Integral of square of Bessel function

vietha
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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


Please help me. It is highly appriciated.
 
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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}=-<br /> a \left( -2\,{{\rm J}_0\left(ba\right)}+\pi \,<br /> {{\rm J}_0\left(ba\right)}{\rm H}_1 \left(ba \right) -\pi \,<br /> {{\rm J}_1\left(ba\right)}{\rm H}_0 \left(ba \right) <br /> \right) /2<br />

added: This is wrong. I for got the square. This is only \int _{0}^{a}\! {{\rm J}_0\left(bx\right)} {dx}
 
Last edited:
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.
 
I looked these up in the book "integrals of bessel functions" by Luke, McGraw-Hill 1962.

vietha said:
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

<br /> \int_0^1 dt \ J_0^2(b t) &amp; = &amp; 2\ J_1(b) \sum_{k=0}^{\infty} <br /> \frac{(-1)^k}{2k+1} J_{2k+1}(b).<br />


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


<br /> \int_0^\infty dt e^{-pt} J_0^2(bt) &amp; = &amp; \frac{k {\mathbf{K}}(k)}{\pi b}<br />

for

<br /> Re(p)&gt;0<br />

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



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


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

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


Are you sure this converges? Given the asymptotic expansion of J_0 I'm skeptical.
 
Thanks jasonRF for the results. I have that book too. Could you tell me in which parts and pages you found that?

Originally Posted by vietha View Post

3. \int_0^\infty J_0^2(bx)\frac{x}{c}dxAre you sure this converges? Given the asymptotic expansion of LaTeX Code: J_0 I'm skeptical.
I made a mistake with the last one. It should be:
\int_0^a J_0^2(bx)\frac{x}{c}dx
 
Last edited:
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.
 
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