# Integral of square of Bessel function

1. Jul 30, 2009

### vietha

Hi there,

I am starting with the Bessel functions and have some problems with it. I am getting stuck with this equation. I could not find this kind of integral in the handbooks.

1. $$\int_0^aJ_0^2(bx)dx$$

Besides of this, I have other equations in similar form but I think this integral is the key to solve others:

2. $$\int_0^\infty J_0^2(bx)e^{-\frac{x}{c}}dx$$

3. $$\int_0^\infty J_0^2(bx)e^{-\frac{x^2}{c}}dx$$

3. $$\int_0^\infty J_0^2(bx)\frac{x}{c}dx$$

2. Jul 31, 2009

### g_edgar

I asked Maple, and got something in terms of the Struve H function.

$$\int _{0}^{a}\! \left( {{\rm J}_0\left(bx\right)} \right) ^{2}{dx}=- a \left( -2\,{{\rm J}_0\left(ba\right)}+\pi \, {{\rm J}_0\left(ba\right)}{\rm H}_1 \left(ba \right) -\pi \, {{\rm J}_1\left(ba\right)}{\rm H}_0 \left(ba \right) \right) /2$$

added: This is wrong. I for got the square. This is only $\int _{0}^{a}\! {{\rm J}_0\left(bx\right)} {dx}$

Last edited: Jul 31, 2009
3. Jul 31, 2009

### vietha

Hi g_edgar,

Thank you for your reply. I tried with Maple too and I got this:

$$a*hypergeom([1/2, 1/2], [1, 1, 3/2], -a^2*b^2)$$

The equation you got must be the result of this integral: $$\int _{0}^{a}\! {{\rm J}_0\left(bx\right)}{dx}$$

I have to search for the generalized hypergeometric function. I have a little knowledge on this.

4. Jul 31, 2009

### jasonRF

I looked these up in the book "integrals of bessel functions" by Luke, McGraw-Hill 1962.

$$\int_0^1 dt \ J_0^2(b t) & = & 2\ J_1(b) \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+1} J_{2k+1}(b).$$

$$\int_0^\infty dt e^{-pt} J_0^2(bt) & = & \frac{k {\mathbf{K}}(k)}{\pi b}$$

for

$$Re(p)>0$$

where
$$k^2 & = & \frac{4 b^2}{p^2 + 4 b^2}$$
and
$${\mathbf{K}}(k)}$$
is the complete elliptic integral of the first kind.

$$\int_0^\infty dt e^{-p^2t^2} J_0^2(bt) & = & \frac{\Gamma(\frac{1}{2})}{2p} \ _3F_3 (\frac{1}{2},1,\frac{1}{2}; 1,1,1 | - \frac{b^2}{p^2} ),$$
$$for Re(p^2)>0$$

Are you sure this converges? Given the asymptotic expansion of $$J_0$$ I'm skeptical.

5. Jul 31, 2009

### vietha

Thanks jasonRF for the results. I have that book too. Could you tell me in which parts and pages you found that?

I made a mistake with the last one. It should be:
$$\int_0^a J_0^2(bx)\frac{x}{c}dx$$

Last edited: Jul 31, 2009
6. Apr 12, 2011

### victor_santos

Hi there,

I found this integral at Gradshteyn:

$$\int_{0}^{1} x\, J_{\nu}(\alpha\,x)J_{\nu}(\beta\,x)\,dx = \frac{\beta J_{\nu-1}(\beta)J_{\nu}(\alpha) - \alpha J_{\nu-1}J_{\nu}(\beta)}{\alpha^2 - \beta^2}.$$

Then, taking the limit $$\beta\rightarrow\alpha$$ you can find

$$\int_{0}^{1}x\,J_{\nu}^2(\alpha\,x) dx = -\frac{1}{2\alpha}\left[J_{\nu-1}(\alpha)J_{\nu}(\alpha) + \alpha J_{\nu-1}^{\prime}(\alpha)J_{\nu}(\alpha) - \alpha J_{\nu-1}(\alpha)J_{\nu}^{\prime}(\alpha)\right].$$

I hope it is useful.