# The Laplace Transform: Transforms of Derivatives

\begin{align*}
\mathcal{L}\{F'(t)\}&=\int_0^\infty e^{-st}F'(t)dt\\
&=e^{-st}F(t)\vert_0^\infty+s\int_0^\infty e^{-st}F(t)dt
\end{align*}
In order to ensure that $\mathcal{L}\{F'(t)\}$ exists, other than requiring that $F'(t)$ is piecewise continuous in every finite interval $[0,T]$ we also need to require that $F(t)$ be of order $e^{\alpha t}$, in terms of the big O notation we write $F(t)=O(e^{\alpha t})$ as $t\to\infty$, meaning $\frac{F(t)}{e^{\alpha t}}$ is bounded for large $t$ i.e. there exists $M>0$ such that $e^{-\alpha t}|F(t)|<M$ for large $t$. With this assumption,
$$|e^{-st}F(t)|<Me^{-(s-\alpha)t}\to 0$$
as $t\to\infty$ provided $s>\alpha$. Consequently, we have

\label{eq:laplace6}
\mathcal{L}\{F'(t)\}=s\mathcal{L}\{F(t)\}-F'(0)

Using \eqref{eq:laplace6}
\begin{align*}
\mathcal{L}\{F^{\prime\prime}(t)\}&=s\mathcal{L}\{F'(t)\}-F'(0)\\
&=s(s\mathcal{L}\{F(t)\}-F'(0))-F'(0)\\
&=s^2\mathcal{L}\{F(t)\}-sF(0)-F'(0)
\end{align*}
Continuing we obtain

\label{eq:laplace7}
\mathcal{L}\{F^{(n)}(t)\}=s^n\mathcal{L}\{F(t)\}-s^{n-1}F(0)-s^{n-2}F'(0)-s^{n-3}F^{\prime\prime}(0)-\cdots -F^{(n-1)}(0)

Some of the transforms we found here can be obtained using \eqref{eq:laplace7} as seen in the next couple of examples below.

Example. Let $F(t)=t$. Then $F'(t)=1$ so \eqref{eq:laplace6} results in $\mathcal{L}\{1\}=s\mathcal{L}\{s\}$ and thereby
$$\mathcal{L}\{t\}=\frac{1}{s^2}$$

Example. Let $F(t)=\sin kt$. Then $F'(t)=k\cos kt$ and $F^{\prime\prime}(t)=-k^2\sin kt$. So using \eqref{eq:laplace7} for $n=2$
$$\mathcal{L}\{F^{\prime\prime}(t)\}=s^2\mathcal{L}\{F(t)\}-sF(0)-F'(0)$$
we have
$$-k^2\mathcal{L}\{\sin kt\}=s^2\mathcal{L}\{\sin kt\}-k$$
that is,
$$\mathcal{L}\{\sin kt\}=\frac{k}{s^2+k^2}$$

Example. Using \eqref{eq:laplace7} we can also prove the formula

\label{eq:laplace8}
\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}

where $n$ is a nonnegative integer. Recall that the formula \eqref{eq:laplace8} appeared here. Let $F(t)=t^n$. Then
\begin{align*}
F(0)&=F'(0)=\cdots=F^{(n-1)}(0)=0\\
F^{(n)}(t)&=n!\\
F^{(n+1)}(t)&=0
\end{align*}
So
$$\mathcal{L}\{F^{(n+1)}(t)\}=s^{n+1}\mathcal{L}\{F(t)\}-s^nF(0)-s^{n-1}F'(0)-\cdots – F^{(n)}(0)$$
reduces to
$$0=s^{n+1}\mathcal{L}\{t^n\}-n!$$
that is,
$$\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$$

Example. For $k>-1$ real,
$$\mathcal{L}\{t^k\}=\int_0^\infty e^{-st}t^kdt$$
with $s>0$. Using the subsitution $x=st$ $\mathcal{L}\{t^k\}$ can be written as

\label{eq:laplace9}
\begin{aligned}
\mathcal{L}\{t^k\}&=\frac{1}{s^{k+1}}\int_0^\infty e^{-x}x^kdx\\
&=\frac{\Gamma(k+1)}{s^{k+1}}
\end{aligned}

\label{eq:gamma}
\Gamma(k+1)=\int_0^\infty e^{-x}x^kdx

is called the Gamma function or factorial function with the argument $k+1$. (There are different ways to define the Gamma function. This definition is due to Euler. For other definitions and for more details please see the reference [1] below.) $\Gamma (k+1)$ is also denoted by $k!$. In fact if $k=n$ is a positive integer,
$$k!=n!=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1$$
When $k=-\frac{1}{2}$ using the substitution $u=\sqrt{x}$ we see that $\Gamma\left(\frac{1}{2}\right)$ is just the Gaussian integral
$$\Gamma\left(\frac{1}{2}\right)=2\int_0^\infty e^{-u^2}du=\sqrt{\pi}$$

References:

[1] Mathematical Methods for Physicists, George Arfken, Third Edition, Academic Press, 1985