# Evaluate integral of Bessel function.

1. Dec 22, 2009

### yungman

I am trying to evaluate$$\int J_{2}(x)dx$$

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

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

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

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

$$J'_{0}(x)=-J_{1}(x)$$ (4)

So far this is the closest:

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

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

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

$$\Rightarrow \frac{1}{2}\int J_{2}(x)dx=-J_{1}(x)+\frac{1}{2}\int J_{0}(x)dx \Rightarrow \int J_{2}(x)dx=-2J_{1}(x)+\int J_{0}(x)dx$$

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

Last edited: Dec 23, 2009
2. Dec 23, 2009

### yungman

3. Dec 23, 2009

### AEM

I'm curious that you are trying to calculate an indefinite integral for $J_2(x)$. 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 $J_0(x)$ 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. Dec 23, 2009

### yungman

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 $$J_{0}$$ would be the last resort, I am just thinking is there any answer using simple J function like those shown.

Thanks

5. Dec 23, 2009

### AEM

I just calculated the integral of $J_0(x)$ 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. Dec 23, 2009

### yungman

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

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

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

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

Thanks
Alan

Last edited: Dec 24, 2009
7. Dec 24, 2009

### yungman

Anyone willing to give me a Christman preasent?!!!

8. Dec 24, 2009

### AEM

Here's a tiny one: $\Gamma(n+1) = n!$ for n integer.

9. Dec 24, 2009

### yungman

This is still in series form, should their be a representation in $$f(J_{p})$$ 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 $$J_{0}(x)$$ and $$J_{1}(x)$$ where $$J_{1}(x)=-J'_{0}(x)$$

Thanks, Merry Christmas
Alan

Last edited: Dec 24, 2009
10. Dec 24, 2009

### Count Iblis

http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html" [Broken] 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.

Last edited by a moderator: May 4, 2017
11. Dec 25, 2009

### yungman

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

Last edited by a moderator: May 4, 2017
12. Dec 25, 2009

### yungman

13. Dec 25, 2009

### AEM

Well, here's what Maple produces when asked to integrate $J_2(x)$

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

According to Maple,

$LommelS1( \mu, \nu, z)$ and $LommelS2( \mu, \nu, z)$

solve the following differential equation:

$z^2 y'' + z y' + (z^2 - \nu^2) y = z^(\mu +1)$

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

14. Dec 25, 2009

### yungman

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. Dec 25, 2009

### AEM

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!!

16. Dec 25, 2009

### Avodyne

17. Dec 25, 2009

### yungman

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