# Laplace transform of $t^{n}f(t)$

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

 $F^{\prime}(s)=-\int_{0}^{\infty}e^{-st}tf(t)\,dt=-\mathcal{L}\{tf(t)\}.$

Differentiating again under the integral sign gives

 $F^{\prime\prime}(s)=+\int_{0}^{\infty}e^{-st}t^{2}f(t)\,dt=\mathcal{L}\{t^{2}f% (t)\}.$

One can continue similarly, and then we apparently have

 $\displaystyle F^{(n)}(s)=(-1)^{n}\int_{0}^{\infty}e^{-st}t^{n}f(t)\,dt=(-1)^{n% }\mathcal{L}\{t^{n}f(t)\}.$ (1)

If this equation is multiplied by $(-1)^{n}$, it gives the

 $\displaystyle\mathcal{L}\{t^{n}f(t)\}=(-1)^{n}F^{(n)}(s)$ (2)

which is true for  $n=1,\,2,\,3,\,\ldots$

Application.  Evaluate the improper integral

 $I:=\int_{0}^{\infty}t^{3}e^{-t}\sin{t}\,dt.$

By the parent entry (http://planetmath.org/LaplaceTransform), we have  $\mathcal{L}\{\sin{t}\}=\frac{1}{1+s^{2}}$.  Using this and (2), we may write

 $\int_{0}^{\infty}t^{3}e^{-st}\sin{t}\,dt=\mathcal{L}\{t^{3}\sin{t}\}=(-1)^{3}% \frac{d^{3}}{ds^{3}}\!\left(\frac{1}{s^{2}+1}\right)=\frac{24(s-s^{3})}{(1+s^{% 2})^{4}}.$

The value of $I$ is obtained by substituting here  $s=1$:

 $I=\frac{24(1-1^{3})}{(1+1^{2})^{4}}=0.$
Title Laplace transform  of $t^{n}f(t)$ LaplaceTransformOfTnft 2013-03-22 18:05:49 2013-03-22 18:05:49 pahio (2872) pahio (2872) 6 pahio (2872) Derivation msc 44A10 TableOfLaplaceTransforms