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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 $\mathcal{F} (\ln(x)\mathcal{F(f)}(x))$ ?

Bertrand
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1 Answers1

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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.

hypernova
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  • $\lim_{\epsilon\to0^+}\int_{\mathbb{R}\setminus\left[x-\epsilon,x+\epsilon\right]}\frac{f(t)}{\left(x-t\right)}{\rm d}t. $ exists but how $\lim_{\epsilon\to0^+}\int_{\mathbb{R}\setminus\left[x-\epsilon,x+\epsilon\right]}\frac{f(t)}{\left|x-t\right|}{\rm d}t. $ can exist ? – Bertrand May 11 '18 at 14:47
  • @Bertrand: You are right. This is a problem. But how do you define $\mathcal{F}(\log(x))$ if you do not take the absolute value of $x$? $\log x$ would not be defined for $x<0$. – hypernova May 11 '18 at 15:02
  • You are right I will add the absolute value to $\ln(|x|)$ in my question. Any idea on where is the problem of this convolution ? Your calculation seems correct but at the end the integral does not converge. – Bertrand May 11 '18 at 15:49
  • @Bertrand: Yes, that seems not to converge. I am thinking about this. I will get to you immediately I have any idea. – hypernova May 11 '18 at 15:58
  • are you sure of the formula for $\mathcal{F}(\ln(|x|)$ ? – Bertrand May 11 '18 at 18:32
  • @Bertrand: Yes, you may refer to Wiki. – hypernova May 11 '18 at 18:50