What is the Laplace transform of the distribution $\delta \left(\cos \left(\frac{1}{x}\right)\right)$?

Question

Trying to find numerocity of the roots of the function $\cos(1/x)$ I stumbled at the following problem: what is the Laplace transform of the distribution $\delta \left(\cos \left(\frac{1}{x}\right)\right)$?

Answer 1

Rewriting, we have that

$$\delta\left(\cos\left(\frac{1}{x}\right)\right) = \sum_{\cos\left(\frac{1}{x_k}\right)=0}\frac{1}{\left|\left[\cos\left(\frac{1}{x}\right)\right]_{x=x_k}^{'}\right|}\delta(x-x_k) = \sum_{k\in\Bbb{Z}}\frac{4}{(\pi+2\pi k)^2}\cdot\delta\left(x-\frac{2}{\pi+2\pi k}\right)$$

Since the Laplace transform is one sided, implicitly the domain of the delta is $[0,\infty)$ and the summation will exclude $k\in\Bbb{Z}^-$. By the linearity of the Laplace transform we obtain

$$\mathcal{L}\left\{\delta\left(\cos\left(\frac{1}{x}\right)\right)\right\} = \sum_{k=0}^\infty \frac{4}{(\pi+2\pi k)^2}\mathcal{L}\left\{\delta\left(x-\frac{2}{\pi+2\pi k}\right)\right\}$$

$$= \sum_{k=0}^\infty \frac{4}{(\pi+2\pi k)^2} \exp\left(\frac{-2s}{\pi+2\pi k}\right)$$