A question about Fourier Transform of a derivative

Question

The Fourier Transform (FT) of the function $x(t)$ is define by $$ \mathcal{F}_x(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt $$ It is well known that $$ \mathcal{F}_{x'}(\omega)=j\omega\mathcal{F}_x(\omega) $$ In order to verify that result, I wanted to compute the FT of $$ y(t)= \begin{cases} 1 & t\in(-1,1)\\ 0 & \hbox{otherwise} \end{cases} $$ using the FT of $$ x(t)= \begin{cases} t & t\in(-1,1)\\ 0 & \hbox{otherwise} \end{cases} $$ By direct computation I got that $$ \mathcal{F}_x(\omega)=2j\cdot\frac{\omega\cos\omega-\sin\omega}{\omega^2} $$ Since $x'=y$ we deduce that $$ \mathcal{F}_y(\omega)=j\omega\cdot2j\cdot\frac{\omega\cos\omega-\sin\omega}{\omega^2}= 2(\mathrm{sinc}(\omega)-\cos\omega) $$ My question: By direct computation I got that $\mathcal{F}_y(\omega)=2\mathrm{sinc}(\omega)$. What is the correct result?

Answer 1

Your functions are not continuous. The "well-known" equality $$ \mathcal{F}_{x'}(\omega)=j\omega\mathcal{F}_x(\omega) $$ is a trivial consequence of integration by parts, when $x$ is differentiable and integrable; this is what makes the first term in the integration by parts $$ \int_{-\infty}^\infty x'(t)\,e^{-i\omega t}\,dt=x(t)\,e^{-i\omega t}\bigg|_{-\infty}^\infty+\frac1{i\omega}\int_{-\infty}^\infty x(t)\,e^{-i\omega t}\,dt $$ vanish. The problem you have is that Barrow's rule $$ \int_a^b f'=f(b)-f (a) $$ requires $f$ to be differentiable everywhere. For instance consider your $x$ and you have $$ \int_{-2}^2 x'(t)\,dt=2\ne 0= x(2)-x(-2). $$