Question #177423

Use Definition 7.1.1.

DEFINITION 7.1.1    Laplace Transform

Let f be a function defined for t ≥ 0.

 Then the integral

ℒ{f(t)} = ∞

e−stf(t) dt0

is said to be the Laplace transform of f, provided that the integral converges.

Find ℒ{f(t)}.

 (Write your answer as a function of s.)

f(t) = {t, 0 ≤ t < 1

{1,  t ≥ 1


Expert's answer

L(f(t))=∫0∞e−stf(t)dt=ℒ(f(t))=\int_0^{\infty}e^{-st}f(t) dt=


=∫01e−sttdt+∫1∞e−stdt=\int_0^1e^{-st}t dt+\int_1^{\infty}e^{-st}dt


∫01e−sttdt=−1se−stt∣01+∫01e−stsdt=\int_0^1e^{-st}tdt=-\frac{1}{s}e^{-st}t|_0^1+\int_0^1\frac{e^{-st}}{s}dt=


=−1se−s−1s2e−st∣01==-\frac{1}{s}e^{-s}-\frac{1}{s^2}e^{-st}|_0^1=


=−1se−s−1s2(e−s−1)=-\frac{1}{s}e^{-s}-\frac{1}{s^2}(e^{-s}-1)


∫1∞e−stdt=−1se−st∣1∞=1se−s\int_1^{\infty}e^{-st}dt=-\frac{1}{s}e^{-st}|_1^{\infty}=\frac{1}{s}e^{-s}


L(f(t))=−1se−s−1s2(e−s−1)+1se−s=ℒ(f(t))=-\frac{1}{s}e^{-s}-\frac{1}{s^2}(e^{-s}-1)+\frac{1}{s}e^{-s}=


=1s2(1−e−s)=\frac{1}{s^2}(1-e^{-s})


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