44A10-LaplaceTransformOfTnft.tex

\documentclass[12pt]{article}
\usepackage{pmmeta}
\pmcanonicalname{LaplaceTransformOfTnft}
\pmcreated{2013-03-22 18:05:49}
\pmmodified{2013-03-22 18:05:49}
\pmowner{pahio}{2872}
\pmmodifier{pahio}{2872}
\pmtitle{Laplace transform of $t^nf(t)$}
\pmrecord{6}{40637}
\pmprivacy{1}
\pmauthor{pahio}{2872}
\pmtype{Derivation}
\pmcomment{trigger rebuild}
\pmclassification{msc}{44A10}
\pmrelated{TableOfLaplaceTransforms}

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\begin{document}
Let
$$\displaystyle F(s) \,:=\, \mathcal{L}\{f(t)\} = \int_0^\infty e^{-st}f(t)\,dt.$$

A differentiation under the integral sign with respect to $s$ yields
$$F'(s) = -\int_0^\infty e^{-st}tf(t)\,dt = -\mathcal{L}\{tf(t)\}.$$
Differentiating again under the integral sign gives
$$F''(s) = +\int_0^\infty e^{-st}t^2f(t)\,dt = \mathcal{L}\{t^2f(t)\}.$$
One can continue similarly, and then we apparently have
\begin{align}
F^{(n)}(s) = (-1)^n\int_0^\infty e^{-st}t^nf(t)\,dt = (-1)^n\mathcal{L}\{t^nf(t)\}.
\end{align}
If this equation is multiplied by $(-1)^n$, it gives the \PMlinkescapetext{formula}
\begin{align}
\mathcal{L}\{t^nf(t)\} = (-1)^nF^{(n)}(s)
\end{align}
which is true for\, $n = 1,\,2,\,3,\,\ldots$\\


\textbf{Application.}\, Evaluate the improper integral
$$I := \int_0^\infty t^3e^{-t}\sin{t}\,dt.$$
By the \PMlinkname{parent entry}{LaplaceTransform}, we have\, $\mathcal{L}\{\sin{t}\} = \frac{1}{1+s^2}$.\, Using this and (2), we may write
$$\int_0^\infty t^3e^{-st}\sin{t}\,dt = \mathcal{L}\{t^3\sin{t}\} = 
(-1)^3\frac{d^3}{ds^3}\!\left(\frac{1}{s^2+1}\right) = \frac{24(s-s^3)}{(1+s^2)^4}.$$
The value of $I$ is obtained by substituting here\, $s = 1$:
$$I = \frac{24(1-1^3)}{(1+1^2)^4} = 0.$$

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