Integral of Bessel Function of the First Kind

[Airsteve0]

Homework Statement

I need to show that the definite integral (from 0 to infinity) of the Bessel function of the first kind (i.e.Jo(x)) goes to 1.

Homework Equations

All of the equations which I was given to do this problem are shown in the picture I have attached. However, I believe the result shown in part (b) is most relevant.

The Attempt at a Solution

I have fully worked through parts (a) and (b) of this question; however, I am unsure about how to attack this last problem. I was wondering if Parseval's theorem may be useful here but that has been my most recent thoughts on it.

Any help would be greatly appreciated.

 

[lanedance]

am i missing something, or what is that theta in the 2nd last formula?

I think you're probably on the right track, parseval's coupled with the eveness of the function should do it

 

[Airsteve0]

that theta is just my prof's way of writing a Heaviside step function

 

[lanedance]

i thought so, but bit confusing with the notation, probably confused other responder stoo... however that should simplify the integral

starting with parseval's and symmetry (even), you have:
[tex] 2\int_0^{\infty}J_0(x)dx = \int_{-\infty}^{\infty}J_0(x)dx = \int_{-\infty}^{\infty}J_0(x)dx
= \int_{-\infty}^{\infty}\tilde{J}_0(k)dk = \int_{-\infty}^{\infty}dk\theta(1-|k|)\frac{2}{\sqrt{1-k^2}}[/tex]

now for a given k, where is the heaviside function 0?

 

[202250]

i thought so, but bit confusing with the notation, probably confused other responder stoo... however that should simplify the integral

starting with parseval's and symmetry (even), you have:
[tex] 2\int_0^{\infty}J_0(x)dx = \int_{-\infty}^{\infty}J_0(x)dx = \int_{-\infty}^{\infty}J_0(x)dx
= \int_{-\infty}^{\infty}\tilde{J}_0(k)dk = \int_{-\infty}^{\infty}dk\theta(1-|k|)\frac{2}{\sqrt{1-k^2}}[/tex]

now for a given k, where is the heaviside function 0?

Doesn't Parseval's relation relate to the Modulus Squared of the function and its Fourier transform?

 

[lanedance]

good point, and sorry for the silly mistake, was running out the door to meet the missus, now as an idea that needs work... starting with what you want to evaluate
[tex] \int_0^{\infty}dx J_0(x) = \int_0^{\infty}dx \int_0^{\infty}dk e^{-ikx} \tilde{J}_0(k) [/tex]

now assuming we can change the order of integration...
[tex] = \int_0^{\infty}dk \int_0^{\infty}dx e^{-ikx} \tilde{J}_0(k)[/tex]

which gives
[tex] = \int_0^{\infty}dk ( e^{-ikx} \tilde{J}_0(k) )|_0^{\infty}[/tex]

the
[tex] = \int_0^{\infty}dk \tilde{J}_0(k)[/tex]

so it's worth seeing what that integrates to and whether the order change is valid...

 

[202250]

good point, and sorry for the silly mistake, was running out the door to meet the missus, now as an idea that needs work... starting what you want to evaluate, is the following
[tex] \int_0^{\infty}dx J_0(x) = \int_0^{\infty}dx \int_0^{\infty}dk e^{-ikx} \tilde{J}_0(k) [/tex]

now assuming we can change the order of integration...
[tex] = \int_0^{\infty}dk \int_0^{\infty}dx e^{-ikx} \tilde{J}_0(k)[/tex]

which gives
[tex] = \int_0^{\infty}dk ( e^{-ikx} \tilde{J}_0(k) )|_0^{\infty}[/tex]

the
[tex] = \int_0^{\infty}dk \tilde{J}_0(k)[/tex]

so it's worth seeing what that integrates to and whether the order change is valid...

I seem to see a problem with the first integration, over x. There should be i and k terms coming down from that integration, sadly. Unless I'm making a silly mistake.

 

[lanedance]

glad you're on the ball tonight... more ideas
[tex] = \int_0^{\infty}dk \int_0^{\infty}dx \\int_0^{\infty} \frac{1}{\pi} ( e^{-ikx} \tilde{J}_0(k) )[/tex]

[tex] = \int_0^{\infty}dk \int_0^{\infty}dx\frac{1}{\pi} (cos(kx) - isin(kx)) \tilde{J}_0(k) )[/tex]

and as we know the function is real, we only need keep the real part for the integral...
[tex] =\frac{1}{\pi} \int_0^{\infty}dk \int_0^{\infty} cos(kx) \tilde{J}_0(k) [/tex]

 

[202250]

glad you're on the ball tonight... more ideas
[tex] = \int_0^{\infty}dk \int_0^{\infty}dx \\int_0^{\infty} \frac{1}{\pi} ( e^{-ikx} \tilde{J}_0(k) )[/tex]

[tex] = \int_0^{\infty}dk \int_0^{\infty}dx\frac{1}{\pi} (cos(kx) - isin(kx)) \tilde{J}_0(k) )[/tex]

and as we know the function is real, we only need keep the real part for the integral...
[tex] =\frac{1}{\pi} \int_0^{\infty}dk \int_0^{\infty} cos(kx) \tilde{J}_0(k) )[/tex]

I think that this might work, but I think the way that we should approach it is to use the fact that we know that the function and the Fourier transform are even, and then we are allowed to do the following:

[tex] \int_0^{\infty}dx J_0(x) = \frac{1}{2} \int_{-\infty}^{\infty}dx J_0(x) [/tex]

Then we can say:

[tex] \frac{1}{2} \int_{-\infty}^{\infty}dx J_0(x) = \frac{1}{2} \int_{-\infty}^{\infty}dx \frac{1}{2\pi} \int_{-\infty}^{\infty}dk e^{ikx} \tilde{J_0} [/tex]

Now, switch the order of integration, and the J-tilde doesn't depend on x, so we can write

[tex] = \frac{1}{4\pi} \int_{-\infty}^{\infty}dk \tilde{J_0} \int_{-\infty}^{\infty}dx e^{ikx} [/tex]

The second integral is the Fourier transform of 1 which is

[tex] 2\pi \delta(k) [/tex]

So integrating over k pulls out the value of J-tilde at k = 0, which is 2 (see above posts).
Then we have the desired result

[tex] \int_0^{\infty}dx J_0(x) = \frac{1}{4\pi} 2*2*\pi = 1 [/tex]

 

[Airsteve0]

Thanks for all the help! That was an interesting problem

 

[mkk123]

Hi All
I need your support for solution for this equation in attached
please I need fast reply for my ask

 

[mkk123]

any body answer me because I relley need this solution or factorial for this eq. in attahced

 

[lanedance]

Hi mkk123 - welcome to PF!

This is an old closed post, no point re-opening. You should post new questions as new questions in the forum - way more people will look at it that way