Integral of Bessel function


by mikel
Tags: bessel, function, integral
mikel
mikel is offline
#1
Nov22-05, 06:07 PM
P: 1
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
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mezarashi
mezarashi is offline
#2
Nov22-05, 06:24 PM
HW Helper
P: 661
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/ConvertI...55.ASP?Res=200

Seems to have what you may be looking for.
saltydog
saltydog is offline
#3
Nov23-05, 03:41 AM
Sci Advisor
HW Helper
P: 1,593
Quote Quote by 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
Have you tried plugging into Mathematica:

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

?

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

[tex]\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][/tex]

Suppose need to first show:

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

Larsiboy
Larsiboy is offline
#4
Dec10-07, 07:06 PM
P: 1

Integral of Bessel function


Use the recurrence relation:

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

to write the integral as

[tex]\int x^3 J_0(x)dx = \int x^3 (\frac{2}{x} J_{1}(x) - J_{2}(x)) dx
= \int (2 x^2 J_{1}(x) - x^3 J_{2}(x) ) dx[/tex]

then use the relation

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

on each of the terms and perform the integration...
sachale
sachale is offline
#5
Dec13-07, 12:27 PM
P: 1
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:

[tex]\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][/tex]

[tex]J_3(aR)=\frac{4}{aR}J_2(aR)-J_1(aR)[/tex]

So
[tex]\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\}=
\frac{R^3}{a}J_1(aR)-\frac{2R^2}{a^2}J_2(aR)[/tex]

I hope i'm right, if not...let's talk about!
Bye
j0rt3g4
j0rt3g4 is offline
#6
Dec6-08, 08:59 PM
P: 1
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:
[tex]
s= ax \quad \quad ds= a dx \quad \quad x=\frac{s}{a}
[/tex]
Take care of limits
[tex]
\textrm{If } \quad x=0 \Rightarrow s=0 \textrm{ and if } x=R \Rightarrow s=aR
[/tex]

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

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

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

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

Replacing this in the integral:
[tex]
\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\} =
\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\}
[/tex]

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

[tex]
\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\} =
\frac{1}{a^4} \left\{ \left( 2 (aR)^2 J_{2}(aR) - 0 \right) - \left( (aR)^3 J_{3}(aR) -0\right) \right\}
[/tex]
So getting the 1/a4 inside.

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

And That's it :) cheers! ... Good look with that work.
samsvl
samsvl is offline
#7
Jan9-09, 08:26 AM
P: 2
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
Thaakisfox is offline
#8
Jan9-09, 10:22 AM
P: 263
We can obtain a recursion equation for it, which in this case can be solved... The answer is not too "nice" though... ;)
Consider:

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

For bessel functions we now the following recursions :

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

So we have:

[tex]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]= [/tex]

[tex]=\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 = [/tex]

[tex]= \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}} [/tex]

Where we used integration by parts twice.

So we obtained the following recursion:

[tex]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}[/tex]

for n=1 we have:

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

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

[tex]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)[/tex]
samsvl
samsvl is offline
#9
Jan12-09, 02:37 AM
P: 2
Hi Thaakisfox

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

Bye,

SamSvL
Rosenheinrich
Rosenheinrich is offline
#10
Apr14-10, 04:23 AM
P: 2
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
Chuck88 is offline
#11
Feb23-12, 06:05 AM
P: 37
Can you turn the formula into tex format?
sabz
sabz is offline
#12
Nov22-12, 02:52 PM
P: 2
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
sabz is offline
#13
Nov22-12, 02:55 PM
P: 2
Hi
I am interested in finding weighting function in Orthogonal Bessel Functions (w(x)) .

[itex]\int[/itex](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
Klungo is offline
#14
Nov22-12, 10:49 PM
P: 136
Off topic:
Im seeing posts from '05 '07 '08 '09 '10 and '12. That's a lot of resurrections.
Woopydalan
Woopydalan is offline
#15
Nov23-12, 12:04 AM
P: 746
To think this thread was created 7 years ago to the day..crazy necro


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