I have a vector field $A_i({\bf r})$, a Fourier transform given by
$$ \tilde A_i({\bf k}) = \int d^3 r~e^{i {\bf r.k}}A_i({\bf r}),$$
and projections given by $\mathcal P_{ij}(\hat r) = \delta_{ij} - \hat r_i \hat r_j$, that acts on $A_i({\bf r})$, giving
$$[\mathcal P.{\bf A}]_i({\bf r}) = A_i({\bf r}) - [\hat r_j A_j({\bf r})]\hat r_i.$$
I'm interested in computing the Fourier transform of the projection, i.e. $\widetilde{ [\mathcal P.{\bf A}]}_i({\bf k})$. I tried obtaining the explicit formula for $\tilde{\mathcal P}_{ij}({\bf k})$ so that I could use convolution
$$ \widetilde{[\mathcal P.{\bf A}]}_i({\bf k}) = \int \frac{d^3 p}{(2\pi)^3}~ \tilde{\mathcal P}_{ij}({\bf p - k}) \tilde A_i({\bf k}). ~~~~~~~~~~~~(1) $$
I got
$$\tilde{\mathcal P}_{ij}({\bf k}) = \int d^3 r ~ \mathcal P_{ij}({\bf r}) e^{i {\bf r.k}} = (2\pi)^3 \delta_{ij} \delta^D({\bf k}) - \int d^3 r ~ (\hat r_i \hat r_j) e^{i {\bf r.k}},$$
but I don't quite know how to represent the last integral as some useful distribution that I would know how to apply in the convolution formula above (similar to the delta function part). I know I can formally represent it as
$$\int d^3 r ~ (\hat r_i \hat r_j) e^{i {\bf r.k}} = (2\pi)^3 \frac{\partial_{k_i}\partial_{k_j}}{\partial^2_{k}}\delta^D({\bf k}),$$
but this does not help me much when I try to apply it in the convolution (1) above for some specific field $A_i({\bf k})$. Is there some useful way I can represent $\tilde{\mathcal P}_{ij}({\bf k})$ that will be practical for computing the convolution (1) above?