📂Odinary Differential Equations

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

Formulas

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

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

Derivation

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

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

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

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

Therefore

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

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

Repeatedly differentiating with respect to $s$ yields the following result.

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

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Examples

1

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

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

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2

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

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

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See Also

• Table of Laplace Transforms