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 xx\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.

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