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I would like to derive the Fourier transform of $f(x)=\ln(x^2+a^2)$, where $a\in \mathbb{R}^+$ by making use of the properties:

\begin{equation} \mathcal{F}[f'(x)]=(ik)\hat{f}(k)\\ \mathcal{F}[-ixf(x)]=\hat{f}'(k) \end{equation} For the Fourier transform I use the definition given by:

\begin{equation} \hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx, k \in \mathbb{R} \end{equation} Until now I found out that by taking the derivative of $f$ and finding the Fourier transform of $f'$ I can then use the relation $\mathcal{F}[f'(x)]=(ik)\hat{f}(k)$ and find $\hat{f}$. The derivative of $f$ would be: \begin{equation} f'(x)=\frac{2x}{x^2+a^2} \end{equation} and by considering $g(x)=1/(x^2+a^2)$, I then have: \begin{equation} f'(x)=2xg(x) \end{equation} Now I know that the Fourier transform of $g$ is given by:

\begin{equation} \hat{g}(k)=\frac{1}{a}\sqrt{\frac{\pi}{2}}e^{-a|k|}, a \in \mathbb{R}, k\in \mathbb{R} \end{equation} Now I must find the Fourier transform of $xg(x)$ which would be given by the derivative of $\hat{g}$ right? But how can this possible since $\hat{g}$ has no derivative?

I think I am really close now but I need that extra tip.

Thank you!

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    $\begingroup$ $\hat{g}$ is differentiable everywhere except at $0$. It is however absolutely continuous, so the almost everywhere defined derivative is what you want. $\endgroup$ – Daniel Fischer Jun 27 '15 at 20:57
  • $\begingroup$ @DanielFischer Are you sure that I can do that? Because I think that I am not supposed to differentiate in this particular case. Is everything else correct until the point I have reached? $\endgroup$ – Mitscaype Jun 27 '15 at 21:01
  • $\begingroup$ I haven't carefully checked whether you made any mistakes so far, and at the moment, I'm not in the mood to do it. Looks okay so far, however. Differentiating $g$ to get the Fourier transform of $f'$ is legitimate here (don't forget that there are some constant factors to take care of). Your problem comes when you divide by $k$ to get the Fourier transform of $f$, because what you get then is not a tempered distribution. And that indicates that something is amiss. $\endgroup$ – Daniel Fischer Jun 27 '15 at 21:09
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    $\begingroup$ @DanielFischer You are right, I do know that $f$ defines a tempered distribution (even though I cannot prove it). And I also know that the Fourier transform of an $f \in \mathcal{S}$ has a Fourier transform that is also defined in the same space. But the Fourier transform here is (finally) given by: \begin{equation} \hat{f}(k)=-\sqrt{2\pi}\frac{e^{-|ak|}}{|k|}\end{equation} which I do not know if it is a "function" of the same space. Anyway, thank you for your assistance! $\endgroup$ – Mitscaype Jun 27 '15 at 21:27
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    $\begingroup$ Minor suggestion to the question (v2): Replace the condition $a\in\mathbb{R}$ with $a>0$ in two places. $\endgroup$ – Qmechanic Jun 28 '15 at 12:03
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If $\left| x \right|^{-1}$ is the distribution defined as $$\left( \left| x \right|^{-1}, \phi \right) = \int_{\left| x \right| < 1} \frac {\phi(x) - \phi(0)} {\left| x \right|} dx + \int_{\left| x \right| > 1} \frac {\phi(x)} {\left| x \right|} dx,$$ then $${\mathcal F}\!\left[ \ln\left( x^2 +a^2 \right) \right] = \frac 1 {\sqrt{2 \pi}} \int_{-\infty}^{\infty} \ln\left( x^2 +a^2 \right) e^{-i k x} dx = -\sqrt{2 \pi} \left( \frac { e^{-a \left| k \right|}} {\left| k \right|} + 2 \gamma \delta(k) \right),$$ where $\gamma$ is Euler's constant. Probably the easiest way to prove it is to compute the inverse transform directly from the definition of $\left| x \right|^{-1}$.

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