
#1
Nov2205, 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 



#2
Nov2205, 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. 



#3
Nov2305, 03:41 AM

Sci Advisor
HW Helper
P: 1,593

[tex]\int_0^R x^3 \text{BesselJ[0,ax]}dx[/tex] ? Wait a minute, let me just flatout 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] 



#4
Dec1007, 07:06 PM

P: 1

Integral of Bessel function
Use the recurrence relation:
[tex]J_{n1}(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_{n1}(x) = \frac{d}{dx}[x^n J_{n}(x)][/tex] on each of the terms and perform the integration... 



#5
Dec1307, 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 electroheating...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 



#6
Dec608, 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_{n1}(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_{n1}(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. 



#7
Jan909, 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 



#8
Jan909, 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^{n1}axJ_0(ax)\; dx = \frac{1}{a}\int x^{n1}\frac{d}{dx}\left(xJ_1(ax)\right)\;dx = \frac{1}{a}\left[x^{n1}\cdot x J_1(ax) \int (n1) x^{n2}\cdot x J_1(ax) \;dx\right]= [/tex] [tex]=\frac{x^n}{a}J_1(ax)+\frac{n1}{a^2}\int x^{n1}\cdot\left(aJ_1(ax)\right)\; dx = \frac{x^n}{a}J_1(ax)+\frac{n1}{a^2}\int x^{n1}\cdot\left(\frac{dJ_0(ax)}{dx}\right)\; dx = [/tex] [tex]= \frac{x^n}{a}J_1(ax) + \frac{n1}{a^2}\left[x^{n1}J_0(ax)\int (n1) x^{n2} J_0(ax)\; dx\right] = \frac{x^n}{a}J_1(ax) +\frac{n1}{a^2}x^{n1}J_0(ax)\frac{(n1)^2}{a^2}\underbrace{\int x^{n2}J_0(ax)\;dx}_{S_{n2}} [/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{n1}{a^2}x^{n1}J_0(ax)\frac{(n1)^2}{a^2}S_{n2}[/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] 



#9
Jan1209, 02:37 AM

P: 2

Hi Thaakisfox
Thanks a lot. This will take me some time to digest !! Bye, SamSvL 



#10
Apr1410, 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.fhjena.de/~rsh/Forschung/Stoer/besint.pdf Perhaps it is still of some use. I am still working to add some more integrals. 



#11
Feb2312, 06:05 AM

P: 37

Can you turn the formula into tex format?




#12
Nov2212, 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 



#13
Nov2212, 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 



#14
Nov2212, 10:49 PM

P: 136

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



#15
Nov2312, 12:04 AM

P: 746

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



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