Solution to "Heat Equation" with Fractional Laplacian in 2 Dimensions Statement of the Problem
We consider the equation:
$ \partial_t u + (- \Delta)^{1/2}u = 0 $
for $ u : \mathbb{R}^2 \rightarrow \mathbb{R} $.
I would like to find a non-trivial solution to this equation, using the Fourier Transform. 
I believe I have used the right methods to find the answer, but think there must be a mistake. I would love for someone to check my results for me.
My Attempt
We first take the FT of the equation with respect to the space variable $x$:
$ \partial_t \hat{u} + |\xi| \hat{u} = 0 $.
The solution to this equation is obviously $ \hat{u}(t,\xi) = e^{-t |\xi|} $.
Then the solution $u$ to our original equation is:
$ u(t,x) = \mathcal{F}[e^{-t |\xi|}](x) $
$ = \int_{\mathbb{R}^2} e^{-t |\xi|} e^{2 \pi i x \cdot \xi} \text{d}\xi $.
We note that this equation is radially symmetric with respect to $x$. That is, 
$ \mathcal{F}[e^{-t |\xi|}](O_2 x) = \mathcal{F}[e^{-t |\xi|}](x) $, where $O_2$ is a rotation in 2 dimensions.
Then we can write:
$ \mathcal{F}[e^{-t |\xi|}](x) = \mathcal{F}[e^{-t |\xi|}](|x|) = \int_{\mathbb{R}^2} e^{-t |\xi|} e^{2 \pi i |x| |\xi|} \text{d}\xi $, which we then rewrite in polar coordinates as:
$ \int_{0}^{2 \pi} \int^{\infty}_{0} e^{-t \rho} e^{2 \pi i |x| \rho} \rho \ \text{d}\rho \text{d}\theta = 2 \pi \int^{\infty}_{0} e^{-t \rho} e^{2 \pi i |x| \rho} \rho \ \text{d}\rho $
$ = 2 \pi \large ( \frac{-(4 \pi^2 x^2 - t^2)}{16 \pi^{16} x^4 + 8 \pi^2 t^2 x^2 + t^4 } + \frac{4 \pi i t |x|}{16 \pi^4 x^4 + 8 \pi^2 t^2 x^2 + t^4} ) $.
This last value was calculated by splitting $ e^{2 \pi i |x| \rho} = \cos(2 \pi i |x| \rho) + i \sin(2 \pi i |x| \rho) $, and then plugging the functions into an integral calculator.
The reason why I suspect that this answer is wrong is that it is a complex number. I have seen here that, since the function $e^{-t|\xi|}$ is real and even with respect to $\xi$, the FT should be real as well. Does this not also apply to the inverse FT?
Have I made a mistake somewhere? Please let me know. Thank you.

The problem is that a radially symmetric function doesn't allow you to ignore the dependence on the angle between $x$ and $\xi$. The integral can be solved without this assumption. Finding the radial symmetry is very cumbersome, but it is possible (full solution below):
We know that
$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \int_{\mathbb{R}^2} e^{-t |\xi|} e^{2 \pi i x\cdot \xi} \text{d}\xi$$
From there, we can then use the definition of dot product $x \cdot \xi = |x||\xi|cos(\phi_{\xi}-\phi_x)$ where $\phi_\xi$ and $\phi_x$ are the angles of the vectors $x$ and $\xi$. Then we have that, in polar coordinates,
$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \int_0^{2\pi} \int_0^{\infty} e^{-|\xi|(t-2\pi i |x|cos(\phi_{\xi}-\phi_x))}|\xi| d|\xi|d\phi_\xi .$$
Solving the improper integral we get
$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \int_0^{2\pi} \frac{1}{[t-2\pi i |x|cos(\phi_{\xi}-\phi_x)]^2} d\phi_\xi ,$$
where $t>0$. Making a change of coordinates $\phi_\xi' = \phi_\xi - \phi_x$, the integral becomes then
$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \int_{-\phi_x}^{2\pi-\phi_x} \frac{1}{[t-2\pi i |x|cos(\phi_{\xi}')]^2} d\phi_\xi' $$
which, by solving on Maxima and considering $0\leq \phi_x \leq 2\pi$, we get, after some simplification,
$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \frac{-\sqrt{4\pi^2 |x|^2+t^2}(4\pi^3 cos^2(\phi_x)t|x|^2 + \pi t^3)}{-64 \pi^6 cos^2(\phi_x)|x|^6 - (32cos^2(\phi_x) + 16)\pi^4 t^2 |x|^4 - (4 cos^2(\phi_x)+8)\pi^2 t^4 |x|^2 - t^6} $$
Changing to cartesian coordinates where $x = (x_1,x_2)$ and simplifying further, we have
$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \frac{-\sqrt{4\pi^2 |x|^2+t^2}(4\pi^2 x_1^2 + t^2) \pi t}{-64 \pi^6 x_1^2 |x|^4 - (32 x_1^2 |x|^2 + 16 |x|^4) \pi^4 t^2 - (4 x_1^2 + 8 |x|^2) \pi^2 t^4 - t^6} $$
Finally, the denominator can be factored:
$$ \mathcal{F}^{-1}[e^{-t|\xi|}] = \frac{-\sqrt{4\pi^2 |x|^2+t^2}(4\pi^2 x_1^2 + t^2) \pi t}{-(4 \pi^2 x_1^2 + t^2)(4 \pi |x|^2 + t^2)^2} $$
Simplifying the fraction we see finally the radial symmetry:
$$ \mathcal{F}^{-1}[e^{-t|\xi|}](x) = \frac{\pi t}{(4 \pi |x|^2 + t^2)^{3/2}} $$
