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Laplace Transform of t^{n}f(t) 📂Odinary Differential Equations

Laplace Transform of t^{n}f(t)

Formulas

Let’s say the Laplace transform of the function f(t)f(t) is L{f(t)}=0estf(t)dt=F(s)\mathcal{L} \left\{ f(t) \right\} = \displaystyle \int _{0} ^\infty e^{-st}f(t)dt = F(s). Then, the Laplace transform of tnf(t)t^{n}f(t) is as follows.

L{tnf(t)}=(1)nF(n)(s) \mathcal{L} \left\{ t^n f(t) \right\} = (-1)^nF^{(n)}(s)

Derivation

First, the Laplace transform of tnf(t)t^nf(t), by definition, is as follows.

0esttf(t)dt \int _{0} ^\infty e^{-st}tf(t) dt

If we look closely at the integral, it can be seen that it is the same as differentiating estf(t)e^{-st}f(t) with respect to ss.

dds(estf(t))=est(t)f(t) \dfrac{d}{ds} \left( e^{-st}f(t) \right) = e^{-st}(-t)f(t)

Therefore

ddsF(s)=0dds(estf(t))dt=0est(t)f(t)dt \dfrac{d}{ds}F(s)=\int_{0}^\infty \dfrac{d}{ds} \left( e^{-st}f(t) \right) dt =\int_{0}^\infty e^{-st}(-t)f(t)dt

    F(s)=L{tf(t)} \implies -F^{\prime}(s) = \mathcal{L} \left\{ tf(t) \right\}

Repeatedly differentiating with respect to ss yields the following result.

F(s)=0est(t)f(t)dt=L{tf(t)}    F(s)=0est(t2)f(t)dt=L{t2f(t)}    F(3)(s)=0est(t3)f(t)dt=L{t3f(t)}    (1)nF(n)(s)=0est(tn)f(t)dt=L{tnf(t)} \begin{align*} && -F^{\prime}(s) &= \int_{0}^\infty e^{-st}(t)f(t)dt=\mathcal{ L} \left\{ tf(t) \right\} \\ \implies && F^{\prime \prime}(s) &= \int_{0}^\infty e^{-st}(t^2)f(t)dt=\mathcal{ L} \left\{ t^2f(t) \right\} \\ \implies && -F^{(3)}(s) &= \int_{0}^\infty e^{-st}(t^3)f(t)dt=\mathcal{ L} \left\{ t^3f(t) \right\} \\ && &\vdots \\ \implies && (-1)^nF^{(n)}(s) &= \int_{0}^\infty e^{-st}(t^n)f(t)dt=\mathcal{ L} \left\{ t^nf(t) \right\} \end{align*}

Examples

1

Let’s say f(t)=tsintf(t)=t\sin t. Since L{sint}=1s2+1\mathcal{ L} \left\{ \sin t \right\}=\dfrac{1}{s^2+1},

L{tsint}=2s(s2+1)2 \mathcal{ L} \left\{ t\sin t \right\}=\dfrac{-2s}{(s^2+1)^2}

2

Let’s say f(t)=teatcosbtf(t)=te^{at}\cos bt. Since L{eatcosbt}=s(sa)2+b2\mathcal{ L} \left\{ e^{at}\cos bt \right\}=\dfrac{s}{(s-a)^2+b^2},

L{teatcosbt}=(sa)2+b22(sa)s((sa)2+b2)4 \mathcal{ L} \left\{ te^{at} \cos bt \right\}=\dfrac{(s-a)^2+b^2-2(s-a)s}{\left( (s-a)^2+b^2 \right)^4}

See Also