Basic confusion with quantum mechanical operators Given a classical observable, $a(x,p),$ Weyl quantization gives the correspondent QM observable as:
$$\langle x | \hat{A} | \phi \rangle=\hbar^{-3}\int \int a \left(\frac{x+y}{2},p\right)\phi(y) e^{2\pi ip(x-y)/\hbar} dy dp.$$
The correspondent inverse transformation (Wigner transform; which if I understand correctly would be the symbol of the pseudo-differential operator defined above, but maybe this is not very important) is:
$$a(x,p)= \int \langle x-\frac{y}{2} \vert \hat{A} \vert x+\frac{y}{2} \rangle e^{-2\pi i yp/\hbar}dy.$$
I wanted to test this in a few simple examples. It is always easy, except for when  $\langle x \vert \hat{A} \vert x^\prime \rangle $ is actually a distribution and cannot be understand as a function. 
A basic case which is confusing me (even ignoring the mathematical subtleties of the matter) is the case in which $\hat{A}$ is $\hat{T},$ the quantum mechanical kinetic operator. If you start with $a(x,p)=\dfrac{p^2}{2},$ it is very easy to see that the first formula above gives the correct $\hat{T},$ i.e, that $\langle x \vert \hat{T} \vert \phi \rangle$ is just the laplacian. But what if want to check the second formula, the inversion of the quantization? Well, in position representation, we have $\langle x \vert \hat{T} \vert x^\prime \rangle=-\dfrac{1}{2}\Delta_{x} \delta(x-x^\prime),$ where $\Delta$ is the laplacian. Therefore, one would expect:
$$\dfrac{p^2}{2}=\int -\dfrac{1}{2}\Delta_{x-\frac{y}{2}} \delta(x-\frac{y}{2}-x-\frac{y}{2}) \exp \left(-2\pi i yp/\hbar \right) dy,$$
but I don't what to do with this; I don't even undertand what is the exact meaning of this.
This can be simplified:
$$-\dfrac{1}{2}\int\Delta_{x-\frac{y}{2}} \ \delta(y) \exp \left(-2\pi i yp/\hbar \right) dy,$$
but I don't know how to continue from this.
 A: Hint: Use Integration by parts. You have $\Delta = \nabla^2$ and you can use
Substitution $z = - \frac{y}{2}$ and then for any functions $X,Y$
$\int (\nabla_z X) Ydz = \int (\nabla_z(XY) - X \nabla_z Y)dz$.
The first term on the right Hand side gives a boundary term and These will always give you Zero.
You do the Integration by part twice and finally you can evaluate the Delta function easily (at $z = 0$).
A: I'll tweak your notations to correct the normalizations and conform with the sound Wikipedia conventions, instead:
The Weyl map is
$$\langle x | \hat{A} | y \rangle=\frac{1}{2\pi\hbar}\int   dp ~
 a \left(\frac{x+y}{2},p\right)  e^{  ip(x-y)/\hbar},$$
with the inverse, Wigner map, 
$$a(x,p)=2 \int dy ~\langle x+y \vert \hat{A} \vert x-y \rangle ~e^{-2 i yp/\hbar}.$$
For $a=p^2$, the Weyl map yields,
$$
\frac{1}{2\pi\hbar}\int   dp ~
 p^2  e^{  ip(x-y)/\hbar}\\
=\frac{-\hbar^2}{2\pi\hbar}\int  dp ~
\partial_x^2 ~  e^{  ip(x-y)/\hbar}\\
=-\hbar^2 \partial_x^2 \delta(x-y)= -\hbar^2\partial_x^2 \langle x| y\rangle  =\langle x| \hat{p}^2|y\rangle ~.
$$
Conversely, in the Wigner map,
$$2 \int dy ~\langle x+y \vert \hat{p}^2 \vert x-y \rangle ~e^{-2 i yp/\hbar}\\
=2 \int dy dp'  ~\langle x+y \vert \hat{p}^2|p'\rangle\langle p' \vert x-y \rangle ~e^{-2 i yp/\hbar}\\
 =2 \int dy dp' ~p'^2 ~\langle x+y \vert  p'\rangle\langle p' \vert x-y \rangle ~e^{-2 i yp/\hbar}\\
 =\frac{2}{2\pi\hbar} \int dy dp' ~p'^2 ~  ~e^{i(-2  yp +2p'y)/\hbar}\\
 =\int dp' ~p'^2 ~ \delta (p-p')=p^2.$$

But for the wrong normalization and exponential factors, etc... your first equation follows from this first equation here for the Weyl map, upon operating on the latter with $\int dy \langle y|\phi\rangle$.
