I'm trying to solve $f'(t)=0$ and $f'(t)=1$ using Fourier transform, but no luck:

a) $f'(t)=0$

$$ f'(t)=0 \Rightarrow jwF(w)=0 \Rightarrow \begin{cases}F(w)=0 ~ \text{if} ~ w \ne 0\\ F(0) = \text{undefined ?} \end{cases} $$

Addendum from comments of @Winther (if I understood correctly): In case we choice $F(0) \ne \infty$ then $f(t)=0$ that is one of the valid solutions; if we choice $F(0) = \infty$ then $f(t)=1$, another valid solution; no idea how to obtain all other possible solutions.

b) $f'(t)=1$

$$ f'(t)=1 \Rightarrow jwF(w)=\delta(w) \Rightarrow \begin{cases}F(w)=0 ~ \text{if} ~ w \ne 0\\ F(0) = \text{impossible! because left part is} =0 \text{ but right is }=\infty \end{cases} $$

After this disaster, I've tried to proof the results (calc the derivative using Fourier transform) that I expect as the correct solutions:

c) $f(t)=c$ (being $c$ in $\mathbb{R}$)

$$f(t)=c \Rightarrow\\ F(w)=c\delta(w) \Rightarrow\\ \mathscr{F}\{f'\}(w) = jwc\delta(w) = 0 \Rightarrow \\ f'(t) = 0 $$


d) $f(t)=t+c$

$$f(t)=t+c\Rightarrow\\ F(w)=\delta^{(1)}(w)+c\delta(w)\Rightarrow\\ \mathscr{F}\{f'\}(w) = jw\delta^{(1)}(w)+jwc\delta(w) = jw\delta^{(1)}(w) = ~ ? $$

( where $\delta^{(1)}(x)=\frac{d}{dx}\delta(x)$ )


  • $\begingroup$ $wF(w) = 0$ has the solution $F(w) = C\delta(w)$ since $w\delta(w) \equiv 0$. $\endgroup$ – Winther Oct 15 '18 at 11:19
  • $\begingroup$ @Winther: thanks for your comment. Yes, what you say is absolutelly true, but I can also defined a $s(w)$ such as $s(w)=0 if w\ne0$, $s(0)=14$ and is also a solution of $wF(w)=0$. Moreover, I do not see how to progress in case "2)" $\endgroup$ – pasaba por aqui Oct 15 '18 at 11:23
  • $\begingroup$ For the first part: yes you can add a finite value like $14$ to $F[0]$ however this does not change the integral when transforming back (only the $\delta$ and it's derivatives are able to do so). Likewise for part 2 you get $F(w) = C\delta(w) + \frac{\delta(w)}{jw}$. $\endgroup$ – Winther Oct 15 '18 at 11:27
  • $\begingroup$ @Winther: sorry, I do not know how to evaluate the inverse Fourier transform of the $s(w)$ I've described in previous comment or, at least, proof that the inverse of $s(w)$ is any constant. Please, if you have time, write an answer. $\endgroup$ – pasaba por aqui Oct 15 '18 at 11:31
  • $\begingroup$ If $s(w) = 0$ except $s(0) = 14$ then the integral equals $\int_{-\epsilon}^{\epsilon} s(w) e^{-jwx}{\rm d}w$ for any $\epsilon > 0$. The integrand is bounded in absolute value by $|s(0)|$ so the integrand is $\ll 2|s(0)|\epsilon$. So it has to be $0$. Maybe later. If I write up something I would like to make it a bit more rigorous than the rough outline in the comments above so only if I get time. $\endgroup$ – Winther Oct 15 '18 at 11:35

In the case $1$, if $F(\omega)=0$ for all $\omega\ne 0$ then $F(\omega)$ can be $k\delta(\omega)$ for any real $k$ therefore$$f(t)={k\over 2\pi}$$Case $2$

First of all, we prove the following interesting property of $\delta(.)$ and $\delta'(.)$: $$\omega\delta'(\omega)=-\delta(\omega)$$notice that $\omega\delta'(\omega)=0\quad,\forall\omega\ne 0$ so we need to prove that $$\int_{0^-}^{0^+}\omega\delta'(\omega)d\omega=-1$$which is obvious because$$\int_{0^-}^{0^+}\omega\delta'(\omega)d\omega=\omega\delta(\omega)|_{0^-}^{0^+}-\int_{0^-}^{0^+}\delta(\omega)d\omega=-1$$which completes our proof. Now we have$$f'(t)=1$$therefore$$F(f')(\omega)=2\pi\delta(\omega)=-2\pi\omega\delta'(\omega)+2\pi j k\omega \delta(\omega)$$therefore$$F(f)(\omega)={F(f')(\omega)\over j\omega}=2\pi j\delta'(\omega)+2\pi k\delta (\omega)$$therefore$$f(t)=t+2\pi k\quad,\quad k\in\Bbb R$$


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