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.

Two points:


*

*We see that your original Fourier integrand
$$e^{-t|\xi|}e^{2\pi i x\cdot\xi}$$
mapping $\xi\to-\xi$ is equivalent to taking a complex conjugate, so the integral must be real. However, when it is manipulated into
$$e^{-t|\xi|}e^{2\pi i |x||\xi|}$$
it has lost this property, and something has gone wrong. I see know reason that the second exponential can be simplified to $e^{2\pi i|x||\xi|}$; perhaps a factor of $\cos(\theta)$ went missing when simplifying the dot product?

*Your use of Euler's identity $e^{ix}=\cos(x)+i\sin(x)$ seems to have imaginary arguments on both sides. In the original identity they differ by a factor of $i$.
