularly in problems involving waves and vibrations. It is named after the German mathematician Friedrich Bessel.

2. What does it mean to evaluate the integral of a Bessel function?

Evaluating the integral of a Bessel function means finding the value of the integral (or area under the curve) of a specific Bessel function over a given range of values. This can be done using various mathematical techniques, such as integration by parts or substitution.

3. What are the applications of Bessel functions?

Bessel functions have many important applications in physics and engineering. They can be used to model the behavior of physical systems that involve waves, such as sound waves, electromagnetic waves, and water waves. They are also used in solving differential equations and in signal processing.

4. How do you evaluate the integral of a Bessel function?

The method for evaluating the integral of a Bessel function depends on the specific function and the range of values given. In general, it involves using mathematical techniques such as integration by parts, substitution, or using known properties of Bessel functions.

5. Can the integral of a Bessel function be evaluated analytically?

Yes, for certain values and ranges, the integral of a Bessel function can be evaluated analytically, meaning it can be expressed in terms of known mathematical functions. However, for more complex functions and ranges, it may require numerical methods to approximate the value of the integral.

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