Does the integral in the formal 2D Fourier transform of the logarithm converge?

Question

If $k$ is a nonzero vector in $\mathbb R^2$, how to interpret this integral: $$\int_{\mathbb R^2}e^{ik\cdot x}\ln{|x|}dx$$ Does it converge and in what sense?

Thanks in advance.

Answer 1

The integral does not converge in any usual sense.

You want the Fourier transform of $\ln |x|$. The Laplacian of $\ln |x|$ is $2\pi \delta_0$, a multiple of the delta function at the origin. The Fourier transform of this $2\pi \delta_0$ is the constant function $2\pi$. On the other hand, taking the Laplacian amounts to multiplying the Fourier transform by $-|k|^2$. So, the Fourier transform of $\ln |x|$ should be $-2\pi |k|^{-2}$.

I say "should be" because $-2\pi |k|^{-2}$ is not locally integrable and therefore does not qualify as a distribution. So the approximation by tempered distributions does not seem to legitimize the result.

However, we can integrate $|k|^{-2}$ against test functions vanishing at the origin. On the $x$ plane, this corresponds to convolution with test functions that have zero integral. So, the transform is valid in the following sense: if $\varphi$ is a Schwartz test function with $\int \varphi=0$, then the convolution $(\ln |x|) *\varphi$ is a tempered distribution, and the Fourier transform of $(\ln |x|) *\varphi$ is $-2\pi |k|^{-2} \hat \varphi$.