Examples

EXAMPLE 10.2.5

Find $F(s) = {\mathcal L} (f(t)),$ where $f(t) = 1, \;\; t \geq 0.$
Solution: $F(s) = \displaystyle\int_0^\infty e^{-s t} dt = \lim\limits_{b \longrightarro... ...aystyle\frac{1}{s} - \lim\limits_{b \longrightarrow \infty} \frac{e^{-sb}}{s}.$
Note that if then

$\displaystyle \lim\limits_{b \longrightarrow \infty} \frac{e^{-sb}}{s} =0.$

Thus,

$\displaystyle F(s) = \frac{1}{s}, \;\; {\mbox{ for }} \;\; s > 0.$

In the remaining part of this chapter, whenever the improper integral is calculated, we will not explicitly write the limiting process. However, the students are advised to provide the details.
Find the Laplace transform of where $f(t) = t, \;\; t \geq 0.$
Solution: Integration by parts gives

$\displaystyle F(s)$ $\displaystyle =$ $\displaystyle \int_0^\infty t e^{-s t} dt = \biggl. \frac{- t e^{-s t}}{s} \biggr\vert _0^\infty + \int_0^\infty \frac{ e^{-s t}}{s} dt$

$\displaystyle =$ $\displaystyle \frac{1}{s^2}\;\; {\mbox{ for }} \;\; s > 0.$
Find the Laplace transform of $f(t) = t^n, \;\; n$ a positive integer.
Solution: Substituting $s t = \tau,$ we get

$\displaystyle F(s)$ $\displaystyle =$ $\displaystyle \int_0^\infty e^{- st} t^n dt$

$\displaystyle =$ $\displaystyle \frac{1}{s^{n+1}} \int_0^\infty e^{-\tau} \tau^n \; d \tau$

$\displaystyle =$ $\displaystyle \frac{n!}{s^{n+1}}\;\; {\mbox{ for }} \;\; s > 0.$
Find the Laplace transform of $f(t) = e^{at}, \; t \geq 0.$
Solution: We have

$\displaystyle {\mathcal L} (e^{at})$ $\displaystyle =$ $\displaystyle \displaystyle \int_0^\infty e^{at} e^{-s t} dt = \int_0^\infty e^{-(s-a) t} dt$

$\displaystyle =$ $\displaystyle \frac{1}{s-a} \;\; {\mbox{ for }} \;\; s > a.$
Compute the Laplace transform of $\cos (a t), \; \; t \geq 0.$
Solution:

$\displaystyle {\mathcal L}(\cos(at))$ $\displaystyle =$ $\displaystyle \int_0^\infty \cos(at) e^{-st} dt$

$\displaystyle =$ $\displaystyle \biggl. \cos(at) \frac{e^{-s t}}{-s} \biggr\vert _0^\infty - \int_0^\infty - a \sin(at) \cdot \frac{e^{-s t}}{-s} dt$

$\displaystyle =$ $\displaystyle \frac{1}{s} - \biggl( \biggl. \frac{a \sin(at)}{s} \frac{e^{-s t}... ..._0^\infty - \int_0^\infty a^2 \frac{\cos(at)}{s} \frac{e^{-s t}}{-s} dt \biggr)$

Note that the limits exist only when Hence,

$\displaystyle \displaystyle \frac{a^2 + s^2}{s^2} \int_0^\infty \cos(at) e^{-st... ...hus }} \;\; {\mathcal L}(\cos(at)) = \frac{s}{a^2 + s^2}; \hspace{.5in} s > 0.$
Similarly, one can show that

$\displaystyle {\mathcal L}(\sin (at)) = \frac{a}{s^2 + a^2}, \;\; s > 0.$
Find the Laplace transform of $f(t) = \displaystyle\frac{1}{\sqrt{t}}, \;\; t > 0.$
Solution: Note that is not a bounded function near (why!). We will still show that the Laplace transform of exists.

$\displaystyle {\mathcal L}(\frac{1}{\sqrt{t}})$ $\displaystyle =$ $\displaystyle \int_0^\infty \frac{1}{\sqrt{t}} e^{-st} dt = \int_0^\infty \frac... ...{\sqrt{\tau}} e^{-\tau} \frac{d\tau}{s} \;\; ({\mbox{ substitute }} \tau = s t)$

$\displaystyle =$ $\displaystyle \frac{1}{\sqrt{s}} \int_0^\infty \tau^{-\frac{1}{2}} e^{-\tau} d \tau = \frac{1}{\sqrt{s}} \int_0^\infty \tau^{\frac{1}{2}-1} e^{-\tau} d \tau.$

Recall that for calculating the integral $\displaystyle \int_0^\infty \tau^{\frac{1}{2}-1} e^{-\tau} d \tau,$ one needs to consider the double integral

$\displaystyle \int_0^\infty \int_0^\infty e^{-(x^2+ y^2)} dx dy = \left( \int_... ...eft( \frac{1}{2} \int_0^\infty \tau^{\frac{1}{2}-1}e^{-\tau} d \tau \right)^2.$

It turns out that

$\displaystyle \displaystyle \int_0^\infty \tau^{\frac{1}{2}-1} e^{-\tau} d \tau = \sqrt{\pi}.$

Thus, $\displaystyle{\mathcal L}(\frac{1}{\sqrt{t}}) = \frac{\sqrt{\pi}}{\sqrt{s}}$ for

We now put the above discussed examples in tabular form as they constantly appear in applications of Laplace transform to differential equations.

$\displaystyle F(s)$	$\displaystyle =$	$\displaystyle \int_0^\infty t e^{-s t} dt = \biggl. \frac{- t e^{-s t}}{s} \biggr\vert _0^\infty + \int_0^\infty \frac{ e^{-s t}}{s} dt$
	$\displaystyle =$	$\displaystyle \frac{1}{s^2}\;\; {\mbox{ for }} \;\; s > 0.$

$\displaystyle F(s)$	$\displaystyle =$	$\displaystyle \int_0^\infty e^{- st} t^n dt$
	$\displaystyle =$	$\displaystyle \frac{1}{s^{n+1}} \int_0^\infty e^{-\tau} \tau^n \; d \tau$
	$\displaystyle =$	$\displaystyle \frac{n!}{s^{n+1}}\;\; {\mbox{ for }} \;\; s > 0.$

$\displaystyle {\mathcal L} (e^{at})$	$\displaystyle =$	$\displaystyle \displaystyle \int_0^\infty e^{at} e^{-s t} dt = \int_0^\infty e^{-(s-a) t} dt$
	$\displaystyle =$	$\displaystyle \frac{1}{s-a} \;\; {\mbox{ for }} \;\; s > a.$

$\displaystyle {\mathcal L}(\cos(at))$	$\displaystyle =$	$\displaystyle \int_0^\infty \cos(at) e^{-st} dt$
	$\displaystyle =$	$\displaystyle \biggl. \cos(at) \frac{e^{-s t}}{-s} \biggr\vert _0^\infty - \int_0^\infty - a \sin(at) \cdot \frac{e^{-s t}}{-s} dt$
	$\displaystyle =$	$\displaystyle \frac{1}{s} - \biggl( \biggl. \frac{a \sin(at)}{s} \frac{e^{-s t}... ..._0^\infty - \int_0^\infty a^2 \frac{\cos(at)}{s} \frac{e^{-s t}}{-s} dt \biggr)$

$\displaystyle {\mathcal L}(\frac{1}{\sqrt{t}})$	$\displaystyle =$	$\displaystyle \int_0^\infty \frac{1}{\sqrt{t}} e^{-st} dt = \int_0^\infty \frac... ...{\sqrt{\tau}} e^{-\tau} \frac{d\tau}{s} \;\; ({\mbox{ substitute }} \tau = s t)$
	$\displaystyle =$	$\displaystyle \frac{1}{\sqrt{s}} \int_0^\infty \tau^{-\frac{1}{2}} e^{-\tau} d \tau = \frac{1}{\sqrt{s}} \int_0^\infty \tau^{\frac{1}{2}-1} e^{-\tau} d \tau.$