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I have the following integral:

$$\displaystyle \int_{\mathbb{R}^2} \left( \int_{\mathbb{R}^2} \frac{J_{1}(\rho |\alpha|)J_{1}(\rho|k- \alpha|)}{|\alpha||k-\alpha|} \ \mathrm{d}\alpha \right)^2 \ \mathrm{d}k,$$

with $\alpha, k \in \mathbb{R}^2$, $\rho$ constant, $J_{\nu}$ the Bessel function of the first kind, $|\cdot|$ the Euclidean norm on $\mathbb{R}^2$.

I want to make the substitution $\alpha = s$, $k = s + t$. If the square wasn't there, this would be easy, and we'd just end up with

$$\displaystyle \int_{\mathbb{R}^2} \int_{\mathbb{R}^2} \frac{J_{1}(\rho |s|)J_{1}(\rho|t|)}{|s||t|} \ \mathrm{d}s \ \mathrm{d}t,$$

since the determinant of the Jacobian corresponding to this change of coordinates is $1$. But the square makes things difficult. What does the integral at the top look like after making that change of coordinates?

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  • $\begingroup$ You could literally square the inner integral and expand it into two integrals inside. $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 13:27
  • $\begingroup$ Do you mean write the integral squared as two copies of itself, one depending on $\alpha$ and the other depending on $\beta$, say? $\endgroup$ – user363087 Feb 18 '17 at 13:28
  • $\begingroup$ Yeah, that's what I was thinking. (didn't really think it all the way through though, so we'll see how it goes) $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 13:30
  • $\begingroup$ I can't get the coordinate change to make sense after doing this, though -- unless I'm missing something. $\endgroup$ – user363087 Feb 18 '17 at 13:35
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IMHO, you should consider apart the transformation of the inside integral.

What follows is a hint, far from a full answer.

2 remarks:

1) $f(\omega)=\pi\dfrac{\sqrt{3}}{2}r^{3/2}\dfrac{J_1(2 \pi \omega r)}{2 \pi \omega r}$ is the Fourier transform of a unit disk with radius $r$.

(http://isi.ssl.berkeley.edu/~tatebe/whitepapers/FT%20of%20Uniform%20Disk.pdf).

2) The inside integral is, up to a multiplicative constant, the convolution of two such transforms, otherwise said, it is a square of convolution.

Thus, the integral can be written, up to a multiplicative constant, as the integral of the Fourier transform of a the area common to two disks with radius $r$, this function being explicitly computable.

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  • $\begingroup$ Thank you for the response, but I'm a little confused by this. The integrals are both over $\mathbb{R}^2$, while the Fourier transform you mention looks to be a function on $\mathbb{R}$ -- does it follow that $\frac{J_{1}(|\alpha|)}{|\alpha|}$ is also the Fourier transform of a unit disc, despite the fact that $\alpha \in \mathbb{R}^2$? I am also not sure how you deduced the last statement. Does this come from the convolution theorem? $\endgroup$ – user363087 Feb 18 '17 at 18:10
  • $\begingroup$ In fact it is 1D convolution along a radial axis, because there is a symmetry of revolution. $\endgroup$ – Jean Marie Feb 18 '17 at 18:20
  • $\begingroup$ I see. Could you explain the last sentence in your answer? I am having trouble understanding how you deduced that from remarks 1) and 2): why does the inner integral correspond to the FT of the intersection of the areas of two discs? Is the computability justified by the convolution theorem here? Incidentally I have a bounty offered at the moment on this very integral. $\endgroup$ – user363087 Feb 18 '17 at 18:51
  • $\begingroup$ I am also not sure why the resulting square of the convolution is integrable: for instance, the Fourier transform you mention (i.e. the Bessel quotient) is not in $L^1$. $\endgroup$ – user363087 Feb 19 '17 at 13:09
  • $\begingroup$ The fact that it is not in $L^1$ (or $L^1 \cap L^2$) means that it has no "classical Fourier Transform". But it has a Fourier Transform in the framework of distributions. $\endgroup$ – Jean Marie Feb 19 '17 at 13:17

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