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Here is one way to do it. Multiplication by $k_ik_j$ on the Fourier transform side corresponds to taking the second derivative $\dfrac{\partial^2 }{\partial x_i \partial x_j}$. Therefore, we can find the inverse Fourier transform of $1/k^4$ and then take the Hessian of that. This is also preferable (to me) because it reduces the task to transforming a scalar function.

Following the computations here, one can see that the transform of a radially symmetric function is radially symmetric. Also, the same computation implies that the transform of a function that is homogeneous of degree $d$ is homogeneous of degree $-(d+n)$ where $n$ is the dimension ($n=3$ here). Therefore, the IFT of $1/k^4$ is of the form $f(x)=c|x|$. Of course, the value of $c$ can be found once we decided on the normalization of the transform*.

The Hessian of $|x|$ is not hard to compute: $D |x| = x|x|^{-1}$, and then by the product rule $$D^2(|x|)=I|x|^{-1}-(x\otimes x)|x|^{-3}\tag1$$ This is exactly what they get for $\mathcal{F}^{-1}(k_ik_j/k^4)$, save for the factor $1/(8\pi)$.

Naturally, I'm ignoring all convergence issues.

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Thank you thank you that way is A LOT easier than what i was trying to do! – Rosie Huggon Feb 27 '13 at 12:25

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