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How do you calculate the inverse Fourier transform of $\frac{k_ik_j}{k^4}$. I know it has to be a matrix of the form $=δ_{ij}A(r)+r_ir_jB(r)$, but how do you calculate the functions A(r) and B(r)?

I am trying to use Fourier transforms to find the Oseen tensor (a solution to Stokes equations).

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It is not appreciated if questions are 'destroyed' in this way, even if the intentions are good. I rolled back to a meaningful version. If you want the question deleted, please leave a comment to that extent and likely that request will be followed. – user9072 May 12 '13 at 12:07

Let me change your notations slightly: you work in three dimensions and you want to compute $$ u_{jk}(x)=\int e^{2i\pi x\cdot \xi} \frac{\xi_j\xi_k}{\vert \xi\vert^4} d\xi, $$ where the integral should not be taken as an $\dots$ integral. You have by homogeneity (in $n$ dimensions the Fourier transform of an homogeneous distribution with degree $\lambda$ is homogeneous with degree $-\lambda-n$) and "radiality"(the Fourier transform of a radial function is also radial) $$ u_{jk}=D_{x_j}D_{x_k}\int e^{2i\pi x\cdot \xi} \frac{1}{\vert \xi\vert^4} d\xi=cD_{x_j}D_{x_k} \vert x\vert , $$ where $c$ is a constant and $D_t=\partial_{t}/2i\pi$. We get $$ j\not=k,\quad u_{jk}=c_1\partial_{x_j}\vert x\vert^{-1}x_k=-c_1 x_jx_k\vert x\vert^{-3}, $$ $$ u_{jj}=c_1\partial_{x_j}\vert x\vert^{-1}x_j=c_1 \bigl(\vert x\vert^{-1} -x_j^2\vert x\vert^{-3}\bigr). $$

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Thank you thank you that way is A LOT easier than what i was trying to do! – sameaspie Feb 27 '13 at 12:26

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