## Question 3

Find, from first principles, $\mathcal{L}\{t^2\}$.

### Solution

Using the definition:

$\mathcal{L}\{t^2\}=\int_0^\infty t^2e^{-st}\,dt$

To ease the steps, first we will evaluate the indefinite integral:

$I= \int t^2e^{-st}\,dt$.

This integral needs to be integrated by parts (this hint would be proffered in an exam situation). Using the LIATE rule, choose $u=t^2$ and $dv=e^{-st}\,dt$. Now

$\frac{du}{dt}=2t\Rightarrow du=2t\,dt$, and

$v=\int dv=\int e^{-st}\,dt=\frac{e^{-st}}{-s}=-\frac{1}{s}e^{-st}$.

In this context $-1/s$ is a constant and it will be handy to have it out the front so it can be easily taken out of an integral ($\int kf(x)=k\int f(x)$.) Hence using the integration by parts formula:

$I=t^2\left(-\frac{1}{s}e^{-st}\right)-\int \left(-\frac{1}{s}e^{-st}2t\,dt\right)=-\frac{1}{s}t^2e^{-st}+\frac{2}{s}\underbrace{\int te^{-st}}_{=:J}$.

Now this second integral will also need integration by parts. By LIATE choose $u=t$ and $dv=e^{-st}\,dt$. Hence

$\frac{du}{dt}=1\Rightarrow du=dt$, and

$v=\int dv=\int e^{-st}\,dt=\frac{e^{-st}}{-s}=-\frac{1}{s}e^{-st}$.

Using the integration by parts formula:

$J=t\left(-\frac{1}{s}e^{-st}\right)-\int\left(-\frac{1}{s}e^{-st}\right)\,dt=-\frac{1}{s}te^{-st}+\frac{1}{s}\int e^{-st}\,dt$

$=-\frac{1}{s}te^{-st}+\frac{1}{s}\left(\frac{e^{-st}}{-s}\right)$.

Therefore,

$I=-\frac{1}{s}t^2e^{-st}+\frac{2}{s}J=-\frac{1}{s}t^2e^{-st}-\frac{2}{s^2}te^{-st}-\frac{2}{s^3}e^{-st}$.

Now putting in the limits $x=0$ to $x= R$:

$\left[I\right]_0^R=\left(-\frac{1}{s}R^2e^{-sR}-\frac{2}{s^2}Re^{-sR}-\frac{2}{s^3}e^{-sR}\right)-$

$\left(-\frac{1}{s}(0)^2e^{-s(0)}-\frac{2}{s^2}(0)e^{s(0)}-\frac{2}{s^3}e^{-s(0)}\right)$

Now three algebraic facts:

1. $e^{-sR}=\frac{1}{e^{sR}}$
2. For any real (or complex) number $x$$x\times 0=0$
3. $e^{-s(0)}=e^{-0}=e^{0}=1$

This yields:

$[I]_0^R=\frac{2}{s^3}-\frac{1}{s}\frac{R^2}{e^{sR}}-\frac{2}{s^2}\frac{R}{e^{sR}}-\frac{2}{s^3}\frac{1}{e^{sR}}$.

Now we have an analysis fact:

Suppose $a,b>0$. Then $e^{ax}$ grows much faster than any $x^b$ so that $\lim_{x\rightarrow \infty}\frac{x^b}{e^{ax}}=0$

(Alternatively if we demonstrate that we know what we are doing we can simply state $e^{-sR}$$Re^{-sR}$ and $R^2e^{-sR}$ tend to zero). Taking the limit as $R\rightarrow \infty$:

$\lim_{R\rightarrow\infty}[I]_0^R=\lim_{R\rightarrow \infty}\left(\frac{2}{s^3}-\frac{1}{s}\underbrace{\frac{R^2}{e^{sR}}}_{\rightarrow 0}-\frac{2}{s^2}\underbrace{\frac{R}{e^{sR}}}_{\rightarrow 0}-\frac{2}{s^3}\underbrace{\frac{1}{e^{sR}}}_{\rightarrow 0}\right)$.

In other words,

$\mathcal{L}\{t^2\}=\frac{2}{s^3}$,

as we would expect, looking at the tables.