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Integral of Bessel function

  1. Nov 22, 2005 #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
     
  2. jcsd
  3. Nov 22, 2005 #2

    mezarashi

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    Last edited: Nov 22, 2005
  4. Nov 23, 2005 #3

    saltydog

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    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]
     
    Last edited: Nov 23, 2005
  5. Dec 10, 2007 #4
    Integrals of Bessel Functions

    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...
     
    Last edited: Dec 10, 2007
  6. Dec 13, 2007 #5
    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
     
    Last edited: Dec 13, 2007
  7. Dec 6, 2008 #6
    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.
     
    Last edited: Dec 6, 2008
  8. Jan 9, 2009 #7
    I am looking for the solution of \int_{0}^{R} x^5 J_0 (ax) dx
    Any ideas about a closed form solution?
    Thanks
     
  9. Jan 9, 2009 #8
    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]
     
  10. Jan 12, 2009 #9
    Hi Thaakisfox

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

    Bye,

    SamSvL
     
  11. Apr 14, 2010 #10
    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.
     
    Last edited: Apr 15, 2010
  12. Feb 23, 2012 #11
    Can you turn the formula into tex format?
     
  13. Nov 22, 2012 #12
    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
     
  14. Nov 22, 2012 #13
    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
     
  15. Nov 22, 2012 #14
    Off topic:
    Im seeing posts from '05 '07 '08 '09 '10 and '12. That's a lot of resurrections.
     
  16. Nov 23, 2012 #15
    To think this thread was created 7 years ago to the day..crazy necro
     
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