Multiplication by t ↔ differentiation

Statement (differentiation in frequency)

If $x(t) \xleftrightarrow{F} X(\omega)$, then

$$t\,x(t) \xleftrightarrow{F} j\frac{dX(\omega)}{d\omega}$$

Proof

Start from the definition:

$$X(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$$

Differentiate both sides with respect to $\omega$. Differentiation can be taken inside the integral, since the limits do not depend on $\omega$:

$$\frac{dX(\omega)}{d\omega} = \int_{-\infty}^{\infty}x(t)\frac{\partial}{\partial\omega}e^{-j\omega t}dt = \int_{-\infty}^{\infty}(-jt)\,x(t)e^{-j\omega t}dt$$

Multiply both sides by $j$, using $j(-j) = 1$:

$$j\frac{dX(\omega)}{d\omega} = \int_{-\infty}^{\infty}t\,x(t)e^{-j\omega t}dt = F\{t\,x(t)\}$$

So multiplying by $t$ in time is the same as differentiating in frequency (and multiplying by $j$). Repeating the step gives $t^n x(t) \leftrightarrow j^n\frac{d^nX(\omega)}{d\omega^n}$.

Example

$e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}$, so

$$t\,e^{-at}u(t) \leftrightarrow j\frac{d}{d\omega}\left(\frac{1}{a + j\omega}\right) = j\cdot\frac{-j}{(a + j\omega)^2} = \frac{1}{(a + j\omega)^2}$$