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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?2011-09-22
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    In any event, you might be interested in [this](http://dx.doi.org/10.1145/1186785.1186790) and [this](http://netlib.org/toms/858).2011-09-22
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    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.2011-09-22

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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.2011-09-22
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    @J.M. Thanks, good point.2011-09-22
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    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.2011-09-22
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    $*$drools$*\text{}$2011-09-22
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    I only remembered [this paper by Adamchik](http://www.cs.cmu.edu/~adamchik/articles/bessel.htm) just now... I still find the use of Meijer $G$ rather unwieldy, but I suppose it's unavoidable here...2011-09-23
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    I guess the solution for case $a=b$ should be $a(\frac3{16}-\frac1{\pi^2})$. You mean no closed form for other cases? I already know it's $0$ if $b>3a$.2011-09-23
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    Sorry, I guess you are wrong. Since $$\int^\infty_0 J^3_1(ax)J_0(ax)/x^2 dx =a \int^\infty_0 J^3_1(x)J_0(x)/x^2 dx = a(\frac3{16}-\frac 1{\pi^2})$$2011-09-23
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    @user16521 Yep, my bad. By the way, do you have any demonstration that it vanishes for $b>3 a$ except for numeric evaluation ? If so, could you hint at it, please.2011-09-23
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    @Sasha I got the case $b>3a$ from the reference book of Gradshteyn and Ryzhik "Table of integrals, series, and products" (1980) formula 6.573 in pp.691. I don't know how to get the answer. The original formula came from Erdelyi, A. et al "Tables of integral transforms" Vol II MaGraw Hill, New York (1954). I am trying to find the old book, maybe can get some idea. I tried an old version of Mathematica, but didn't work. Don't know how about new one or Maple.2011-09-24
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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.