[Math] Fourier Transform of $\ln(f(t))$

fourier analysisfunctional-analysisreal-analysis

I want to compute Fourier transform of $\ln(f(t))$ maybe in a sense of distributions?
Where we can assume that:

  1. $f(t) > 0$
  2. $f(t) \in L^1$
  3. $f(t)$ is continuous
  4. $\lim_{t \to \infty} f(t)=0$ and $\lim_{t \to -\infty} f(t)=0$
  5. Denote Fourier transform of $f(t)$ by $\mathcal{F}(f(t))=F(\omega)$

I am also fine with some other restrictions. For example: $f(t)$ is analytical (i.e. all derivatives exist)
My main question are:

1) is the above set of condition enough to guarantee existence of Fourier transform? If not, what are the requirements?

2) what is the Fourier transform?

Edit Based on the suggestion of Mattos

\begin{align*}
&\int_{-\infty}^\infty \ln(f(t)) e^{-i \omega t} dt= \ln(f(t))\frac{e^{-j\omega t}}{-j\omega} \Big|_{t=-\infty}^{t=\infty}-\int_{-\infty}^\infty \frac{e^{-j\omega t}}{-j\omega} \frac{f'(t)}{f(t)} dt\\
& =\ln(f(t))\frac{e^{-jwt}}{-jw} \Big|_{t=-\infty}^{t=\infty} +\frac{1}{j \omega}\mathcal{F} \left(\frac{f'(t)}{f(t)} \right)
\end{align*}

But, now how to compute $\ln(f(t))\frac{e^{-jwt}}{-jw} \Big|_{t=-\infty}^{t=\infty}$.

Possible Solution to 2):

I might have solution but not sure. It relays on using the following property.

\begin{align*}
\mathcal{F} \left( \frac{d}{dt} g(t) \right)=(j\omega) \mathcal{F}(g(t))
\end{align*}

Now take $g(t)=\ln(f(t))$ so we have
\begin{align*}
\mathcal{F} \left( \frac{d}{dt} \ln(f(t)) \right)=(j\omega) \mathcal{F}(\ln(f(t)))\\
\mathcal{F} \left( \frac{f'(t)}{f(t)}\right)=(j\omega) \mathcal{F}(\ln(f(t)))\\
\frac{\mathcal{F} \left( \frac{f'(t)}{f(t)}\right)}{(j\omega)}= \mathcal{F}(\ln(f(t)))
\end{align*}

So, of course this solution requires that $f'(t)$ exists.
Is this reasoning correct? Is there any technicality that I missed?

Here is the example I tried

Let $f(t)=e^{-t^2}$ then $\ln(f(t))=-t^2$,
\begin{align*}
\mathcal{F}(\ln(f(t)))=\mathcal{F}(-t^2)=2\pi\delta^{(2)}(\omega)
\end{align*}

On the other hand,
\begin{align*}
\frac{1}{j \omega} \mathcal{F} \left( \frac{f'(t)}{f(t)} \right)&=\frac{1}{j \omega} \mathcal{F} \left( \frac{-2te^{-t^2}}{e^{-t^2}} \right)=\frac{1}{j \omega} \mathcal{F} \left( -2t \right)=-2 \frac{1}{j \omega} (2 \pi j) \delta^{(1)}(\omega)\\
&= -(4 \pi ) \frac{1}{ \omega}\delta^{(1)}(\omega)=(2 \pi ) \delta^{(2)}(\omega)
\end{align*}

the last equality uses $\omega \delta^{(2)}(\omega)=-2\delta^{(1)}(\omega)$

So, this is an example when my approach works.

Thank you for any help, I really appreciate it.

Best Answer

For suitable $f$, we have $$ \mathcal{F}(f')(\xi)=2\pi i\xi\mathcal{F}(f)(\xi) $$ Therefore, if one exists, it would be $$ \mathcal{F}(f'/f)(\xi)=2\pi i\xi\mathcal{F}(\log(f))(\xi) $$ which would lead to $$ \mathcal{F}(\log(f))(\xi)=\frac1{2\pi i\xi}\mathcal{F}(f'/f)(\xi) $$ if $f'/f$ has a Fourier Transform.

To show a bit more care, we can make the preceding a bit more rigorous, by letting $\phi(x)=e^{-\lambda x^2}$ and sending $\lambda\to0^+$: $$ \begin{align} &\int_{-\infty}^\infty\log(f(x))\phi(x)e^{-2\pi ix\xi}\,\mathrm{d}x\\ &=\frac1{2\pi i\xi}\int_{-\infty}^\infty\left(\frac{f'(x)}{f(x)}\phi(x)+\log(f(x))\phi'(x)\right)e^{-2\pi ix\xi}\,\mathrm{d}x\\ &\to\frac1{2\pi i\xi}\int_{-\infty}^\infty\frac{f'(x)}{f(x)}\,e^{-2\pi ix\xi}\,\mathrm{d}x\\ \end{align} $$

Related Question