Find the Laplace transform of f(t)=t^{3} using two methods (definition and table of common transforms).

## 1. Apply the Laplace trasnform definition

Let F(s) be the Laplace transform of f(t) which is defined as follows:

\displaystyle F(s)=\int_{0}^{\infty}e^{-st}t^{3}dt

i.e.,

\displaystyle F(s)=\lim_{b\to\infty}\int_{0}^{b}e^{-st}t^{3}dt

## 2. Integration by parts I

To evaluate the integral above, we will apply the integration by parts:

\displaystyle\int udv=uv-\int v du

Let u=t^{3}, and dv=e^{-st}dt then:

du=3t^{2}dt, \displaystyle v=-\frac{e^{-st}}{s}

and so:

\begin{aligned} & \int_{0}^{b}e^{-st}t^{3}dt=\left.t^{3}\cdot\left(-\frac{e^{-st}}{s}\right)\right|_{t=0}^{t=b}-\int_{0}^{b}-\frac{e^{-st}}{s}\cdot3t^{2}dt \\ & \int_{0}^{b}e^{-st}t^{3}dt=\left.t^{3}\cdot\left(-\frac{e^{-st}}{s}\right)\right|_{t=0}^{t=b}+\frac{3}{s}\int_{0}^{b}e^{-st}t^{2}dt \end{aligned}

## 3. Integration by parts II

Let us denote the second integral as I. To calculate it, we will integrate by parts again: let u=t^{2} and dv=e^{-st}dt, thus

du=2t\:dt and \displaystyle v=-\frac{e^{-st}}{s}

Hence

\begin{aligned} & I=\int_{0}^{b}e^{-st}t^{2}dt=\left.t^{2}\cdot\left(-\frac{e^{-st}}{s}\right)\right|_{t=0}^{t=b}-\int_{0}^{b}-\frac{e^{-st}}{s}2tdt \\ & I=\int_{0}^{b}e^{-st}t^{2}dt=\left.t^{2}\cdot\left(-\frac{e^{-st}}{s}\right)\right|_{t=0}^{t=b}+\frac{2}{s}\int_{0}^{b}e^{-st}tdt \end{aligned}

## 4. Integration by parts III

To the integral

\displaystyle II=\int_{0}^{b}e^{-st}tdt

as we apply the integration by parts, we get

\displaystyle II=\int_{0}^{b}e^{-st}tdt=\left.t\cdot\left(-\frac{e^{-st}}{s}\right)\right|_{t=0}^{t=b}-\int_{0}^{b}-\frac{e^{-st}}{s}dt

in which we did u=t\Rightarrow du=dt and again

dv=e^{-st}dt\Rightarrow v=-\frac{e^{-st}}{s}

Thus,

\displaystyle II=-b\cdot\frac{e^{-sb}}{s}+0\cdot\frac{e^{-s\cdot0}}{s}+\frac{1}{s}\int_{0}^{b}e^{-st}dt

## 5. Simplify the result

Since

\begin{aligned} \int_{0}^{b}e^{-st}dt & = -\left.\frac{e^{-st}}{s}\right|_{t=0}^{t=b} \\ & = -\frac{e^{-sb}}{s}+\frac{e^{-s\cdot0}}{s} \\ & = -\frac{e^{-sb}}{s}+\frac{1}{s} \end{aligned}

therefore

\begin{aligned} II & =-b\cdot\frac{e^{-sb}}{s}+\frac{1}{s}\left(-\frac{e^{-sb}}{s}+\frac{1}{s}\right) \\ & =-b\cdot\frac{e^{-sb}}{s}-\frac{e^{-sb}}{s^{2}}+\frac{1}{s^{2}} \end{aligned}

which gives us

\begin{aligned} I &=-b^{2}\cdot\frac{e^{-sb}}{s}+0^{2}\cdot\frac{e^{-s\cdot0}}{s}+\frac{2}{s}\cdot II \\ &=-b^{2}\cdot\frac{e^{-sb}}{s}+\frac{2}{s}\left(-b\cdot\frac{e^{-sb}}{s}-\frac{e^{-sb}}{s^{2}}+\frac{1}{s^{2}}\right) \\ &=-b^{2}\cdot\frac{e^{-sb}}{s}-\frac{2be^{-sb}}{s^{2}}-\frac{2e^{-sb}}{s^{3}}+\frac{2}{s^{3}} \end{aligned}

Note that

\begin{aligned} \int_{0}^{b}e^{-st}t^{3}dt & = -\frac{b^{3}e^{-sb}}{s}+\frac{0^{3}e^{-s\cdot0}}{s}+\frac{3}{s}I \\ \int_{0}^{b}e^{-st}t^{3}dt & = -\frac{b^{3}e^{-sb}}{s}+\frac{3}{s}I \end{aligned}

## 6. Calculate the limit

Thus, as we calculate the limit in both sides as b goes to \infty, we get

F(s)=-\lim_{b\to\infty}\frac{b^{3}e^{-sb}}{s}+\frac{3}{s}\lim_{b\to\infty}I

Now, let us evaluate \displaystyle\lim_{b\to\infty}I:

\displaystyle \lim_{b\to\infty}I =-\frac{1}{s}\cdot\lim_{b\to\infty}b^{2}e^{-sb}-\frac{2}{s^{2}}\cdot\lim_{b\to\infty}be^{-sb}-\frac{2}{s^{3}}\cdot\lim_{b\to\infty}e^{-sb}+\frac{2}{s^{3}}

To evaluate the limits above, we use the L’Hopital’s rule:

\begin{aligned} \lim_{b\to\infty}b^{2}e^{-sb} & =\lim_{b\to\infty}\frac{b^{2}}{e^{sb}} \\ & =\lim_{b\to\infty}\frac{2b}{se^{sb}} \\ & =\lim_{b\to\infty}\frac{2}{e^{sb}(1+s^{2})} & =0 \end{aligned}

for s>0. Proceeding in the same way

\begin{aligned} \lim_{b\to\infty}be^{-sb} & =\lim_{b\to\infty}\frac{b}{e^{sb}} \\ & =\lim_{b\to\infty}\frac{1}{se^{sb}} \end{aligned}

which is equal to zero if s>0. And since

\displaystyle\lim_{b\to\infty}e^{-sb}=0

for positive values of s, we may conclude that

\displaystyle \lim_{b\to\infty}I=\frac{2}{s^{3}}

From L’Hopital’s rule, we may conclude that

\displaystyle \lim_{b\to\infty}\frac{b^{3}e^{-sb}}{s}=0

Hence

\displaystyle F(s)=\frac{3}{s}\cdot\frac{2}{s^{3}}

## 7. Result

The Laplace transform of f(t)=t^{3} using the definition is given by:

\displaystyle F(s)=\frac{6}{s^{4}}

## Method 02: Using the table of common Laplace transforms

From the second line in the table of Laplace transforms, we have:

\displaystyle \mathscr{L}\{t^{n}\}=\frac{n!}{s^{n+1}}

It is clear in our case n=3. Hence:

\displaystyle \begin{aligned} & \mathscr{L}\{t^{3}\}=\frac{3!}{s^{3+1}} \ \Rightarrow & \mathscr{L}\{t^{3}\}=\frac{6}{s^{4}} \end{aligned}

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