Question 8.12: Using the time integration property to derive a transform pa......

Derive the same pair from \operatorname{u}(t)\!\overset{\mathcal L}{\longleftrightarrow } 1/s,\ \sigma\gt 0 using the time integration property instead.

\int\limits_{-\infty }^{t}{} u(\tau )d\tau =\left \{ \begin{matrix} \int_{0^{-}}^{t}d\tau=\tau, & t\geq0 \\ 0, & t\lt 0 \end{matrix} \right \} =t\mathbf{u}(t)_{.}

Therefore

t\mathbf{u}(t) \overset{\mathcal L}{\longleftrightarrow } \ {\frac{1}{s}}\times{\frac{1}{s}}={\frac{1}{s^{2}}},\ \sigma\gt 0.

Successive integrations of u(t) yield

t\mathbf{u}(t),\;{\frac{t^{2}}{2}}\mathbf{u}(t),\;{\frac{t^{3}}{6}}\mathbf{u}(t)

and these can be used to derive the general form

{\frac{t^{n}}{n!}}\mathbf{u}(t) \overset{\mathcal L}{\longleftrightarrow } {\frac{1}{s^{n+1}}},\ \sigma\gt 0.