How Do You Integrate x^3 J_0(ax) Over 0 to R?

[mikel]

Hello,

I am a geologist working on a fluid mechanics problem. Solving the PDE for my problem, this Bessel integral arises:

\int_{0}^{R} x^3 J_0 (ax) dx

where J_0 is the Bessel function of first kind, and a is a constant.

I haven't found the solution in any table or book, and due to my limited background in applied mathematics I don't know how to integrate it by myself.

Does anybody know the solution?

Thanks a lot in advance

 

[mezarashi]

Yes, in fact, the Bessel function cannot be described using a finite number of 'elementary functions'. That's what the tables are for. Let me see if I can find one for you.

http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP?Res=200

Seems to have what you may be looking for.

 

[saltydog]

Hello,
I am a geologist working on a fluid mechanics problem. Solving the PDE for my problem, this Bessel integral arises:
\int_{0}^{R} x^3 J_0 (ax) dx
where J_0 is the Bessel function of first kind, and a is a constant.
I haven't found the solution in any table or book, and due to my limited background in applied mathematics I don't know how to integrate it by myself.
Does anybody know the solution?
Thanks a lot in advance

Have you tried plugging into Mathematica:

\int_0^R x^3 \text{BesselJ[0,ax]}dx

?

Wait a minute, let me just flat-out ask how does one verify that:

\int_0^R x^3 J_0(ax)dx=\frac{1}{a^2}\left[R^2\left\{2J_2(aR)-aRJ_3(aR)\right\}\right]

Suppose need to first show:

\int x^3J_0(x)dx=x^2\left[2J_2(x)-xJ_3(x)\right]

 

[Larsiboy]

Integrals of Bessel Functions

Use the recurrence relation:

J_{n-1}(x) = \frac{2n}{x} J_{n}(x) - J_{n+1}(x)

to write the integral as

\int x^3 J_0(x)dx = \int x^3 (\frac{2}{x} J_{1}(x) - J_{2}(x)) dx<br /> = \int (2 x^2 J_{1}(x) - x^3 J_{2}(x) ) dx

then use the relation

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

on each of the terms and perform the integration...

 

[sachale]

May be it's too late, for you...but I've just found the same integral studying heat transmission in electro-heating...so don't become mad, i think it's easy:

\int_0^R x^3 J_0(ax)dx=\frac{1}{a^2}\left[R^2\left\{2J_2(aR)-aRJ_3(aR)\right\}\right]

J_3(aR)=\frac{4}{aR}J_2(aR)-J_1(aR)

So
\int_0^R x^3 J_0(ax)dx=\frac{1}{a^2}R^2\left\{2J_2(aR)-(aR)\left[\frac{4}{aR}J_2(aR)-J_1(aR)\right]\right\}=\frac{1}{a^2}R^2\left\{-2J_2(aR)+(aR)J_1(aR)\right\}=<br /> \frac{R^3}{a}J_1(aR)-\frac{2R^2}{a^2}J_2(aR)

I hope I'm right, if not...let's talk about!
Bye

 

[j0rt3g4]

Well I was searching for the normalitation of the bessel funtions, and I found this... It was interesting for me so I realice this integral. First of all, I will make a change:
<br /> s= ax \quad \quad ds= a dx \quad \quad x=\frac{s}{a}<br />
Take care of limits
<br /> \textrm{If } \quad x=0 \Rightarrow s=0 \textrm{ and if } x=R \Rightarrow s=aR<br />

So this: multiplied for 1 = a/a \int_{0}^{R} x^3 J_0 (ax) dx \Rightarrow \frac{1}{a}\int_{0}^{R} x^3 J_0 (ax) adx

Became this:\frac{1}{a^4} \int_0^{aR} s^3 J_0 (s) ds

Next Step is use J_{n-1}(x) = \frac{2n}{x} J_{n}(x) - J_{n+1}(x)

Taking n=1 \Rightarrow J_{0}(s) = \frac{2}{s} J_{1}(s) - J_{2}(s)

Replacing this in the integral:
<br /> \frac{1}{a^4} \int_0^{aR} s^3 J_0 (s) ds = \frac{1}{a^4} \left\{ \int_0^{aR} s^3 \left( \frac{2}{s} J_{1}(s) - J_{2}(s) \right) ds \right\} =<br /> \frac{1}{a^4} \left\{ 2\int_0^{aR} s^2 J_{1}(s) ds - \int_0^{aR} s^3 J_{2}(s) ds \right\}<br />

In this time I use:
<br /> \int \frac{d}{ds}[s^n J_{n}(s)] = \int s^n J_{n-1}(s) = \left[s^n J_{n}(s)\right]<br />

<br /> \frac{1}{a^4} \left\{ 2\int_0^{aR} s^2 J_{1}(s) ds - \int_0^{aR} s^3 J_{2}(s) ds \right\} =\frac{1}{a^4} \left\{2 s^2 J_{2}(s)\left|_0^{aR} \right. - s^3 J_{3}(s)\left|_0^{aR} \right. \right\} =<br /> \frac{1}{a^4} \left\{ \left( 2 (aR)^2 J_{2}(aR) - 0 \right) - \left( (aR)^3 J_{3}(aR) -0\right) \right\}<br />
So getting the 1/a4 inside.

<br /> \left[ <br /> \frac{2}{a^2} R^2 J_2(aR) -\frac{1}{a} R^3 J_3(aR)<br /> \right]<br />

And That's it :) cheers! ... Good look with that work.

 

[samsvl]

I am looking for the solution of \int_{0}^{R} x^5 J_0 (ax) dx
Any ideas about a closed form solution?
Thanks

 

[Thaakisfox]

We can obtain a recursion equation for it, which in this case can be solved... The answer is not too "nice" though... ;)
Consider:

S_n=\int x^n J_0(ax)\; dx
In your case n=5.

For bessel functions we now the following recursions :

\frac{d}{dx}\left(xJ_1(ax)\right)=axJ_0(ax)
and
\frac{d}{dx}J_0(ax)=-aJ_1(ax)

So we have:

S_n=\int x^n J_0(ax)\; dx= \frac{1}{a}\int x^{n-1}axJ_0(ax)\; dx = \frac{1}{a}\int x^{n-1}\frac{d}{dx}\left(xJ_1(ax)\right)\;dx = \frac{1}{a}\left[x^{n-1}\cdot x J_1(ax) -\int (n-1) x^{n-2}\cdot x J_1(ax) \;dx\right]=

=\frac{x^n}{a}J_1(ax)+\frac{n-1}{a^2}\int x^{n-1}\cdot\left(-aJ_1(ax)\right)\; dx = \frac{x^n}{a}J_1(ax)+\frac{n-1}{a^2}\int x^{n-1}\cdot\left(\frac{dJ_0(ax)}{dx}\right)\; dx =

= \frac{x^n}{a}J_1(ax) + \frac{n-1}{a^2}\left[x^{n-1}J_0(ax)-\int (n-1) x^{n-2} J_0(ax)\; dx\right] = \frac{x^n}{a}J_1(ax) +\frac{n-1}{a^2}x^{n-1}J_0(ax)-\frac{(n-1)^2}{a^2}\underbrace{\int x^{n-2}J_0(ax)\;dx}_{S_{n-2}}

Where we used integration by parts twice.

So we obtained the following recursion:

S_n=\frac{x^n}{a}J_1(ax) +\frac{n-1}{a^2}x^{n-1}J_0(ax)-\frac{(n-1)^2}{a^2}S_{n-2}

for n=1 we have:

S_1=\int xJ_0(ax)\;dx = \frac{1}{a}\int axJ_0(ax)\; dx = \frac{x}{a} J_1(ax)

Using the the recursion for n=5 will be after manipulation:

S_5 = \int x^5 J_0(ax) \; dx = \frac{4x^4}{a^2}\left[1-\frac{8}{a^2x^2}\right]J_0(ax) +\frac{x^5}{a}\left[1-\frac{16}{a^2x^2}+\frac{48}{a^4x^4}\right]J_1(ax)

 

[samsvl]

Hi Thaakisfox

Thanks a lot. This will take me some time to digest !

Bye,

SamSvL

 

[Rosenheinrich]

To solve some problems I was looking for integrals of Bessel functions.
In the end I decided to make my own table.
It can be found here:

http://www.fh-jena.de/~rsh/Forschung/Stoer/besint.pdf

Perhaps it is still of some use.
I am still working to add some more integrals.

 

[Chuck88]

Can you turn the formula into tex format?

 

[sabz]

Hi
I am interested in finding weighting function in Orthogonal Bessel Functions (w(x)) .

integral|[0,c](w(x)*J(V)*J(W))
I have considered many books but they do not explained how can we obtain it.
I know there are 2 different bessel functions in this case and each have special weighting function in orthogonal condition.
Could anybody help me please?
Best regards

 

[sabz]

Hi
I am interested in finding weighting function in Orthogonal Bessel Functions (w(x)) .

\int(w(x)*J(V)*J(W))dx
I have considered many books but they do not explained how can we select or obtain it.
I know there are 2 different bessel functions in this case and each have special weighting function in orthogonal condition.
Could anybody help me please?
Best regards

 

[Klungo]

Off topic:
Im seeing posts from '05 '07 '08 '09 '10 and '12. That's a lot of resurrections.

 

[member 392791]

To think this thread was created 7 years ago to the day..crazy necro