$\left(\nabla^2\frac{1}{|\mathbf{x}|}\right)e^{-i\mathbf{q}\cdot\mathbf{x}}=(-4\pi\delta^3(\mathbf{x}))e^{-i\mathbf{q}\cdot\mathbf{x}}$

Question

I am studying the following calculation: \begin{align} \int d^3\mathbf{x}\frac{1}{|{\mathbf{x}}|}e^{-i\mathbf{q}\cdot\mathbf{x}} & = \frac{-1}{|\mathbf{q}|^2}\int d^3\mathbf{x}\frac{1}{|\mathbf{x}|}\nabla^2\left(e^{-i\mathbf{q}\cdot\mathbf{x}}\right)= \\ &= \frac{-1}{|\mathbf{q}|^2}\int d^3\mathbf{x}\left(\nabla^2\frac{1}{|\mathbf{x}|}\right)e^{-i\mathbf{q}\cdot\mathbf{x}}\\ &=\frac{-1}{|\mathbf{q}|^2}\int d^3\mathbf{x}\left(-4\pi\delta^3(\mathbf{x})\right)e^{-i\mathbf{q}\cdot\mathbf{x}} = \\ &= \frac{4\pi}{|\mathbf{q}|^2} \end{align} I don't understand the following two equalities by what they are justified, there is some intermediate step? $$\frac{1}{|\mathbf{x}|}\nabla^2\left(e^{-i\mathbf{q}\cdot\mathbf{x}}\right)=\left(\nabla^2\frac{1}{|\mathbf{x}|}\right)e^{-i\mathbf{q}\cdot\mathbf{x}}$$ $$\left(\nabla^2\frac{1}{|\mathbf{x}|}\right)e^{-i\mathbf{q}\cdot\mathbf{x}}=(-4\pi\delta^3(\mathbf{x}))e^{-i\mathbf{q}\cdot\mathbf{x}}$$

Answer 1

The calculation $\int d^3x\frac{1}{|x|}\nabla^2e^{-iq\cdot x}=\int d^3x\left(\nabla^2\frac{1}{|x|}\right)e^{-iq\cdot x}$ uses integrration by parts twice, not the first identity you've conjectured, which is false. Your second conjectured equation is correct (proof here).