Calcul of an expression with Fourier and logarithm 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:" Fourier transform of $\log x$ $ |x|^{s} $ and $\log|x| $ " I found 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 $\mathcal{F} (\ln(x)\mathcal{F(f)}(x))$ ?
 A: For Fourier transforms, especially when applying the convolution theorem, a divergent $\left|x\right|^{-1}$ could appear and could be accepted, on condition that you take its Cauchy principal value, i.e.,
$$
\int_{-\infty}^{\infty}\frac{f(t)}{\left|x-t\right|}{\rm d}t:=\lim_{\epsilon\to0^+}\int_{\mathbb{R}\setminus\left[x-\epsilon,x+\epsilon\right]}\frac{f(t)}{\left|x-t\right|}{\rm d}t.
$$
Say you want to simplify the Fourier transform of
$$
\hat{f}(x)\log\left|x\right|,
$$
where $\hat{f}$ denotes the Fourier transform of $f$. According to the convolution theorem,
\begin{align}
\mathcal{F}(\hat{f}\log\left|\cdot\right|)(x)&=\left(\hat{\hat{f}}*\widehat{\log\left|\cdot\right|}\right)(x)\\
&=\left(f(-\cdot)*\left(-\frac{1}{2}\frac{1}{\left|\cdot\right|}-\gamma\delta(\cdot)\right)\right)(x)\\
&=-\left(f(-\cdot)*\left(\frac{1}{2}\frac{1}{\left|\cdot\right|}+\gamma\delta(\cdot)\right)\right)(x)\\
&=-\frac{1}{2}\oint_{\mathbb{R}}\frac{f(-t)}{\left|x-t\right|}{\rm d}t-\gamma f(-x)\\
&=-\frac{1}{2}\oint_{\mathbb{R}}\frac{f(t)}{\left|x+t\right|}{\rm d}t-\gamma f(-x)\\
&=-\frac{1}{2}\lim_{\epsilon\to 0^+}\int_{\mathbb{R}\setminus\left[-x-\epsilon,-x+\epsilon\right]}\frac{f(t)}{\left|x+t\right|}{\rm d}t-\gamma f(-x),
\end{align}
where $\oint$ means to take the Cauchy principal value. Here we take $e^{2\pi ix\xi}$ as the Fourier transform factor.
