Laplace Transform of t^{n}f(t)
📂Odinary Differential EquationsLaplace Transform of t^{n}f(t)
Let’s say the Laplace transform of the function f(t) is L{f(t)}=∫0∞e−stf(t)dt=F(s). Then, the Laplace transform of tnf(t) is as follows.
L{tnf(t)}=(−1)nF(n)(s)
Derivation
First, the Laplace transform of tnf(t), by definition, is as follows.
∫0∞e−sttf(t)dt
If we look closely at the integral, it can be seen that it is the same as differentiating e−stf(t) with respect to s.
dsd(e−stf(t))=e−st(−t)f(t)
Therefore
dsdF(s)=∫0∞dsd(e−stf(t))dt=∫0∞e−st(−t)f(t)dt
⟹−F′(s)=L{tf(t)}
Repeatedly differentiating with respect to s yields the following result.
⟹⟹⟹−F′(s)F′′(s)−F(3)(s)(−1)nF(n)(s)=∫0∞e−st(t)f(t)dt=L{tf(t)}=∫0∞e−st(t2)f(t)dt=L{t2f(t)}=∫0∞e−st(t3)f(t)dt=L{t3f(t)}⋮=∫0∞e−st(tn)f(t)dt=L{tnf(t)}
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Examples
1
Let’s say f(t)=tsint. Since L{sint}=s2+11,
L{tsint}=(s2+1)2−2s
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2
Let’s say f(t)=teatcosbt. Since L{eatcosbt}=(s−a)2+b2s,
L{teatcosbt}=((s−a)2+b2)4(s−a)2+b2−2(s−a)s
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See Also