The following integral identity for a modified Bessel function of the second kind is well known $$ 2 K_0(x) =\int_{-\infty}^{\infty}\frac{e^{-i t x}}{\sqrt{1 + t^2}}\operatorname{d}t. $$ I was wondering if was any known closed form solution, or approximation of, the following (seemingly natural - although perhaps the change in dimension needs to be accounted for) multivariate generalisation for $\mathbf{x}, \mathbf{t} \in \mathbb{R}^n$ $$ I(\mathbf{x}):=\int_{\mathbb{R}^n}\frac{e^{-i\mathbf{t}^T\mathbf{x}}}{\sqrt{1 + \mathbf{t}^T\mathbf{t}}} \operatorname{d}\mathbf{t}. $$

I have tried for the case where $n=2$, which after switching to polar coordinates I have $$ \int_{\mathbb{R}}\int_{\mathbb{R}}\frac{e^{-i(x_1t_1 + x_2 t_2)}}{\sqrt{1+t_1^2 + t_2^2}} \operatorname{d}t_1\operatorname{d}t_2= \int_{0}^{\infty}\int_0^{2\pi} \frac{r}{\sqrt{1+r^2}}e^{-ir(x_1 \cos \theta + x_2 \sin \theta)} \operatorname{d} \theta \operatorname{d}r. $$ For the inner integral I have \begin{align} \int_{0}^{2\pi}e^{-ir(x_1 \cos \theta + x_2 \sin \theta)} \operatorname{d}\theta &= \int_0^{2\pi}\cos(-r(x_1\cos\theta+x_2\sin\theta))\operatorname{d}\theta \\ &= 2\pi J_0\left(r\sqrt{x_1^2 + x_2^2} \right). \end{align} I then have to evaluate \begin{align} \int_0^{\infty}\frac{r}{\sqrt{1+r^2}}J_0\left(r\cdot \sqrt{x_1^2 + x_2^2}\right)\operatorname{d}r, \end{align} which I'm not immediately certain even converges.

Any help with finishing this special case, if possible, and tackling the more general case would be much appreciated.


According to Maple,

$$ \int_0^\infty \frac{r}{\sqrt{1+r^2}} J_0(r s)\; dr = \frac{e^{-s}}{s} \ \text{for}\ s > 0$$

  • $\begingroup$ Great thank you, I had been accidently dropping the square on the $r$ - with that in mind I'm going to look over it all again for mistakes of that nature but this was very useful $\endgroup$ – Nadiels Mar 24 '18 at 0:06

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