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.

Laplace transform of tn⁢f⁢(t)