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


*

*$f(t) > 0$

*$f(t) \in L^1$ 

*$f(t)$ is continuous 

*$\lim_{t \to \infty} f(t)=0$ and  $\lim_{t \to -\infty} f(t)=0$

*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. 
 A: 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 $\lambda\to0^+$:
$$
\begin{align}
&\int_{-\infty}^\infty\log(f(x))\phi(x)e^{-2\pi ix\xi}\,\mathrm{d}x\\
&=\frac1{2\pi i\xi}\int_{-\infty}^\infty\left(\frac{f'(x)}{f(x)}\phi(x)+\log(f(x))\phi'(x)\right)e^{-2\pi ix\xi}\,\mathrm{d}x\\
&\to\frac1{2\pi i\xi}\int_{-\infty}^\infty\frac{f'(x)}{f(x)}\,e^{-2\pi ix\xi}\,\mathrm{d}x\\
\end{align}
$$
A: To supplement the answer already given, i will use the trick of adding a small real part to the exponent $i\omega t \rightarrow i\omega t-\delta|t|$ so that the integral is convergent for large $t$. 
We then have to calculate the following expression:
$$
\lim_{\delta\rightarrow0}\int_0^{\infty}e^{i\omega t-\delta t}\log (g(t))+\lim_{\delta\rightarrow0}\int_{-\infty}^0e^{i\omega t+\delta t}\log (g(t))
$$
Partial integration yields 
$$
\lim_{\delta\rightarrow0}\big{(}\underbrace{\frac{1}{i\omega-\delta}[e^{i\omega t-\delta t}\log (g(t))]_0^{\infty}}_{(1)}-\underbrace{\frac{1}{i\omega-\delta}\int_0^{\infty}e^{i\omega t-\delta t}\frac{g'(t)}{g(t)}}_{(2)}+\\
\underbrace{\frac{1}{i\omega+\delta}[e^{i\omega t+\delta t}\log (g(t))]_{-\infty}^0}_{(3)}-\underbrace{\frac{1}{i\omega+\delta}\int_{-\infty}^0e^{i\omega t+\delta t}\frac{g'(t)}{g(t)}}_{(4)}\big{)}
$$
Now assuming that the FT of $(g'/g)$ exists, we can perform the limit after plugging in the endpoints of the integration domain.
We see that $(1)+(3)=0$ and $(2)+(4)=$
$$ 
\frac{-1}{i\omega}\int_{-\infty}^{\infty}e^{i\omega t}\frac{g'(t)}{g(t)}=\frac{-1}{i\omega}\text{FT}\big{[}\frac{g'(t)}{g(t)}\big{]}
$$ 
which is the same as in robs answer, by identifying $i\omega \rightarrow -2 \pi i \xi$
Please note that this crucially depends on the order in which we take limit and integration. For some reason this seems not to work out if we want to interpret this FT in the sense of distributions by applying Dirac's identity, which is at least surprising to me because i think this is the more general way to look at it.
