Fourier tranform of vector and dot product in three dimentions I have a problem finding the Inverse Fourier transform of the
\begin{equation}
\dfrac{1}{k^2+a^2}\dfrac{1}{k^2}e^{i\textbf{k}\cdot \bf{r}}(\textbf{k}\cdot \bf{v}) \bf{k}
\end{equation}
where $\bf{r}$ is the radius vector, $\bf{v}$ is the velocity,$\bf{k}$ is the variable in Fourier space and $a$ is a contant.
Any suggestion?
 A: Let $F(k)=\frac{1}{k^2(k^2+a^2)}$.  We can write $F(k)$ as
$$\begin{align}
F(k)&=\frac{1}{k^2(k^2+a^2)}\\\\
&=\frac{1}{a^2}\left(\frac{1}{k^2}-\frac{1}{k^2+a^2}\right)
\end{align}$$
Then, the Fourier Transform of $F$ can be written 
$$\begin{align}
\mathscr{F}\{F\}(\vec r)&=\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty F(k)\,e^{i\vec k \cdot \vec r}\,dk_x\,dk_y\,dk_z\\\\
&=\frac1{a^2}\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i\vec k \cdot \vec r}}{k^2}\,dk_x\,dk_y\,dk_z-\frac1{a^2}\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i\vec k \cdot \vec r}}{k^2+a^2}\,dk_x\,dk_y\,dk_z
\end{align}$$
In THIS ANSWER, I evaluated the Fourier Transform of $\frac{1}{k^2+a^2}$ as $f(r;a)=2\pi^2\frac{e^{-|a|r}}{r}$.
so that 
$$\begin{align}
\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\left(\frac{(\vec v\cdot \vec k)\vec k}{k^2(k^2+a^2)}\right)\,e^{i\vec k \cdot \vec r}\,dk_x\,dk_y\,dk_z&=-(\vec v\cdot \nabla)\nabla\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{e^{i\vec k \cdot \vec r}}{k^2(k^2+a^2)}\,dk_x\,dk_y\,dk_z\\\\
&=\frac{1}{a^2}(\vec v\cdot \nabla)\nabla(f(r;a)-f(r;0))
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{\large\mathbb{R}^{3}}{1 \over k^{2} + a^{2}}
\,{1 \over k^{2}}\,\expo{\ic\mathbf{k}\cdot\mathbf{r}}
\pars{\mathbf{k}\cdot\mathbf{v}}\mathbf{k}\,\,\dd^{3}\mathbf{k} =
-\ic\nabla\pars{\mathbf{v}\cdot\int_{\large\mathbb{R}^{3}}
{1 \over k^{2} + a^{2}}\,{1 \over k^{2}}\,\expo{\ic\mathbf{k}\cdot\mathbf{r}}
\mathbf{k}\,\,\dd^{3}\mathbf{k}}
\\[5mm] = &\
-\nabla\pars{\mathbf{v}\cdot\nabla\int_{\large\mathbb{R}^{3}}
{1 \over k^{2} + a^{2}}\,{1 \over k^{2}}\,\expo{\ic\mathbf{k}\cdot\mathbf{r}}
\dd^{3}\mathbf{k}}
\\[5mm] = &\
-\nabla\bracks{\mathbf{v}\cdot\nabla\int_{0}^{\infty}
{1 \over k^{2} + a^{2}}\,{1 \over k^{2}}\,
\pars{\int\expo{\ic\mathbf{k}\cdot\mathbf{r}}
\,{\dd\Omega_{\mathbf{k}} \over 4\pi}}4\pi k^{2}\,\dd k}
\\[5mm] = &\
-4\pi\nabla\bracks{\mathbf{v}\cdot\nabla\int_{0}^{\infty}{1 \over k^{2} + a^{2}}
{\sin\pars{kr} \over kr}\,\dd k} =
-4\pi\nabla\braces{\mathbf{v}\cdot\nabla\bracks{r\int_{0}^{\infty}{1 \over \xi^{2} + \pars{ar}^{2}}
{\sin\pars{\xi} \over \xi}\,\dd\xi}}
\\[5mm] = &\
-4\pi a\nabla_{\mathbf{R}}\braces{\mathbf{v}\cdot\nabla_{\mathbf{R}}
\bracks{R\int_{0}^{\infty}{1 \over \xi^{2} + R^{2}}
{\sin\pars{\xi} \over \xi}\,\dd\xi}}\,,\qquad\qquad
\mbox{where}\ \mathbf{R} \equiv a\mathbf{r}
\end{align}


With
  $\ds{\mrm{f}\pars{z} \equiv {1 - \expo{-z} \over z}}$,

\begin{align}
&\int_{\large\mathbb{R}^{3}}{1 \over k^{2} + a^{2}}
\,{1 \over k^{2}}\,\expo{\ic\mathbf{k}\cdot\mathbf{r}}
\pars{\mathbf{k}\cdot\mathbf{v}}\mathbf{k}\,\,\dd^{3}\mathbf{k}=
-2\pi^{2}a\,\nabla_{\mathbf{R}}\bracks{\mathbf{v}\cdot\nabla_{\mathbf{R}}
\mrm{f}\pars{R}} =
-2\pi^{2}a\,\nabla_{\mathbf{R}}\bracks{\mathbf{v}\cdot
\mrm{f}'\pars{R}\,{\mathbf{R} \over R}}
\\[5mm] = &\
-2\pi^{2}a\,\sum_{i}\hat{x}_{i}\,\partiald{}{x_{i}}
\sum_{j}v_{j}\bracks{\mrm{f}'\pars{R}{x_{j} \over R}} =
-2\pi^{2}a\,\sum_{ij}\hat{x}_{i}\,v_{j}\,\partiald{}{x_{i}}
\bracks{x_{j}\,{\mrm{f}'\pars{R} \over R}}
\\[5mm] = &\
-2\pi^{2}a\,\sum_{ij}\hat{x}_{i}\,v_{j}
\braces{\delta_{ij}\,{\mrm{f}'\pars{R} \over R} + x_{j}\,{x_{i} \over R}\,
\totald{}{R}\bracks{\mrm{f}'\pars{R} \over R}}
\\[5mm] = &\
\bbox[15px,#ffe,border:1px dotted navy]{\ds{%
-2\pi a^{2}\,{\mrm{f}'\pars{R} \over R}\,\mathbf{v} -
2\pi a^{2}\braces{{1 \over R}\,\totald{}{R}\bracks{\mrm{f}'\pars{R} \over R}}
\pars{\mathbf{v}\cdot\mathbf{R}}
\mathbf{R}}}
\end{align}
