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.
 A: 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:
$$ \mathcal{F}^{-1}[e^{-t|\xi|}](x) = \frac{\pi t}{(4 \pi |x|^2 + t^2)^{3/2}} $$
A: 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 $i$.
A: Here's a solution 'by hand', copied from my handwritten class notes. Oddly, the radial symmetry isn't used. We first use a tiny bit of complex analysis and the magic of Fubini's theorem to arrive at the following lemma:

Lemma(Subordination Principle.) $$e^{-\beta}=\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-i \beta^2 / 4 u} d u.$$
  Proof: First note that
  \begin{align}
\frac{2}{\pi} \int_{0}^{\infty} \frac{\cos \beta x}{1+x^{2}} d x
&=\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{e^{i \beta x}}{1+x^2} d x
\\
&=2 \pi i \operatorname{Res}_{z=i}\left(\frac{1}{\pi} \frac{e^{i 
\beta x}}{1+x^{2}}\right)
\\
&= e^{-\beta}.
\end{align}
  Therefore:
  \begin{align}
e^{-\beta}
&=\frac{2}{\pi } \int_{0}^{\infty} \color{red}{\frac{1}{1+x^2}} \cos \beta x d x
\\
&=\frac{2}{\pi }\int_0^\infty \cos\beta x \color{red}{\int_0^\infty e^{-u} e^{-x^2 u} du}  dx
\\
&\overset{\smash{Fubini}}{=}\frac{2}{\pi }\int_{0}^{\infty} e^{-u} \int_{0}^{\infty} \cos \beta x e^{-x^{2} u} d x d u
\\
&=\frac{2}{\pi }\int_{0}^{\infty} e^{-u}  \frac{1}{2} \int_{-\infty}^{\infty} e^{-x^{2} u} e^{i \beta x} d x d u
\\
&\overset{x=2\pi y}= 2\int_{0}^{\infty} e^{-u}  \int_{-\infty}^{\infty} e^{-4 \pi^{2} u y^{2}} e^{-2 \pi i \beta y} d y d u
\\
&\overset{\substack{\text{FT of}\\ \text{Gaussian}}}=2\int_{0}^{\infty} e^{-u}  \frac{1}{2\sqrt{\pi u}} e^{-\beta^{2} / 4 u} d u
\\
&=\frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^{2} / 4 u} d u .
\end{align}

I used the following two properties of the inverse Fourier transform, which I state for $n$ dimensions (which depends on your convention)-
\begin{align}
\mathcal F^{-1}(e^{-\pi |x|^2}) (\xi) = e^{-\pi |\xi|^2}\\
\mathcal F^{-1}(f(\lambda x))(\xi) = \frac1{\lambda^{n}} (\mathcal F^{-1}f)\left(\frac\xi\lambda\right)
\end{align}
Now lets compute the inverse Fourier transform of $e^{-2\pi |\xi|}$  (I think  you actually want this factor of $2\pi$), in dimension $n$:
\begin{align}
\int_{\mathbb R^n} \color{blue}{e^{-2\pi |\xi|} }e^{2\pi i x \xi} d\xi 
&= \frac1{\sqrt\pi}\int_{\mathbb R^n}
\color{blue}{\int_0^\infty
\frac{e^{-u}}{\sqrt{u}} e^{-4\pi^2|\xi|^{2} / 4 u} d u 
}
 e^{2\pi i x \xi} d\xi
\\
&\overset{Fubini}= \frac1{\sqrt\pi} \int_0^\infty e^{-u}{\sqrt u} \int_{\mathbb R^n} e^{-\pi ^2 |\xi|^2/u} e^{2\pi i x\xi} d\xi du
\\
&\overset{\substack{\text{FT of}\\ \text{Gaussian}}}= \frac1{\sqrt\pi} \int_0^\infty \frac{e^{-u}}{\sqrt u} \frac{u^{n/2}}{\pi^{n/2}} e^{-u |x|^2} du
\\
&\overset{v = (1+|x|^2)u}= \frac1{\pi^{\frac{n+1}2}}\int_0^\infty v^{\frac{n-1}2} e^{-v} dv \frac1{(1+|x|^2)^{\frac{n-1}2}} \frac1{1+|x|^2}
\\
&= \frac{\Gamma\left({\frac{n+1}2}\right)}{\pi^{\frac{n+1}2}}\frac1{(1+|x|^2)^{\frac{n+1}2}}
\end{align}
Having solved the problem for $t=1$, we now note that 
\begin{align}
 \mathcal F^{-1}(e^{-2\pi t|\xi|})(x) 
&= \int_{\mathbb R^n} e^{-2\pi t|\xi|} e^{2\pi i x \xi} d\xi 
\\
&\overset{\eta = t\xi}= \int_{\mathbb R^n} e^{-2\pi |\eta|}  e^{2\pi i (x/t) \eta} d\eta t^{-n} 
\\
&= \frac1{t^n}\mathcal F^{-1}(e^{-2\pi|\xi|})(x/t)
\\
&=\frac{\Gamma\left({\frac{n+1}2}\right) t}{\pi^{\frac{n+1}2}(t^2+|x|^2)^{\frac{n+1}2}}
\end{align}
For $n=2$ we have 
$\Gamma(3/2)=\frac{\sqrt\pi}2$, so
$$ \mathcal F^{-1}(e^{-2\pi t|\xi|})(x) = \frac{t}{2\pi (t^2+|x|^2)^{3/2}}$$
If you want $ \mathcal F^{-1}(e^{-t|\xi|})(x)$ instead here it is for reference- we simply sub in $t=T/2\pi$,
$$  \mathcal F^{-1}(e^{- T|\xi|})(x) =\frac{T}{(2\pi)^2(\frac{T^2}{(2\pi)^2}+|x|^2)^{3/2}} = \frac{T}{(2\pi)^{2-3}(T^2+|2\pi x|^2)^{3/2}} = \frac{2\pi T}{(T^2+|2\pi x|^2)^{3/2}} $$
and I seem to have discovered that one of us is off by a factor of 2...
