What it the fourier transform of laplacian and shifted funtion? I'm looking for the Fourier transform of $\nabla^2f(\vec{r}-\vec{a})$
I can assume that the 3D Fourier transform of $f(\vec{r})$ is $\tilde{f}(\vec{q})$ and the vector $\vec{a}$ is a const vector.
Thanks.
 A: Denote the Fourier transformation by $\def\F{\mathscr F}\F$, we have
\begin{align*}
   \F[\Delta f(\cdot -a)](\xi) &= \def\c{\frac 1{(2\pi)^{3/2}}}\c \int_{\def\R{\mathbf R}\R^3} \Delta f(x-a)\exp(-i\def\<#1>{\left<#1\right>}\<x,\xi>)\, dx\\
          &=-\c \int_{\R^3} \sum_j \partial_j f(x-a)(-i\xi_j)\exp(-i\<x,\xi>)\, dx\\
          &= -\c \sum_{j} \xi_j^2 \int_{\R^3} f(x-a)\exp(-i\<x,\xi>)\, dx\\
          &= -\<\xi,\xi> \cdot \c \int_{\R^3} f(x)\exp(-i\<x+a, \xi>)\, dx\\
          &= -\<\xi,\xi> \exp(-i\<a,\xi>)\F f(\xi)
\end{align*}
A: Let the Fourier transform, $\tilde{f}(\vec{q})$, of $f(\vec r)$ be given by
$$\tilde{f}(\vec{q})=\int f(\vec r)\,e^{i\vec q\cdot \vec r} \,d^3\vec r$$ 
Then, the Fourier transform of $\nabla^2f(\vec r-\vec a)$ is given by
$$\begin{align}
\int \nabla^2f(\vec r-\vec a)\,e^{i\vec q\cdot \vec r} \,d^3\vec r&=-\int \nabla f(\vec r-\vec a)\cdot \,\nabla (e^{i\vec q\cdot \vec r}) \,d^3\vec r \tag 1\\\\
&=\int f(\vec r-\vec a)\,\nabla^2(e^{i\vec q\cdot \vec r}) \,d^3\vec r \tag 2\\\\
&=\int  f(\vec r-\vec a)\,(-|\vec q|^2)\,e^{i\vec q\cdot \vec r}\,d^3\vec r \tag 3\\\\
&=-|\vec q|^2\,e^{i\vec q\cdot \vec a}\,\tilde{f}(\vec{q})\tag 4
\end{align}$$
In arriving at $(1)$, we integrated by parts by making use of the product rule $\nabla \cdot (\phi \vec F)=\phi \nabla \cdot \vec F+\nabla \phi \cdot \vec F$ along with the Divergence Theorem.  We also assumed that $r^2 \nabla f$ goes to zero as $r \to \infty$.
In going from $(1)$ to $(2)$, we integrated by parts again.
In going from $(2)$ to $(3)$, we evaluated the Laplacian of the exponential term.
And in going from $(3)$ to $(4)$, we made a simple change of variables and carried out the integral. 
