Frequency Differentiation Property of The Laplace Transform

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If $F(s)$ is the Laplace transform of $f(t)$, then
$$F(s)=\int_{0}^{\infty} f(t) e^{-s t} d t$$
Taking the derivative with respect to $s$,
$$\frac{d F(s)}{d s}=\int_{0}^{\infty} f(t)\left(-t e^{-s t}\right) d t=\int_{0}^{\infty}(-t f(t)) e^{-s t} d t=\mathcal{L}[-t f(t)]$$
and the frequency differentiation property becomes
$$\mathcal{L}[t f(t)]=-\frac{d F(s)}{d s} \tag{1}$$
Repeated applications of this equation lead to
$$\mathcal{L}\left[t^{n} f(t)\right]=(-1)^{n} \frac{d^{n} F(s)}{d s^{n}}$$
For example, we know from Example 1 that $\mathcal{L}\left[e^{-a t}\right]=1 /$ $(s+a)$. Using the property in Eq. (1),
$$\mathcal{L}\left[t e^{-a t}\right]=-\frac{d}{d s}\left(\frac{1}{s+a}\right)=\frac{1}{(s+a)^{2}}$$
Note that if $a=0$, we obtain $\mathcal{L}[t]=1 / s^{2}$ as in ramp function, and repeated applications of Eq. (1) will yield
$$\mathcal{L}\left[t^{n}\right]=\frac{n !}{s^{n+1}}$$

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