# Use the definition of Laplace Transforms to show that: displaystyle{L}{leftlbrace{t}^{n}rightrbrace}=frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},ldots

Question
Laplace transform
Use the definition of Laplace Transforms to show that:
\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}},{n}={1},{2},{3},\ldots\)

2021-02-13
Step 1
Given:
The laplace transformation,
$$\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}$$
Step 2
Proof:
Consider $$L(t^n)$$ and simplify it as follows.
$$\displaystyle{L}{\left\lbrace f{{\left({t}\right)}}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$
$$\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}}{\left({t}^{n}\right)}{\left.{d}{t}\right.}$$
Apply integration by parts, $$\displaystyle{u}{v}'={u}{v}-\int{u}'{v}$$
Here,
$$\displaystyle{u}={t}^{n}{d}{v}={e}^{{-{s}{t}}}$$
$$\displaystyle{d}{u}={n}{t}^{{{n}+{1}}}\ \ \ {v}=\frac{{-{e}^{{-{s}{t}}}}}{{s}}$$
$$\displaystyle{L}{\left\lbrace{t}^{n}\right\rbrace}={{\left[\frac{{-{t}^{n}\cdot{e}^{{-{s}{t}}}}}{{s}}\right]}_{{0}}^{\infty}}-{\int_{{0}}^{\infty}}{n}{t}^{{{n}-{1}}}{\left(\frac{{{e}^{{-{s}{t}}}}}{{s}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle={0}+\frac{n}{{s}}{\int_{{0}}^{\infty}}{t}^{{{n}-{1}}}{\left({e}^{{-{s}{t}}}\right)}{\left.{d}{t}\right.}$$
$$\displaystyle=\frac{n}{{s}}{L}{\left({t}^{{{n}-{1}}}\right)}$$
Step 3
Now substitute the values 1, 2, 3... for n.
When n=1
$$\displaystyle{L}{\left\lbrace{t}^{1}\right\rbrace}=\frac{1}{{s}}{L}{\left({1}\right)}$$
$$\displaystyle=\frac{1}{{s}}{\left(\frac{1}{{s}}\right)}$$
$$\displaystyle=\frac{1}{{s}^{2}}$$
When n=2
$$\displaystyle{L}{\left\lbrace{t}^{2}\right\rbrace}=\frac{2}{{s}}{\left({L}{\left({t}^{1}\right)}\right)}$$
$$\displaystyle=\frac{2}{{s}}{\left(\frac{1}{{s}^{2}}\right)}$$
$$\displaystyle=\frac{{{2}!}}{{s}^{3}}$$
Step 4
When n = 3,
$$\displaystyle{L}{\left\lbrace{t}^{3}\right\rbrace}=\frac{3}{{s}}{L}{\left({t}^{{{3}-{1}}}\right)}$$
$$\displaystyle=\frac{3}{{s}}{\left({L}{\left({t}^{2}\right)}\right)}$$
$$\displaystyle=\frac{3}{{s}}{\left(\frac{2}{{s}^{3}}\right)}$$
$$\displaystyle=\frac{6}{{{s}^{{{3}+{1}}}}}$$
$$\displaystyle=\frac{{{3}!}}{{{s}^{{{3}+{1}}}}}$$
Step 5
Thus, for n = n,
$$\displaystyle{L}{\left({t}^{n}\right)}=\frac{{{n}!}}{{{s}^{{{n}+{1}}}}}$$
Hence it is proved.

### Relevant Questions

Find the inverse Laplace transform $$f{{\left({t}\right)}}={L}^{ -{{1}}}{\left\lbrace{F}{\left({s}\right)}\right\rbrace}$$ of each of the following functions.
$${\left({i}\right)}{F}{\left({s}\right)}=\frac{{{2}{s}+{1}}}{{{s}^{2}-{2}{s}+{1}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}+{2}}}{{{s}^{2}-{3}{s}+{2}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
$${\left({i}{i}{i}\right)}{F}{\left({s}\right)}=\frac{{{3}{s}^{2}+{4}}}{{{\left({s}^{2}+{1}\right)}{\left({s}-{1}\right)}}}$$
Hint – Use Partial Fraction Decomposition and the Table of Laplace Transforms.
Find the laplace transform by definition.
a) $$\displaystyle{L}{\left\lbrace{2}\right\rbrace}$$
b) $$\displaystyle{L}{\left\lbrace{e}^{{{2}{t}}}\right\rbrace}$$
c) $$\displaystyle{L}{\left[{e}^{{-{3}{t}}}\right]}$$
In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative $$\frac{dy}{dt}$$ also appears. Consider the following initial value problem, defined for t > 0:
$$\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{4}{\int_{{0}}^{{t}}}{y}{\left({t}-{w}\right)}{e}^{{-{4}{w}}}{d}{w}={3},{y}{\left({0}\right)}={0}$$
a) Use convolution and Laplace transforms to find the Laplace transform of the solution.
$${Y}{\left({s}\right)}={L}{\left\lbrace{y}{\left({t}\right)}\right)}{\rbrace}-?$$
b) Obtain the solution y(t).
y(t) - ?
Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)
$${L}{\left\lbrace{e}^{{-{t}}}\cdot{e}^{t} \cos{{\left({t}\right)}}\right\rbrace}$$
Use the definition of Laplace Transforms to find $$L\left\{f(t)\right\}$$
$$f(t)=\begin{cases}-1 & 0\leq t <1\\1 & t\geq 1\end{cases}$$
Then rewrite f(t) as a sum of step functions, $$u_c(t)$$, and show that by taking Laplace transforms, this yields the same answer as your direct computation.
Find the Laplace transforms of the following
$${L}{\left\lbrace\frac{{{e}^{{-{6}{t}}} \sinh{{\left({2}{t}\right)}}}}{{{e}^{{-{6}{t}}} \cosh{{\left({2}{t}\right)}}}}\right\rbrace}$$
Part II
29.[Poles] (a) For each of the pole diagrams below:
(i) Describe common features of all functions f(t) whose Laplace transforms have the given pole diagram.
(ii) Write down two examples of such f(t) and F(s).
The diagrams are: $$(1) {1,i,-i}. (2) {-1+4i,-1-4i}. (3) {-1}. (4)$$ The empty diagram.
(b) A mechanical system is discovered during an archaeological dig in Ethiopia. Rather than break it open, the investigators subjected it to a unit impulse. It was found that the motion of the system in response to the unit impulse is given by $$w(t) = u(t)e^{-\frac{t}{2}} \sin(\frac{3t}{2})$$
(i) What is the characteristic polynomial of the system? What is the transfer function W(s)?
(ii) Sketch the pole diagram of the system.
(ii) The team wants to transport this artifact to a museum. They know that vibrations from the truck that moves it result in vibrations of the system. They hope to avoid circular frequencies to which the system response has the greatest amplitude. What frequency should they avoid?
Let f(t) be a function on $$\displaystyle{\left[{0},\infty\right)}$$. The Laplace transform of fis the function F defined by the integral $$\displaystyle{F}{\left({s}\right)}={\int_{{0}}^{\infty}}{e}^{{-{s}{t}}} f{{\left({t}\right)}}{\left.{d}{t}\right.}$$ . Use this definition to determine the Laplace transform of the following function.
$$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}-{t}&{0}<{t}<{1}\\{0}&{1}<{t}\end{matrix}\right.}$$
a)$$f(t)=1+2t$$ b)$$f(t) =\sin \omega t \text{Hint: Use Euler’s relationship, } \sin\omega t = \frac{e^(j\omega t)-e^(-j\omega t)}{2j}$$
c)$$f(t)=\sin(2t)+2\cos(2t)+e^{-t}\sin(2t)$$
$$f(t)=2+2(e^{-t}-1)u_1(t)$$