# The Laplace Transform: Convolution

Definition. The convolution $F\ast G$ of $F(t)$ and $G(t)$ is defined by
$$F(t)\ast G(t)=\int_0^t F(\tau)G(t-\tau)d\tau$$

We introduce Convolution Theorem without a proof. For those who are interested a proof can be found in [1] of the References below.

Theorem. Let $\mathcal{L}\{F(t)\}=f(s)$ and $\mathcal{L}\{G(t)\}=g(s)$. Suppose that $F(t)$ and $G(t)$ are piecewise continuous and are of order $e^{\alpha t}$ as $t\to\infty$. Then $\mathcal{L}\{F(t)\ast G(t)\}$ exists when $s>\alpha$ and it is $f(s)g(s)$. Equivalently,
$$\mathcal{L}^{-1}\{f(s)g(s)\}=F(t)\ast G(t)$$

Example. Let $F(t)=t$ and $G(t)=e^{at}$. Then $\mathcal{L}\{F(t)\}=\frac{1}{s^2}$ and $\mathcal{L}\{G(t)\}=\frac{1}{s-a}$. By Convolution Theorem
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{s^2}\frac{1}{s-a}\right\}&=t\ast e^{at}\\
&=\int_0^t\tau e^{a(t-\tau)}d\tau\\
&=\frac{1}{a^2}(e^{at}-at-1)
\end{align*}
Note that partial fractions can be also used to obtain the result.

When $G(t)=F(t)$, we have the formula
$$[f(s)]^2=\mathcal{L}\{F(t)\ast F(t)\}$$

Example.
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{(s^2+k^2)^2}\right\}&=\frac{1}{k^2}\mathcal{L}^{-1}\left\{\frac{k^2}{(s^2+k^2)^2}\right\}\\
&=\frac{1}{k^2}\int_0^t\sin kt\ast \sin kt\\
&=\frac{1}{k^2}\int_0^t\sin k\tau\sin k(t-\tau)d\tau\\
&=\frac{1}{k^3}\{\sin kt-kt\cos kt\}
\end{align*}

Theorem (Properties of Convolution).

1. $F(t)\ast G(t)=G(t)\ast F(t)$
2. $F(t)\ast[G(t)+H(t)]=F(t)\ast G(t)+F(t)\ast H(t)$
3. $F(t)\ast [kG(t)]=k[F(t)\ast G(t)]$ where $k$ is a constant.
4. $F(t)\ast[G(t)\ast H(t)]=[F(t)\ast G(t)]\ast H(t)$

Proof. We prove only the part 1.
\begin{align*}
F(t)\ast G(t)&=\int_0^t F(\tau)G(t-\tau)d\tau\\
&=\int_0^tF(t-\lambda)G(\lambda)d\lambda\ (\lambda=t-\tau)\\
&=G(t)\ast F(t)
\end{align*}

\begin{align*}
\frac{1}{s}f(s)&=\mathcal{L}\{F(t)\ast 1\}\\
&=\mathcal{L}\left\{\int_0^t F(\tau)d\tau\right\}\\
\frac{1}{s^2}f(s)&=\mathcal{L}\{F(t)\ast 1\ast 1\}\\
&=\mathcal{L}\{(F(t)\ast 1)\ast 1\}\\
&=\mathcal{L}\left\{\int_0^t(F(\tau)\ast 1)d\tau\right\}\\
&=\mathcal{L}\left\{\int_0^t\int_0^{\tau}F(\lambda)d\lambda d\tau\right\}
\end{align*}
Therefore we have the following theorem.

Theorem.
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{s}f(s)\right\}&=\int_0^tF(\tau)d\tau\\
\mathcal{L}^{-1}\left\{\frac{1}{s^2}f(s)\right\}&=\int_0^t\int_0^{\tau}F(\lambda)d\lambda d\tau
\end{align*}

Convolution Theorem can be used for solving non-homogeneous linear differential equations.

Example. Find the general solution of the differential equations
$$\ddot{X}(t)+k^2X(t)=F(t)$$

Solution. Let $\mathcal{L}\{X(t)\}=x(s)$. Then the transformed equation is
$$s^2x(s)-sX(0)-\dot{X}(0)+k^2x(s)=f(s)$$
So we have
$$x(s)=\frac{1}{k}\frac{k}{s^2+k^2}f(s)+\frac{s}{s^2+k^2}X(0)+\frac{1}{k}\frac{k}{s^2+k^2}\dot{X}(0)$$
Therefore by Convolution Theorem the general solution is given by

\begin{aligned}
X(t)&=\frac{1}{k}(\sin kt)\ast F(t)+X(0)\cos kt+\frac{\dot{X}(0)}{k}\sin kt\\
&=\frac{1}{k}\int_0^t\sin k(t-\tau)F(\tau)d\tau+C_1\cos kt+C_2\sin kt
\end{aligned}\label{eq:conv}

Let us redo the problem in the last example in here using \eqref{eq:conv}.

Example. Find the general solution of the differential equation
$$\ddot{x}+4x=8\tan t,\ -\frac{\pi}{2}<x<\frac{\pi}{2}$$

Solution. Using the formula \eqref{eq:conv} with $k=2$
\begin{align*}
x(t)&=\frac{1}{2}\int_0^t\sin 2(t-\tau)(8\tan\tau)d\tau+C_1\cos 2t+C_2\sin 2t\\
&=4\int_0^t\sin 2(t-\tau)\tan\tau d\tau+C_1\cos 2t+C_2\sin 2t\\
&=C_1\cos 2t+C_2\sin 2t-4t\cos 2t+4\sin 2t\ln(\cos t)
\end{align*}

Convolution Theorem can be also used for solving integral equations as shown in the following example.

Example. Solve the integral equation
$$X(t)=at+\int_0^tX(\tau)\sin(t-\tau)d\tau$$

Solution. $X(t)=at+X(t)\ast \sin t$ so its transformed equation is
$$x(s)=\frac{a}{s^2}+x(s)\frac{1}{s^2+1}$$
Hence we have
\begin{align*}
x(s)&=\frac{a}{s^2}\frac{s^2+1}{s^2}\\
&=a\left(\frac{1}{s^2}+\frac{1}{s^4}\right)
\end{align*}
Therefore the solution is given by
$$X(t)=a\left(t+\frac{1}{6}t^3\right)$$

References:

[1] Ruel V. Churchill, Operational Mathematics, McGraw-Hill, 1958

# The Laplace Transform: Further Properties

Let $\mathcal{L}\{F(t)\}=f(s)$ i.e. $f(s)=\int_0^\infty e^{-st}F(t)dt$. Then

\label{eq:laplace11}
e^{-bs}f(s)=\int_0^\infty e^{-s(t+b)}F(t)dt

where $b$ is a positive constant. Using the substitution $\tau=t+b$ \eqref{eq:laplace11} can be written as
$$e^{-bs}f(s)=\int_b^\infty e^{-s\tau}F(\tau-b)d\tau$$
Define a function $F_b(\tau)$ by

F_b(\tau)=\left\{\begin{aligned}
0,\ &0<\tau<b\\
F(\tau-b),\ &\tau>b
\end{aligned}\right.\label{eq:laplace12}

Then
$$e^{-bs}f(s)=\int_0^\infty e^{-s\tau}F_b(tau)d\tau$$

Theorem. If $\mathcal{L}\{F(t)\}=f(s)$ then for any positive constant $b$,
$$e^{-bs}f(s)=\mathcal{L}\{F_b(t)\}$$
where $F_b(t)$ is defined by \eqref{eq:laplace12}.

Let
S_b(t)=\left\{\begin{aligned} 0,\ &0<t<b\\ 1,\ &t>b \end{aligned}\right.
for any positive constant $b$. Define $S_0(t)=1$ for $t>0$. Then $S_b(t)$ is called a \emph{unit step function}. Using this unit step function, $F_b(t)$ can be written as
$$F_b(t)=S_b(t)F(t-b)$$

Example.
\begin{align*}
F_b(t)&=S_b(t)\sin k(t-b)\\
&=\mathcal{L}^{-1}\left\{\frac{ke^{-bs}}{s^2+k^2}\right\}
\end{align*}

\begin{align*}
\mathcal{L}\{F(at)\}&=\int_0^\infty e^{-st}F(at)dt\ (a>0)\\
&=\frac{1}{a}\int_0^\infty e^{-\frac{s}{a}\tau}F(\tau)d\tau\ (\tau=at)\\
&=\frac{1}{a}f\left(\frac{s}{a}\right)
\end{align*}

Theorem. If $\mathcal{L}\{F(t)\}=f(s)$ when $s>\alpha$, then
$$\mathcal{L}\{F(at)\}=\frac{1}{a}f\left(\frac{s}{a}\right)\ (s>a\alpha,\ a>0)$$

Example. Given that
$$\frac{s}{s^2+1}=\mathcal{L}\{\cos t\}$$
\begin{align*}
\frac{s}{s^2+k^2}&=\frac{1}{k^2}\frac{s}{\left(\frac{s}{k}\right)^2+1}\\
&=\frac{1}{k}\frac{\frac{s}{k}}{\left(\frac{s}{k}\right)^2+1}\\
&=\mathcal{L}\{\cos kt\}
\end{align*}

With the translation of transform (equation (1) in here) and the above theorem the effect of a general linear substitution for $s$ can be shown in the following formula.
\begin{align*}
f(as-b)&=f\left[a\left(s-\frac{b}{a}\right)\right]\\
&=\mathcal{L}\left\{\frac{1}{a}e^{\frac{b}{a}t}F\left(\frac{t}{a}\right)\right\}\ (a>0)
\end{align*}

# The Laplace Transform: Solving Differential Equations

In this note we study how the Laplace transform can be used to solve differential equations. For now we only consider linear differential equations with constant coefficients. But later on we will see that the Laplace transform is also a powerful tool to solve linear differential equations with variable coefficients.

Example. Find the general solution of the differential equation
$$\ddot{X}(t)+k^2X(t)=0$$

Solution. Using the linearity transform the differential equation.

\label{eq:laplace10}
\mathcal{L}\{\ddot{X}(t)\}+k^2\mathcal{L}\{X(t)\}=0

By the transform of derivative formula
$$\mathcal{L}\{\ddot{X}(t)\}=s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)$$
Let $X(0)=A$ and $\dot{X}(0)=B$. Then \eqref{eq:laplace10} becomes an algebraic equation of $\mathcal{L}\{X(t)\}$
$$(s^2+k^2)\mathcal{L}\{X(t)\}-As-B=0$$
Thus
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{As+B}{s^2+k^2}\\
&=A\frac{s}{s^2+k^2}+\frac{B}{k}\frac{k}{s^2+k^2}
\end{align*}
Hence,
$$X(t)=A\cos kt+B’\sin kt$$
where $B’=\frac{B}{k}$.

Example. Find the solution of the initial value problem
$$\ddot{X}(t)-\dot{X}(t)-6X(t)=2$$
with $X(0)=1$ and $\dot{X}(0)=0$.

Solution. The transform of the differential equation, with the aid of the transform of derivative formula, is
$$[s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)]-[s\mathcal{L}\{X(t)\}-X(0)]-6\mathcal{L}\{X(t)\}=\frac{2}{s}$$
This simplifies to
\begin{align*}
(s^2-s-6)\mathcal{L}\{X(t)\}&=\frac{2}{s}+s-1\\
&=\frac{s^2-s+2}{s}
\end{align*}
So
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{s^2-s+2}{s(s^2-s-6)}\\
&=\frac{s^2-s+2}{s(s-3)(s+2)}\\
&=\frac{A}{s}+\frac{B}{s-3}+\frac{C}{s+2}
\end{align*}
This results in the equation
$$A(s-3)(s+2)+Bs(s+2)+Cs(s-3)=s^2-s+2$$
For $s=0$, we obtain $-6A=2$ i.e $A=-\frac{1}{3}$. For $s=3$, we obtain $15B=8$ i.e. $B=\frac{15}{8}$. For $s=-2$, we obtain $10C=8$ i.e $C=\frac{4}{5}$.
Hence,
$$\mathcal{L}\{X(t)\}=-\frac{1}{3}\frac{1}{s}+\frac{15}{8}\frac{1}{s-3}+\frac{4}{5}\frac{1}{s+2}$$
and therefore
$$X(t)=-\frac{1}{3}+\frac{15}{8}e^{3t}+\frac{4}{5}e^{-2t}$$

Example. Solve the problem
$$\dddot{X}(t)-2\ddot{X}(t)+5\dot{X}(t)=0$$
with $X(0)=0$, $\dot{X}(0)=1$, and $X\left(\frac{\pi}{8}\right)=1$.

Solution. The transformed equation, with the aid ofthe transform of derivative formula, is
\begin{align*}
[s^3\mathcal{L}\{X(t)\}-s^2X(0)-s\dot{X}(0)-\ddot{X}(0)]&-2[s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)]\\&
+5[s\mathcal{L}\{X(t)\}-X(0)]=0
\end{align*}
This simplifies, with the initial conditions $X(0)=0$ and $\dot{X}(0)=1$, to the equation
$$(s^3-2s^2+5s)\mathcal{L}\{X(t)\}-s+2-\ddot{X}(0)=0$$
Let $\ddot{X}(0)=C$. Then
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{C-2+s}{s^3-2s^2+5s}\\
&=\frac{C-2+s}{s(s^2-2s^+5)}\\
&=\frac{C-2+s}{s[(s-1)^2+4]}\\
&=\frac{C-2}{5}\left[\frac{1}{s}-\frac{s-1}{(s-1)^2+4}\right]+\frac{C+3}{10}\frac{2}{(s-1)^2+4}
\end{align*}
Hence
$$X(t)=\frac{C-2}{5}+e^t\left(\frac{C+3}{10}\sin 2t-\frac{C-2}{5}\cos 2t\right)$$
The condition $X\left(\frac{\pi}{8}\right)=1$ results in the equation
$$1=\frac{C-2}{5}+\frac{e^{\frac{\pi}{8}}}{10\sqrt{2}}(-C+7)$$ whose solution is $C=7$. Therefore,
$$X(t)=1+e^t(\sin 2t-\cos 2t)$$

# The Laplace Transform: The Inverse Transforms

Let $\mathcal{L}^{-1}\{f(s)\}$ denote a function whose Laplace transform is $f(s)$ i.e. if $\mathcal{L}\{F(t)\}=f(s)$ then $F(t)=\mathcal{L}^{-1}\{f(s)\}$. For example,
$$\mathcal{L}^{-1}\left\{\frac{1}{s-k}\right\}=e^{kt},\ \mathcal{L}^{-1}\left\{\frac{k}{s^2+k^2}\right\}=\sin kt$$
$F(t)$ is called the inverse transform of $f(s)$. There is a question that must be addressed. Is the inverse transform of $f(s)$ unique? The answer is not really. For example, we know that $F_1(t)=e^{kt}$ is an inverse transform of $\frac{1}{s-k}$. Define
F_2(t)=\left\{\begin{aligned} e^{kt},\ &0<t<2,\ t>2\\ 1,\ &t=2 \end{aligned}\right.
The transform of $F_2(t)$ is
\begin{align*}
\mathcal{L}\{F_2(t)\}&=\int_0^\infty e^{-st}F_2(t)dt\\
&=\int_0^2 e^{-st}e^{kt}dt+\int_2^\infty e^{-st}e^{kt}dt\\
&=\frac{1}{s-k}
\end{align*}
So are we in trouble then? Not really. We introduce the following theorem without proof.

Theorem. If two functions $F_1(t)$ and $F_2(t)$ have the same Laplace transform, then $F_2(t)=F_1(t)+N(t)$ where $N(t)$ satisfies
$$\int_0^T N(t)dt=0$$
for every positive $T$.

Such a function $N(t)$ is called a null function. In the above example,
N(t)=\left\{\begin{aligned} 1-e^{2k},\ &=t=2\\ 0,\ &t\ne 2 \end{aligned}\right.
In view of this theorem the inverse transform is essentially unique because a null function is not important in the applications. For this reason this theorem is called the uniqueness of the inverse transform.

Due to the linearity of $\mathcal{L}$, we have
\begin{align*}
\mathcal{L}\{AF(t)+BG(t)\}&=A\mathcal{L}\{F(t)\}+B\mathcal{L}\{G(t)\}\\
&=Af(s)+Bg(s)
\end{align*}
Now
\begin{align*}
\mathcal{L}^{-1}\{Af(s)+Bg(s)\}&=AF(t)+BG(t)\\
&=A\mathcal{L}^{-1}\{f(s)\}+B\mathcal{L}^{-1}\{g(s)\}
\end{align*}
so $\mathcal{L}^{-1}$ is also linear.

Suppose that the Laplace transform of $F(t)$ converges when $s>k$.
$$f(s)=\int_0^\infty e^{-st}F(t)dt\ (s>k)$$
Substitute $s-a$ for $s$. Then
\begin{align*}
f(s-a)&=\int_0^\infty e^{-(s-a)t}F(t)dt\\
&=\int_0^\infty e^{-st}e^{at}F(t)dt\\
&=\mathcal{L}\{e^{at}F(t)\}
\end{align*}
Therefore

\mathcal{L}\{e^{at}F(t)\}=f(s-a),\ s>a+k

Example.
$$\mathcal{L}\{e^{at}t^n\}=\frac{n!}{(s-a)^{n+1}},\ s>a$$

Example.
$$\mathcal{L}\{e^{-at}\cos kt\}=\frac{s+a}{(s+a)^2+k^2},\ s>-a$$

Partial fractions often play a useful role in finding $\mathcal{L}^{-1}$ as seen in the following examples.

Example. Find $\mathcal{L}^{-1}\left\{\frac{s+1}{s^2+2s}\right\}$

Solution. $s^2+2s=s(s+1)$ so let
$$\frac{s+1}{s^2+2s}=\frac{A}{s}+\frac{B}{s+2}$$
Then we have
\begin{align*}
s+1&=A(s+2)+Bs\\
&=(A+B)s+2A
\end{align*}
This means $A+B=1$ and $2A=1$ i.e. $A=B=\frac{1}{2}$. Hence
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{s+1}{s^2+2s}\right\}&=\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{2}\right\}+\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}\\
&=\frac{1}{2}+\frac{1}{2}e^{-2t}
\end{align*}

Example. Find $\mathcal{L}^{-1}\left\{\frac{a^2}{s(s+a)^2}\right\}$ ($a\ne 0$).

Solution. Let
$$\frac{a^2}{s(s+a)^2}=\frac{A}{s}+\frac{B}{s+a}+\frac{C}{(s+a)^2}$$
Then we have
$$a^2=A(s+a)^2+Bs(s+a)+Cs$$
For $s=0$, $a^2=Aa^2$ i.e. $A=1$. For $s=-a$, $a^2=-Ca$ i.e. $C=-a$. For $s=a$, $a^2=3a^2+2a^2B$ which is reduced to $-2a^2=2a^2B$ i.e. $B=-1$. Hence
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{a^2}{s(s+a)^2}\right\}&=\mathcal{L}^{-1}\left\{\frac{1}{2}\right\}-\mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\}-a\mathcal{L}^{-1}\left\{\frac{1}{(s+a)^2}\right\}\\
&=1-e^{-at}-ate^{-at}
\end{align*}

Example. Find $\mathcal{L}^{-1}\left\{\frac{s}{(s^2+a^2)(s^2+b^2}\right\}$ ($a^2\ne b^2$).

$$\frac{s}{(s^2+a^2)(s^2+b^2)}=\frac{As+B}{s^2+a^2}+\frac{Cs+D}{s^2+b^2}$$
However one can write
\begin{align*}
\frac{s}{(s^2+a^2)(s^2+b^2)}&=\frac{s}{a^2-b^2}\left[\frac{s^2+a^2-(s^2+b^2)}{(s^2+a^2)(s^2+b^2)}\right]\\
&=\frac{1}{b^2-a^2}\left[\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}\right]
\end{align*}
Hence
$$\mathcal{L}^{-1}\left\{\frac{s}{(s^2+a^2)(s^2+b^2)}\right\}=\frac{1}{b^2-a^2}(\cos at-\cos bt)$$

Example. Find $F(t)$ is $f(s)=\frac{5s+3}{(s-1)(s^2+2s+5)}$.

Solution. Let
$$\frac{5s+3}{(s-1)(s^2+2s+5)}=\frac{A}{s-1}+\frac{Bs+C}{s^2+2s+5}$$
Then we obtain $A=1$, $B=-1$ and $C=2$.
\begin{align*}
f(s)&=\frac{1}{s-1}-\frac{s-2}{s^2+2s+5}\\
&=\frac{1}{s-1}-\frac{s-2}{(s+1)^2+4}\\
&=\frac{1}{s-1}-\frac{s+1-3}{(s+1)^2+2^2}\\
&=\frac{1}{s-1}-\frac{s+1}{(s+1)^2+2^2}+\frac{3}{2}\frac{2}{(s+1)^2+2^2}
\end{align*}
Hence
$$F(t)=e^t-e^{-t}\cos 2t+\frac{3}{2}e^{-t}\sin 2t$$

# 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