I recently came across the following remarkable identity, due to Hardy:

$$\displaystyle \int_{-\infty}^{\infty} \frac{J_{\mu}(a(z+t))}{(z+t)^{\mu}} \frac{J_{\nu}(a(\zeta + t))}{(\zeta + t)^{\nu}} \ \mathrm{d}t = \frac{\Gamma(\mu + \nu)\Gamma(\frac{1}{2})}{\Gamma(\mu + \frac{1}{2})\Gamma(\nu + \frac{1}{2})} \times \left( \frac{2}{a}\right)^{\frac{1}{2}} \frac{J_{\mu + \nu - \frac{1}{2}}(a(z - \zeta))}{(z - \zeta)^{\mu + \nu - \frac{1}{2}}},$$

where $J_{\nu}$ denotes the Bessel function of the first kind, $\mathrm{Re}(\mu + \nu) > 0$, $a > 0$, and $z, \zeta \in \mathbb{R}$. If anyone is interested in further details, it can be found in "A Treatise on the Theory of Bessel Functions", page 422. Some of the book (including the page referenced) can be found here.

In my research, I am dealing with an integral which looks identical to this one, with the exception that it is in $d$ dimensions. In particular, the integral I am concerned with is

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

where $|\cdot|$ denotes the Euclidean norm on $\mathbb{R}^d$, and $\rho$ does not depend on $k \neq 0$ nor on $\alpha$. The obvious co-ordinate change is to take $|\alpha| = \tau$ and to assume without loss of generality that $|k| = r$, but this creates unwanted trigonometric terms in the argument of one of the Bessel functions, as well as destroying the symmetry of the two integrals. Can anyone see some way of relating Hardy's identity to my integral?

  • 1
    $\begingroup$ Everything looks kind of radial, so maybe a polar change of coordinates may help? then the integral over the radius will hopefully look like the one-dimensional case? Or at least you may be able to find an upperbound. $\endgroup$ – Martingalo Feb 8 '17 at 0:19
  • $\begingroup$ I have tried putting $|\alpha| = \tau$ and $|k| = r$, but it seems to make the second factor quite messy, and I am not sure how to make it look like Hardy's identity from there. $\endgroup$ – user363087 Feb 8 '17 at 0:25

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