Inverse Fourier Tranform of $e^{-|\xi|^{2s}}$ is in $L^p$ Problem
We wish to show that $ \mathcal{F}^{-1}[ e^{-|\xi|^{2s}}] \in L^{p}(\mathbb{R^2}) $, for all $ p \in [1,\infty], \ s\in(0,1) $.
That is, we wish to show that $ (\int_{\mathbb{R}^2} | \int_{\mathbb{R}^2} e^{2 \pi i x\cdot\xi} e^{-|\xi|^{2s}} \text{d}\xi |^p \text{d}x )^{1/p} < \infty $, for $p \in [1,\infty)$.
My Attempt
My thinking was that we could simply observe that $ e^{-|\xi|^{2s}} \leq 1 $, and then only consider:
$ (\int_{\mathbb{R}^2} | \int_{\mathbb{R}^2} e^{2 \pi i x\cdot\xi} \text{d}\xi |^p \text{d}x )^{1/p} = (\int_{\mathbb{R}^2} | \int_{-\infty}^{\infty} e^{2 \pi i x_1 \xi_1} \text{d}\xi_1 \int_{-\infty}^{\infty} e^{2 \pi i x_2 \xi_2} \text{d}\xi_2  |^p \text{d}x )^{1/p} $
We then use the fact that $ \int_{-\infty}^{\infty} e^{2 \pi i x_1 \xi_1} \text{d}\xi_1 = \frac{1}{2 \pi} \delta(x_1) $.
Now, I believed that, by the usual properties of the Dirac-Delta Distribution, we could say that:
$ \int_{-\infty}^{\infty} \frac{1}{2 \pi} \delta(x_1) \text{d}x_1 = \frac{1}{2\pi} $
and thus $ (\int_{-\infty}^{\infty} |\frac{1}{2 \pi} \delta(x_1)| \text{d}x_1)^{1/p} = \frac{1}{2\pi} $.
However, after some research online, it's come to my attention that this is NOT true, and in fact the dirac delta distribution is not in $L^p$, for any $p$.
As such, I am now lost as to how I can prove the above claim. Any hints/advice appreciated. Thank you.
Update
The case $p = \infty$ is actually quite easy to prove, so I only need help with the case $p \in [1,\infty)$.
 A: Let $f(\xi)=e^{-|\xi|^{2s}}$, where $0<s<1$. Since $f\in L^q(\mathbb{R}^n)$ for any $q\in[1,\infty]$, we have 
$$
F=\mathcal{F}^{-1}(f)\in L^p(\mathbb{R}^n)
\qquad\textrm{for}\quad
p\in[2,\infty],
$$ 
by the Hausdorff-Young inequality. 
So far, we have only used the decay of $f$. Now we want to exploit the regularity of $f$, which is related to the decay of $F$. We start by noting that
$$
|x|^2F=c\mathcal{F}^{-1}(\Delta f),
$$
where $c$ is a constant (involving $i$, $\pi$ etc.). Compute the Laplacian of $f$ as
$$
\Delta f(\xi)=e^{-|\xi|^{2s}}(c_1|\xi|^{4s-2}+c_2|\xi|^{2s-2}),
$$
where $c_1$ and $c_2$ are again constants. Since $s>0$, it is easy to see that $\Delta f\in L^{1+\varepsilon}(\mathbb{R}^2)$ for some $\varepsilon>0$. This gives us $|x|^2F\in L^q(\mathbb{R}^2)$ for some $q<\infty$ by Hausdorff-Young, and thus 
$$
F\in L^p(\mathbb{R}^2)
\qquad\textrm{for}\quad
p\geq1.
$$
Let me clarify the last step a bit further. Since $F$ is smooth, it suffices to show that 
$$
\|F(x)\|_{L^1(Q)}<\infty
\qquad\textrm{with}\quad
Q=\{|x|>1\}.
$$
Thus we compute 
$$
\int_Q|F(x)|dx=\int_Q|F(x)||x|^2|x|^{-2}dx
\leq
\||x|^2F\|_{L^q(Q)}\||x|^{-2}\|_{L^{r}(Q)},
$$
where $r$ is the Hölder conjugate of $q$. The crucial point is that $r>1$, and hence $|x|^{-2}\in L^r(Q)$, because $q<\infty$.
