I want to calculate / simplify:

$$\mathcal{F} (\ln(|x|)\mathcal{F(f)}(x))=\mathcal{F} (\ln(|x|)) \star f$$

where $\mathcal{F}$ is the Fourier transform ($\mathcal[f](\xi)=\int_{\mathbb R}f(x)e^{ix\xi}\,dx$) and where $f$ is an even function.

Looking here: wiki, we find that

$$\mathcal{F}[\log|x|](\xi)=-2\pi\gamma\delta(\xi)-\frac\pi{|\xi|},$$

so we should have:

$$\mathcal{F} (\ln(|x|)) \star f = (-2\pi\gamma\delta(x)-\frac\pi{|x|}) \star f(x) $$ $$ = -2\pi\gamma f(x)- \pi \int_{-\infty}^{\infty} \frac{f(t)}{|x-t|}) dt $$

but the integral of the second term does not converge... whereas the term $\mathcal{F} (\ln(x)\mathcal{F(f)}(x))$ is well defined providing the function $f$ is of rapide decrease near zero and infinity. So where is the problem ? and what is finally the "simplified expression" of $\mathcal{F} (\ln(x)\mathcal{F(f)}(x))$ ? We cannot use this distribution in a convolution product with a function?

I already post this on Stackexchange but did not receive an answer.