Fourier Transform - Dirac Delta How can I compute this Fourier transform?
$$\int \frac{d^3q}{(2\pi)^3}(\vec q \cdot \vec a)(\vec q \cdot \vec b)\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }\,$$
My idea was to write it as
$$-( \vec a \cdot \vec \nabla)(\vec b \cdot \vec \nabla)\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }$$
and use the fact that $\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }=\frac{1}{4\pi r}$, together with $\Delta \frac{1}{r} = -4\,\pi \,\delta(\vec r)$ but is is not exactly the same.
 A: So the other answers are correct only for $x≠0$. In the sense of distributions, however, there is indeed a Dirac delta appearing in the result.
The first steps in your approach are correct. both functions $\frac{1}{q^2}$ and $\frac{(q·a)\,(q\cdot b)}{q^2}$ are indeed locally integrable functions and bounded outside a compact set, so are tempered distributions: one can take their Fourier transform in the sense of distributions (remark however that these functions are not in $L^1$ or in $L^2$, so one cannot take the Fourier transform in the usual sense).
In the sense of distributions (writing with abuse of notation the Fourier transform as an integral)
$$
\frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q = -(a\cdot\nabla)(b\cdot\nabla)\, \frac{1}{4π|x|}.
$$
Taking a first derivative of $1/|x|$, one obtains $-x/|x|^3$ (even in the sense of distributions) which is still a locally integrable function. However, taking two derivatives leads to a non locally integrable function, and so one has to be careful as you suspect. A formula that generalizes the formula of the Laplacian $-\Delta \frac{1}{4\pi|x|} = \delta_0$ is the Formula of the Hessian (where $\nabla^2 = \nabla\nabla$)
$$
\nabla^2\left(\frac{1}{4\pi|x|}\right) = \frac{1}{4π} \,\mathrm{pv.}\,\frac{3x\otimes x - |x|^2 \,\mathrm{Id}}{|x|^5} - \frac{1}{3}\, \delta_0 \,\mathrm{Id}
$$
where pv. denotes the principal value, see e.g. p.55 here. (In particular, if you sum the coordinates $(j,j)$ of this matrix, you obtain the formula for the Laplacian). Now remark that $(a\cdot\nabla)(b\cdot\nabla) = (a\otimes b):\nabla^2$ (if you prefer coordinates, $\sum_{ij} a_i\partial_i\, b_j\partial_j = \sum_{ij} a_ib_j\,\partial_i\partial_j$) and so
$$
\begin{align}
\frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q &= -(a\otimes b):\nabla^2\left(\frac{1}{4\pi|x|}\right).
\\
&=  \frac{1}{4π} \,\mathrm{pv.}\,\frac{|x|^2 (a·b) - 3\,(a\cdot x)(b\cdot x)}{|x|^5} + \frac{a\cdot b}{3}\, \delta_0
\end{align}
$$
The physicist usually do not write the principal value, and so in shorter notations (with $r=|x|$)
$$
\boxed{\frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q =  \frac{1}{4πr^3} \left(a·b - 3\,(a\cdot \tfrac{x}{r})(b\cdot \tfrac{x}{r})\right) + \frac{a\cdot b}{3}\, \delta_0}
$$
A: I would go with spherical polar coordinates. Perhaps with $\vec{r}=(r,0,0)$, then the dot products can be evaluated using trigonometry functions of $\phi$ and $\theta$. It would eventually look like a Dirac delta modulated by the dot products among $\vec{a}$, $\vec{b}$ and $\vec{r}$.
Edit:
It is indeed more straightforward with your original idea.
$-(a\cdot\nabla)(b\cdot\nabla) \Delta^{-1} \delta(\vec{r})=-(a\cdot\nabla)(b\cdot\nabla) \frac{1}{4\pi r}=b\cdot\nabla(a\cdot \frac{\vec{r}}{r^3})=\frac{1}{4\pi r^3}[\vec{a}\cdot\vec{b}-3(\vec{a}\cdot\frac{\vec{r}}{r}) (\vec{b}\cdot\frac{\vec{r}}{r})].$
So it's the mutual energy between two dipoles a and b.
I would answer the above comment here (I cannot make comments as I'm new to the site): The Dirac delta is there at $r=0$, it's just that it is also nonzero at other places.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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In order to deal with possible issues related to integrals convergence, I'll include the $\ds{\underline{\it parameter}\,\,\, \beta}$. Its relevance is discussed at the very end.
\begin{align}
&\bbox[5px,#ffd]{{1 \over \pars{2\pi}^{3}}
\iiint_{\mathbb{R}^{3}}\pars{\vec{q}\cdot\vec{a}}
\pars{\vec{q}\cdot\vec{b}}
{\expo{\ic\vec{q}\cdot\vec{r}} \over
q^{2 + \beta}\,}\,\,\dd^{3}\vec{q}}
\\[5mm] = &\
{1 \over \pars{2\pi}^{3}}\,\vec{a}\cdot\bracks{%
\iiint_{\mathbb{R}^{3}}\vec{q}\,\vec{q}\,
{\expo{\ic\vec{q}\cdot\vec{r}} \over q^{2}}{\dd^{3}\vec{q} \over q^{\beta}}}\cdot\vec{b}
\\[5mm] = &\
{1 \over \pars{2\pi}^{3}}\,\vec{a}\cdot\bracks{%
\int_{0}^{\infty}\pars{%
\int_{\Omega_{\vec{q}}}\hat{q}\,\hat{q}\,
\expo{\ic\vec{q}\cdot\vec{r}}\,\dd\Omega_{\vec{q}}}
q^{2 - \beta}\,\,\dd q}\cdot\vec{b}\label{1}\tag{1}
\end{align}
Lets perform the angular integration where, for simplicity, $\ds{\vec{r}}$ is chosen along the $\ds{\hat{z}}$-axis. Later on, the $\ds{\vec{r}}$-general direction can be suitable restored:
\begin{align}
&\int_{\Omega_{\vec{q}}}\hat{q}\,\hat{q}\,
\expo{\ic\vec{q}\cdot\vec{r}}\,\dd\Omega_{\vec{q}} =
\int_{0}^{2\pi}\int_{0}^{\pi}
\hat{q}\,\hat{q}\expo{\ic qr\cos\pars{\theta}}\,
\sin\pars{\theta}\,\dd\theta\,\dd\phi
\\[5mm] = &\
\int_{0}^{\pi}
\bracks{\sin^{2}\pars{\theta}\,\pi\,\hat{x}\,\hat{x} +
\sin^{2}\pars{\theta}\,\pi\,\hat{y}\,\hat{y} +
\cos^{2}\pars{\theta}\pars{2\pi}\hat{z}\,\hat{z}}\ \times
\\ &\ \phantom{\,\,\int_{0}^{\pi}}
\,\,\,\expo{\ic qr\cos\pars{\theta}}\,\,\sin\pars{\theta}
\,\dd\theta
\\ = &\
\pi\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}}
\int_{-1}^{1}\pars{1 - \xi^{2}}\expo{\ic qr\xi}\,\dd\xi +
2\pi\,\hat{z}\,\hat{z}
\int_{-1}^{1}\xi^{2}\expo{\ic qr\xi}\,\dd\xi
\\ = &\
4\pi\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}}
\on{f}\pars{qr} +
4\pi\,\hat{z}\,\hat{z}\,\on{g}\pars{qr}
\\[5mm] &\ \mbox{where}\quad
\left\{\begin{array}{rcl}
\ds{\on{f}\pars{\xi}} & \ds{\equiv} &
\ds{\sin\pars{\xi} - \xi\cos\pars{\xi} \over \xi^{3}}
\\[2mm]
\ds{\on{g}\pars{\xi}} & \ds{\equiv} &
\ds{\pars{\xi^{2} - 2}\sin\pars{\xi} + 2\xi\cos\pars{\xi} \over \xi^{3}}
\end{array}\right.
\end{align}

(\ref{1}) becomes
\begin{align}
&\bbox[5px,#ffd]{{1 \over \pars{2\pi}^{3}}
\iiint_{\mathbb{R}^{3}}\pars{\vec{q}\cdot\vec{a}}
\pars{\vec{q}\cdot\vec{b}}
{\expo{\ic\vec{q}\cdot\vec{r}} \over q^{2 + \beta}}\,\dd^{3}\vec{q}}
\\[5mm] = &\
{1 \over 2\pi^{2}}\,{1 \over r^{3 - \beta}\,}
\,\,\vec{a}\,\cdot
\\[2mm] &\
\bracks{\pars{\hat{x}\,\hat{x} + \hat{y}\,\hat{y}}
\int_{0}^{\infty}\on{f}\pars{\xi}\,\xi^{2 - \beta}\,
\,\dd\xi +
\hat{z}\,\hat{z}
\int_{0}^{\infty}\on{g}\pars{\xi}\,\xi^{2 - \beta}\,
\,\dd\xi}
\cdot\vec{b}
\end{align}
The integrations are given by
\begin{align}
&\int_{0}^{\infty}\on{f}\pars{\xi}\,\xi^{2 - \beta}\,
\,\dd\xi =
{\Gamma\pars{2 - \beta}\sin\pars{\pi\beta/2} \over \beta}
\,,\quad\,\,\,\,\,\, 0 < \Re\pars{\beta} < 1
\\[5mm] &\
\int_{0}^{\infty}\on{g}\pars{\xi}\,\xi^{2 - \beta}\,
\,\dd\xi =
-\,{\Gamma\pars{3 - \beta}\sin\pars{\pi\beta/2} \over \beta}
\,,\quad 1 < \Re\pars{\beta} < 3
\end{align}
The first integral $\ds{\to {\pi \over 2}}$ as $\ds{\beta \to 0^{+}}$ while such a limit is not allowed for the second integral -unless we assume an analytical continuation- which we left as a discussion for the OP.
