MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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 $$ j\not=k,\quad u_{jk}=c_1\partial_{x_j}\vert x\vert^{-1}x_k=-c_1 x_jx_k\vert x\vert^{-3}, $$ $$ u_{jj}=c_1\partial_{x_j}\vert x\vert^{-1}x_j=c_1 \bigl(\vert x\vert^{-1} -x_j^2\vert x\vert^{-3}\bigr). $$

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.