Let ft be piecewise continuous on 0 infinity and of exponent

Let f(t) be piecewise continuous on [0, infinity) and of exponential order. (a) Show that there exist constants K and a such that |f(t)lesstahnoreqalto Ke^at for all Greterthanorequalto 0. (b) By using the definition of the transform and estimating the integral with the help of part (a), prove that

Solution

Solution :

a) Since f is piecewise continuous, f consists of finitely many continuous pieces (f1, ..., fn).

By hypothesis, each piece is of exponential order. So for each k = 1, ..., n, there exist constants K₁, ..., K_n and ₁, ..., _n such that |fk(t)| K_k * e^{(k * t)} on their respective intervals of definition.

Let K = max{K₁, ..., K_n} and = max{₁, ..., _n}.
Then, it immediately follows that |f(t)| K e^(t) for all t 0.

b) |L{f(s)}| = |(t = 0 to ) e^(-st) f(t) dt|, by definition of L
................. (t = 0 to ) |e^(-st) f(t)| dt
.................= (t = 0 to ) e^(-st) |f(t)| dt
................. (t = 0 to ) e^(-st) * K e^(t)dt, by part a
.................= (t = 0 to ) Ke^(-(s-)t)dt
.................= Ke^(-(s-)t)/-(s-) { for t = 0 to }
.................= K/(s - ), assuming that s > .

In summary, |L{f(s)}| K/(s - ), for all s > .
Since lim(s) ±K/(s - ) = 0, we conclude from the Squeeze Theorem that lim(s) L{f(s)} = 0.