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Is there somebody who knows the solution for the integral $$\int_0^\infty\frac{J^3_1(ax)J_0(bx)}{x^2} dx$$ where $a>0,b>0$ and $J(\cdot)$ the bessel function of the first kind with integer order?

Reference, or solution from computer programs all are welcome. Thanks!

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Are you interested in a closed form (doesn't seem likely to me), or in a numerical method for computing this? –  J. M. Sep 22 '11 at 14:16
In any event, you might be interested in this and this. –  J. M. Sep 22 '11 at 14:28
Thank you! It's interesting! But I need closed form. I already know it's 0 when $b>3a$. I want other cases. I wonder if Maple or Mathematica works for it. I have not these softwares on hand. –  user16521 Sep 22 '11 at 14:51
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The integral can be approached by using Mellin convolution technique. Let's first start with a simple case of $a=b$. In this case we compute Mellin transform of $J_1(x)^2$ and $J_0(x) J_1(x)$: $$ \mathcal{M}_s( J_1(x)^2 ) = \int_0^\infty x^{s} J_1(x)^2 \frac{\mathrm{d} x}{x} = \frac{\Gamma \left(\frac{1}{2}-\frac{s}{2}\right) \Gamma \left(\frac{s}{2}+1\right)}{2 \sqrt{\pi } \Gamma \left(1-\frac{s}{2}\right) \Gamma \left(2-\frac{s}{2}\right)} \qquad \text{for} \qquad -2 < \mathrm{Re}(s) < 1 $$ and $$ \mathcal{M}_s( J_1(x) J_0(x) ) = \frac{\Gamma \left(1-\frac{s}{2}\right) \Gamma \left(\frac{s+1}{2}\right)}{2 \sqrt{\pi } \Gamma \left(\frac{3}{2}-\frac{s}{2}\right)^2} \qquad \text{for} \qquad -1 < \mathrm{Re}(s) < 2 $$ Now $$ \begin{eqnarray} \int_0^\infty x^{-2} J_1(a x)^3 J_0(a x) \mathrm{d} x &=& \frac{a}{2} \frac{1}{2 \pi i} \int_{-1 - i \infty}^{-1 + i \infty} \frac{\Gamma \left(\frac{1}{2}-\frac{s}{2}\right) \Gamma \left(\frac{s}{2}+1\right) \Gamma \left(\frac{s}{2}+\frac{3}{2}\right) \Gamma \left(-\frac{s}{2}\right)}{2 \pi \Gamma \left(1-\frac{s}{2}\right) \Gamma \left(2-\frac{s}{2}\right) \Gamma \left(\frac{s}{2}+2\right)^2} \mathrm{d} s \\ &=& \frac{a}{2 \pi i} \int_{-\frac{1}{2} - i \infty}^{-\frac{1}{2} + i \infty} \frac{\Gamma \left(\frac{1}{2}-s\right) \Gamma \left(s+1\right) \Gamma \left(s+\frac{3}{2}\right) \Gamma \left(-s\right)}{2 \pi \Gamma \left(1-s\right) \Gamma \left(2-s\right) \Gamma \left(s+2\right)^2} \mathrm{d} s \\ &=& \frac{a}{2 \pi} G_{4,4}^{2,2}\left(1\left| \begin{array}{c} \frac{1}{2},1,2,2 \\ 1,\frac{3}{2},0,-1 \\ \end{array} \right.\right) = a \left( \frac{3 }{16} - \frac{1}{\pi^2} \right) \end{eqnarray} $$

Here $G_{4,4}^{2,2}(1)$ denotes Meijer's G-function.

Now when $a \not= b$, Mellin transform of $J_1(a x) J_0(b x)$ is no longer a ratio of $\Gamma$-functions: $$ \left. \mathcal{M}_s( J_1( a x) J_0(b x)) \right\vert_{-1 < \mathrm{Re}(s) <2} = \left\{ \begin{array}{cc} \frac{a 2^{s-1} b^{-s-1} \Gamma \left(\frac{s+1}{2}\right) \, _2F_1\left(\frac{s+1}{2},\frac{s+1}{2};2;\frac{a^2}{b^2}\right)}{\Gamma \left(\frac{1}{2}-\frac{s}{2}\right)} & a < b \\ \frac{2^{s-1} a^{-s} \Gamma \left(\frac{s+1}{2}\right) \, _2F_1\left(\frac{s-1}{2},\frac{s+1}{2};1;\frac{b^2}{a^2}\right)}{\Gamma \left(\frac{3}{2}-\frac{s}{2}\right)} & a > b \end{array} \right. $$ Notice that Gauss hypergeometric function $ {}_2F_1$ can be represented by its defining sum, whose terms are ratios of Gamma functions themselves. Continuing this way will produce sum representation of the integral.

Integrals of this form has been discussed by W.N. Bailey in 1936, see link.

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Interesting. You might want to mention that you used Meijer's $G$-function in the fifth line. –  J. M. Sep 22 '11 at 15:20
@J.M. Thanks, good point. –  Sasha Sep 22 '11 at 15:27
I'll note a few special cases: if $s$ is odd, ${}_2 F_1$ turns to a rational function, while if $s$ is even, one gets rational functions multiplied by complete elliptic integrals. –  J. M. Sep 22 '11 at 15:32
$*$drools$*\text{}$ –  anon Sep 22 '11 at 21:53
I only remembered this paper by Adamchik just now... I still find the use of Meijer $G$ rather unwieldy, but I suppose it's unavoidable here... –  J. M. Sep 23 '11 at 3:15
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For $a=b=1$ ... $$\int_0^\infty \frac{\mathrm{J}_1(x)^3 \mathrm{J}_0(x)}{x^2}\,dx = \frac{3}{16} - \frac{1}{\pi^2}$$ I'll leave the others to the reader.

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