It is possible to calculate the Laplacian of a function by its integral representation? More precisely
Since $-\partial_{x}^2(x^2)=-2$ and $-\partial_{x}^2 u(x)=\mathcal{F}^{-1}(\xi^2 \widehat{u}(\xi))(x)$ ($u$ adecuate) Then $$\int_{\mathbb{R}} e^{ix\xi} \xi^2 \widehat{\xi^2} d\xi=-2?$$ Or the above is not true since $x^2$ is not a suitable function, for example, it is not Schwartz, and that does not allow a Fourier representation for the Laplacian operator.
Let $u(x) = x^2$. The distributional Fourier transform of $u$ is $$ \mathcal{F}\{u\} = \mathcal{F}\{x^2 \mathbf{1}\} = (i\partial)^2 \mathcal{F}\{\mathbf{1}\} = (i\partial)^2 (2\pi \delta) = -2\pi \delta'', $$ where $\mathbf{1}$ is the constant function with value $1$ everywhere.
The distributional Fourier transform of the second derivative of $u$ is $$ \mathcal{F}\{u''\}(\xi) = (i\xi)^2 \mathcal{F}\{u\}(\xi) = (i\xi)^2 (-2\pi \delta'')(\xi) = 2\pi \xi^2\delta''(\xi) = 2\pi \cdot 2\delta(\xi) = \mathcal{F}\{2\mathbf{1}\}(\xi). $$
We can of course turn this around to get more what you want: $$ \partial^2 u = \partial^2 \mathcal{F}^{-1}\{-2\pi\delta''\} = \mathcal{F}^{-1}\{(ix)^2 \cdot (-2\pi\delta'')\} = 2\pi \mathcal{F}^{-1}\{x^2 \delta''\} \\ = 2\pi \mathcal{F}^{-1}\{2\delta\} = 2\pi \cdot \frac{1}{2\pi}\mathcal{F}\{2\delta\}(-x) = 2\mathbf{1}. $$
Showing $\xi^2\delta''(\xi)=2\delta(\xi)$ that is used above: $$ \langle x^2\delta'', \varphi \rangle = \langle \delta'', x^2\varphi \rangle = \langle \delta, (x^2\varphi)'' \rangle = \langle \delta, 2\varphi + 4x\varphi' + x^2\varphi'' \rangle \\ = 2\varphi(0) + 4\cdot 0 \cdot\varphi'(0) + 0^2 \cdot \varphi''(0) = 2\varphi(0) = \langle 2\delta, \varphi \rangle. $$