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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