Question 5.2.1: Compute the Laplace transform of f (t ) = t .

By definition,

L[t]=\int_{0}^{∞}{te^{−st} dt }             (5.2.6)

Replacing the improper integral with a limit and integrating by parts, we observe that

L[t]=\underset{r\rightarrow \infty }{\lim } \int_{0}^{r}{te^{−st} dt }

=\underset{r\rightarrow \infty }{\lim }[−\frac {1}{s}(t + \frac {1}{s})e^{−st}\overset{r}{\underset{0}{\mid }}]

=\underset{r\rightarrow \infty }{\lim }[−\frac {1}{s}(r + \frac {1}{s})e^{−sr} + \frac {1}{s}(0+ \frac {1}{s})e^{0}]

=\underset{r\rightarrow \infty }{\lim }[−\frac {r}{s}e^{−sr} − \frac {1}{s^{2}} e^{−sr} + \frac {1}{s^{2}}]             (5.2.7)

By L’Hopital’s Rule,^{1} we know that re^{−sr} → 0 as r →∞ for each s > 0. Combined with the fact that e^{−sr} →0 as r→∞, it follows from (5.2.7) that

L[t] = F(s) = \frac {1}{s^{2}}                  (5.2.8)

1: \underset{r\rightarrow \infty }{\lim }\frac {r}{e^{sr}}=\underset{r\rightarrow \infty }{\lim }\frac {1}{se^{sr}}= 0.