# Prove that Laplace density (standard) lies in Sobolev space of order $\leq 3/2$

I'm trying to prove, that the function $f(x) = e^{-|x|}$ lies in $H^{s}(\mathbb{R}^2)$ for $s\leq \frac{3}{2}$.

Therefore I calculate the functions sobolev norm $$\|f\|_s^2 = \frac{1}{4\pi^2}\int_{\mathbb{R}^2}(1+\| u\|)^s |\mathscr{F}f(u)|^2du$$

Since for its fourier transform holds $\mathscr{F}f(u) = (1+\|u\|^2)^{-1}$, we consider $$\|f\|_s^2 = \int_{\mathbb{R}^2}(1+\|u\|^2)^{s-2}du$$.

Now I want to show two things: for $s=\frac{3}{2}$, this integral is not finite (and therefore obviously not for bigger $s$ and secondly, that this is finite for $s<\frac{3}{2}$.

I proved the first part as following:

$$\|f\|_s^2 = \int_{\mathbb{R}^2}(1+\|u\|^2)^{-\frac{1}{2}}du \\ \geq \int_{\mathbb{R}^2}\frac{1}{1+\|u\|}du \geq \int_{\mathbb{R}^2}\frac{1}{\sum_{i=1}^2|u_i|}du > \infty$$, using $(a+b)^\frac{1}{2} \leq a^\frac{1}{2} + b^\frac{1}{2}$ and the behaviour of $\frac{1}{x}$ for $x\rightarrow 0$.

For the second point I still miss any idea of how to calculate an upper bound for the integral. I hope there is somebody seeing the point I'm missing actually in this case.

This yields $$\int_0^{2\pi}\int_0^\infty r\;(1+r^2)^{s-2}drd\theta$$
. Assuming $s<2$ (otherwise, the integral would obviously diverge) and using $r\leq 1+r$, we find $$\|f\|_s^2 \leq 2\pi \int_0^\infty (1+r)^{2s-3}dr$$, which converges for all $s < 1$.
This is correct for the two-dimensional case, unfortunately the bound $3/2$ was for the one-dimensional case, I realized this at the very end.
Therefore we get the norm estimate $\|f\|_s^2 \leq 2\pi \int_0^\infty (1+r)^{2s-3}dr = \frac{\pi}{1-s} \; \forall s<1$, which answers my question (even if we're not sure, whether this bound is tight).