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7.4 Operational Properties II: Class Notes

7.4a

Derivatives of Transforms

The formulas we saw before $ℒ\{y\prime \}=sY(s)-y(0)$ and $ℒ\{y″\}=s^{2}Y(s)-sy(0)-y\prime (0)$ are transforms of derivatives. Now we will consider derivatives of transforms.

Consider ${\displaystyle F(s)={\int }_0^{\infty }{e}^{-st}f(t)dt}$. Differentiating with respect to $s$ yields

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{d}{ds}[F(s)]}& {\displaystyle =\frac{d}{ds}[{\int }_0^{\infty }{e}^{-st}f(t)dt]}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={\int }_0^{\infty }\frac{\partial }{\partial s}\left(e^{-st}\right)f(t)dt=-{\int }_0^{\infty }{e}^{-st}tf(t)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle =-ℒ\left\{tf(t)\right\}}\hfill \end{array}}$$

The derivative of $e^{-st}$ with respect to $s$ is $-t{e}^{-st}$, which brings out the minus sign and the extra factor $t$.

${\displaystyle ℒ\{tf(t)\}=-\frac{d}{ds}[F(s)]}$

Example

Evaluate the given Laplace transform.

$ℒ\left\{t\sin t\right\}$

Solution

The factor $t$ means a negative derivative of the transform of $\sin t$:

$${\displaystyle ℒ\{t\sin t\}=-\frac{d}{ds}\left(\frac{1}{s^2+1}\right)=\frac{2s}{(s^2+1{)}^2}}$$

$ℒ\left\{t{e}^{-3t}\sin t\right\}$

Solution

Now $e^{-3t}$ is a shift (replace $s$ by $s+3$ in the transform of $\sin t$), and $t$ is a negative derivative:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{t{e}^{-3t}\sin t\}}& {\displaystyle =-\frac{d}{ds}\left[\frac{1}{{(s+3)}^2+1}\right]=-\frac{d}{ds}{[(s+3{)}^2+1]}^{-1}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{2(s+3)}{{[(s+3{)}^2+1]}^2}}\hfill \end{array}}$$

Two factors of $t$. Apply the rule twice:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{t^{2}f(t)\}=ℒ\{t·tf(t)\}}& {\displaystyle =-\frac{d}{ds}\left[ℒ\{tf(t)\}\right]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{d^2}{d{s}^2}[F(s)]}\hfill \end{array}}$$

(Each factor of $t$ brings one derivative and one minus sign; the two minus signs cancel.)

In general, we have

Theorem 7.4.1: Derivatives of Transforms

If $F(s)=ℒ\{f(t)\}$ and $n=1,2,3,\dots$, then ${\displaystyle ℒ\{t^{n}f(t)\}=(-1{)}^n\frac{d^n}{d{s}^n}F(s)}$.

Example

Use Theorem 7.4.1 to evaluate the given Laplace transform.

$ℒ\{t^2\cos t\}$

Solution

Here $n=2$ and ${\displaystyle F(s)=\frac{s}{s^2+1}}$. The first derivative (quotient rule) is ${\displaystyle \frac{1-s^2}{(s^2+1{)}^2}}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{t^2\cos t\}}& {\displaystyle =(-1{)}^2\frac{d^2}{d{s}^2}\left(\frac{s}{s^2+1}\right)=\frac{d}{ds}\left(\frac{1-s^2}{(s^2+1{)}^2}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{-2s(s^2+1){}^2+(s^2-1)·2(s^2+1)·2s}{(s^2+1{)}^4}}\hfill \end{array}}$$

Cancel one factor of $s^2+1$ from every term, so the denominator becomes $(s^2+1{)}^3$:

$${\displaystyle =\frac{-2s(s^2+1)+4s(s^2-1)}{(s^2+1{)}^3}=\frac{2s^3-6s}{(s^2+1{)}^3}}$$

Example

Use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix C as needed.

$$y\prime -y=t{e}^t\sin t,y(0)=0$$

Solution

The right side is the previous kind of transform, with the shift $s-1$ from $e^t$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle sY(s)-Y(s)}& {\displaystyle =-\frac{d}{ds}\left[\frac{1}{{(s-1)}^2+1}\right]=\frac{2(s-1)}{{[(s-1{)}^2+1]}^2}}\hfill \\ \hfill {\displaystyle (s-1)Y(s)}& {\displaystyle =\frac{2(s-1)}{{[(s-1{)}^2+1]}^2}}\hfill \\ \hfill {\displaystyle Y(s)}& {\displaystyle =\frac{2}{{[{(s-1)}^2+1]}^2}}\hfill \end{array}}$$

Entry 25 of the table in Appendix C (with $k=1$) gives ${\displaystyle ℒ\{\sin t-t\cos t\}=\frac{2}{(s^2+1{)}^2}}$. Here s is shifted to $s-1$, so multiply by $e^t$:

7.4b

$$y″+y=f(t),y(0)=1,y\prime (0)=0,\text{where }f(t)=\{\begin{array}{cc}{\displaystyle 1,}\hfill & {\displaystyle 0\le t<\pi /2}\hfill \\ {\displaystyle \sin t,}\hfill & {\displaystyle t\ge \pi /2}\hfill \end{array}$$

Solution

Write $f(t)$ with unit step functions, as a sum of two pieces:

$$\begin{array}{cc}\hfill {\displaystyle f(t)}& {\displaystyle =\{\begin{array}{cc}{\displaystyle 1,}\hfill & {\displaystyle 0\le t<\pi /2}\hfill \\ {\displaystyle 0,}\hfill & {\displaystyle t\ge \pi /2}\hfill \end{array}+\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<\pi /2}\hfill \\ {\displaystyle \sin t,}\hfill & {\displaystyle t\ge \pi /2}\hfill \end{array}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =1-𝒰(t-\pi /2)+\sin t𝒰(t-\pi /2)}\hfill \end{array}$$

$$y″+y=1-(1-\sin t)𝒰(t-\pi /2)$$

The factor $1-\sin t$ is not written in terms of $t-\pi /2$, so use the alternative form of the second translation theorem: replace $t$ by $t+\pi /2$.

$${\displaystyle s^{2}Y(s)-s+Y(s)=\frac{1}{s}-e^{-\pi s/2}ℒ\{1-\sin (t+\pi /2)\}}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle (s^2+1)Y(s)}& {\displaystyle =s+\frac{1}{s}-e^{-\pi s/2}(\frac{1}{s}-\frac{s}{s^2+1})}\hfill \end{array}}$$

Since ${\displaystyle s+\frac{1}{s}=\frac{s^2+1}{s}}$, dividing by $s^2+1$ gives

$${\displaystyle Y(s)=\frac{1}{s}-e^{-\pi s/2}[\frac{1}{s(s^2+1)}-\frac{s}{(s^2+1{)}^2}]}$$

First piece. Partial fractions:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{1}{s(s^2+1)}=\frac{A}{s}+\frac{Bs+C}{s^2+1}}& {\displaystyle \Rightarrow 1=A(s^2+1)+(Bs+C)s}\hfill \\ \hfill {\displaystyle 1=A{s}^2+A+B{s}^2+Cs}& {\displaystyle \Rightarrow 1=(A+B)s^2+Cs+A}\hfill \end{array}}$$

$s^2$ terms: $A+B=0\Rightarrow B=-1$. $s$ terms: $C=0$. Constant: $A=1$. So ${\displaystyle \frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}}$.

Second piece. Partial fractions ${\displaystyle \frac{As+B}{s^2+1}+\frac{Cs+D}{(s^2+1{)}^2}}$ just gives the term back (with $C=1$ and the other constants 0), so instead use ${\displaystyle ℒ\{tf(t)\}=-\frac{d}{ds}[F(s)]}$, that is, ${\displaystyle tf(t)={ℒ}^{-1}\{-\frac{d}{ds}[F(s)]\}}$. Since

$${\displaystyle -\frac{1}{2}\int \frac{2s}{(s^2+1{)}^2}ds=\frac{1/2}{s^2+1},\text{we have}\frac{s}{(s^2+1{)}^2}=-\frac{d}{ds}\left(\frac{1/2}{s^2+1}\right)}$$

and ${\displaystyle \frac{1/2}{s^2+1}}$ is the transform of ${\displaystyle \frac{1}{2}\sin t}$, so ${\displaystyle {ℒ}^{-1}\left\{\frac{s}{(s^2+1{)}^2}\right\}=\frac{1}{2}t\sin t}$.

$${\displaystyle Y(s)=\frac{1}{s}-e^{-\pi s/2}[\frac{1}{s}-\frac{s}{s^2+1}+\frac{d}{ds}\left(\frac{1/2}{s^2+1}\right)]}$$

Use the inverse form of the second translation theorem: replace $t$ by $t-\pi /2$.

$${\displaystyle y(t)=1-𝒰(t-\pi /2)[1-\cos (t-\pi /2)-\frac{1}{2}(t-\pi /2)\sin (t-\pi /2)]}$$

Simplify with $\cos (t-\pi /2)=\sin t$ and $\sin (t-\pi /2)=-\cos t$:

7.4c

Convolution

Definition

The convolution of $f$ and $g$ is ${\displaystyle f\ast g={\int }_0^{t}f(\tau )g(t-\tau )d\tau }$

($\tau$ is the Greek letter tau.) The result is a function of $t$. It can be shown that $f\ast g=g\ast f$.

Tip: Don’t confuse convolution with composition, $(f\circ g)(x)=f(g(x))$, the composition of $f$ with $g$.

Example

Find $\cos 2t\ast e^t$.

Solution

With $f(t)=\cos 2t$ and $g(t)=e^t$, $g(t-\tau )=e^{t-\tau }$:

$${\displaystyle \cos 2t\ast e^t={\int }_0^t\cos 2\tau e^{t-\tau }d\tau =I}$$

Integrate by parts with $u=\cos 2\tau ,dv=e^{t-\tau }d\tau$, so $du=-2\sin 2\tau d\tau ,v=-e^{t-\tau }$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle I}& {\displaystyle ={-e^{t-\tau }\cos 2\tau |}_0^t+{\int }_0^t(-2\sin 2\tau )e^{t-\tau }d\tau }\hfill \\ \hfill {\displaystyle }& {\displaystyle =-\cos 2t+e^t+{\int }_0^t(-2\sin 2\tau )e^{t-\tau }d\tau }\hfill \end{array}}$$

Integrate by parts with $u=-2\sin 2\tau ,dv=e^{t-\tau }d\tau$, so $du=-4\cos 2\tau d\tau ,v=-e^{t-\tau }$.

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle I}& {\displaystyle =-\cos 2t+e^t+{2\sin 2\tau e^{t-\tau }|}_0^t-4{\int }_0^t\cos 2\tau e^{t-\tau }d\tau }\hfill \\ \hfill {\displaystyle }& {\displaystyle =-\cos 2t+e^t+2\sin 2t-4I}\hfill \end{array}}$$

The last integral is $I$ again. Solve for $I$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 5I}& {\displaystyle =e^t-\cos 2t+2\sin 2t}\hfill \\ \hfill {\displaystyle I}& {\displaystyle =\frac{1}{5}{e}^t-\frac{1}{5}\cos 2t+\frac{2}{5}\sin 2t}\hfill \end{array}}$$

Example

Find $ℒ\{\cos 2t\ast e^t\}$.

Solution

Use the convolution from the previous example:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{\cos 2t\ast e^t\}}& {\displaystyle =ℒ\{\frac{1}{5}{e}^t-\frac{1}{5}\cos 2t+\frac{2}{5}\sin 2t\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{5}(\frac{1}{s-1}-\frac{s}{s^2+4}+\frac{4}{s^2+4})}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{5}·\frac{s^2+4+(-s+4)(s-1)}{(s-1)(s^2+4)}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{5}·\frac{5s}{(s-1)(s^2+4)}=\frac{s}{s^2+4}·\frac{1}{s-1}}\hfill \end{array}}$$

Notice: $ℒ\{\cos 2t\ast e^t\}=ℒ\{\cos 2t\}ℒ\left\{e^t\right\}$.

In general,

Theorem: Convolution Theorem

If $f(t)$ and $g(t)$ are piecewise continuous on $[0,\infty )$ and of exponential order, then $ℒ\{f\ast g\}=ℒ\{f(t)\}ℒ\{g(t)\}=F(s)G(s)$.

Proof

$${\displaystyle ℒ\left\{f\ast g\right\}={\int }_0^{\infty }{e}^{-st}{\int }_0^{t}f(\tau )g(t-\tau )d\tau dt}$$

Note that $𝒰(t-\tau )=\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<\tau }\hfill \\ {\displaystyle 1,}\hfill & {\displaystyle t\ge \tau }\hfill \end{array}$, so $𝒰(t-\tau )=0$ if $\tau >t$. Multiplying by $𝒰(t-\tau )$ lets the inner integral run to $\infty$ (the extra part, ${\displaystyle {\int }_t^{\infty }}$, is zero):

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{f\ast g\}}& {\displaystyle ={\int }_0^{\infty }{e}^{-st}[{\int }_0^{\infty }f(\tau )g(t-\tau )𝒰(t-\tau )d\tau ]dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle ={\int }_0^{\infty }f(\tau )\left[{\int }_0^{\infty }{e}^{-st}g(t-\tau )𝒰(t-\tau )dt\right]d\tau }\hfill \end{array}}$$

After switching the order of integration, the inner integral is $ℒ\{g(t-\tau )𝒰(t-\tau )\}$ = $e^{-s\tau }G(s)$ by the second translation theorem:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle }& {\displaystyle ={\int }_0^{\infty }f(\tau )e^{-s\tau }G(s)d\tau }\hfill \\ \hfill {\displaystyle }& {\displaystyle =G(s){\int }_0^{\infty }f(\tau )e^{-s\tau }d\tau =G(s)F(s)✓}\hfill \end{array}}$$

Example

Find the Laplace transform of $f\ast g$ using the Convolution Theorem. Do not evaluate the convolution integral before transforming.

$ℒ\{t^3\ast t\sin 3t\}$

Solution

For $t\sin 3t$, use ${\displaystyle -\frac{d}{ds}[F(s)]}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{t^3\ast t\sin 3t\}}& {\displaystyle =ℒ\left\{t^3\right\}ℒ\{t\sin 3t\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{6}{s^4}\left(-\frac{d}{ds}\left(\frac{3}{s^2+9}\right)\right)=\frac{6}{s^4}·\frac{6s}{(s^2+9{)}^2}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{36}{s^3(s^2+9{)}^2}}\hfill \end{array}}$$

${\displaystyle ℒ\{{\int }_0^t\sin \tau \cos (t-\tau )d\tau \}}$

Solution

The integral is $\sin t\ast \cos t$ (compare with the definition ${\displaystyle f\ast g={\int }_0^{t}f(\tau )g(t-\tau )d\tau }$):

$${\displaystyle ℒ\{\sin t\ast \cos t\}=ℒ\{\sin t\}ℒ\{\cos t\}=\frac{1}{s^2+1}·\frac{s}{s^2+1}=\frac{s}{(s^2+1{)}^2}}$$

7.4d

${\displaystyle ℒ\{{\int }_0^t{e}^{-\tau }\cos \tau d\tau \}}$

Solution

Here $f(\tau )=e^{-\tau }\cos \tau$ and $g(t-\tau )=1$, so the integral is $e^{-t}\cos t\ast 1$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{e^{-t}\cos t\ast 1\}}& {\displaystyle =ℒ\{e^{-t}\cos t\}ℒ\left\{1\right\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{s+1}{(s+1{)}^2+1}·\frac{1}{s}=\frac{s+1}{s[(s+1{)}^2+1]}}\hfill \end{array}}$$

The transform of an integral

When $g(t)=1$, $ℒ\{g(t)\}=G(s)=1/s$, the convolution theorem gives

${\displaystyle ℒ\{{\int }_0^{t}f(\tau )d\tau \}=\frac{F(s)}{s}}$, which in inverse form is ${\displaystyle {\int }_0^{t}f(\tau )d\tau ={ℒ}^{-1}\left\{\frac{F(s)}{s}\right\}}$.

That is, ${\displaystyle {\int }_0^{t}f(\tau )d\tau =f\ast 1}$: dividing a transform $F(s)$ by $s$ integrates $f$ from 0 to $t$.

Example

Evaluate the given inverse transform.

${\displaystyle {ℒ}^{-1}\left\{\frac{1}{s(s-a{)}^2}\right\}}$

Solution

No partial fractions are needed. Write it as ${\displaystyle \frac{1}{s}}$ times ${\displaystyle F(s)=\frac{1}{(s-a{)}^2}}$, and ${\displaystyle ℒ\left\{t{e}^{at}\right\}=\frac{1}{(s-a{)}^2}}$, so $f(\tau )=\tau e^{a\tau }$:

$${\displaystyle {ℒ}^{-1}\{\frac{1}{s}·\frac{1}{(s-a{)}^2}\}={\int }_0^t\tau e^{a\tau }d\tau }$$

Integrate by parts with $u=\tau$, $dv=e^{a\tau }d\tau$, so $du=d\tau$, ${\displaystyle v=\frac{1}{a}{e}^{a\tau }}$:

$${\displaystyle ={(\frac{\tau }{a}{e}^{a\tau }-\frac{1}{a^2}{e}^{a\tau })|}_0^t=\frac{1}{a}t{e}^{at}-\frac{1}{a^2}{e}^{at}+\frac{1}{a^2}}$$

Example

Use the Laplace transform to solve the given integral equation.

$${\displaystyle f(t)+2{\int }_0^{t}f(\tau )\cos (t-\tau )d\tau =4e^{-t}+\sin t}$$

Solution

The integral is $f\ast \cos t$, so its transform is ${\displaystyle F(s)·\frac{s}{s^2+1}}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle F(s)+2F(s)\frac{s}{s^2+1}}& {\displaystyle =\frac{4}{s+1}+\frac{1}{s^2+1}}\hfill \\ \hfill {\displaystyle F(s)(1+\frac{2s}{s^2+1})}& {\displaystyle =\frac{4(s^2+1)+(s+1)}{(s+1)(s^2+1)}}\hfill \\ \hfill {\displaystyle F(s)·\frac{s^2+2s+1}{s^2+1}}& {\displaystyle =\frac{4s^2+s+5}{(s+1)(s^2+1)}}\hfill \\ \hfill {\displaystyle F(s)}& {\displaystyle =\frac{4s^2+s+5}{(s+1{)}^3}}\hfill \end{array}}$$

Partial fractions, with a repeated linear factor:

$${\displaystyle \frac{4s^2+s+5}{(s+1{)}^3}=\frac{A}{s+1}+\frac{B}{(s+1{)}^2}+\frac{C}{(s+1{)}^3}\Rightarrow 4s^2+s+5=A(s+1{)}^2+B(s+1)+C}$$

$s=-1$: $C=4-1+5=8$. $s^2$: $A=4$. $s$: $2A+B=1$, so $B=-7$.

$${\displaystyle F(s)=\frac{4}{s+1}-\frac{7}{(s+1{)}^2}+\frac{8}{(s+1{)}^3}=\frac{4}{s+1}-\frac{7}{(s+1{)}^2}+4·\frac{2!}{(s+1{)}^3}}$$

Example

Use the Laplace transform to solve the given integrodifferential equation.

$${\displaystyle \frac{dy}{dt}+6y(t)+9{\int }_0^{t}y(\tau )d\tau =1,y(0)=0}$$

Solution

The integral is $1\ast y(t)$, so its transform is ${\displaystyle \frac{1}{s}Y(s)}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle sY(s)+6Y(s)+\frac{9}{s}Y(s)}& {\displaystyle =\frac{1}{s}}\hfill \\ \hfill {\displaystyle s^{2}Y(s)+6sY(s)+9Y(s)}& {\displaystyle =1}\hfill \\ \hfill {\displaystyle (s^2+6s+9)Y(s)=1\Rightarrow Y(s)}& {\displaystyle =\frac{1}{(s+3{)}^2}}\hfill \end{array}}$$

7.4e

Theorem: Transform of a Periodic Function

If $f(t)$ is piecewise continuous on $[0,\infty )$, of exponential order, and periodic with period $T$, then ${\displaystyle ℒ\{f(t)\}=\frac{1}{1-e^{-sT}}{\int }_0^T{e}^{-st}f(t)dt}$.

Proof

$${\displaystyle ℒ\{f(t)\}={\int }_0^T{e}^{-st}f(t)dt+{\int }_T^{\infty }{e}^{-st}f(t)dt}$$

In the second integral, let $t=u+T$, so $dt=du$ and $u=t-T$. The limits change: $t=T$ becomes $u=0$, and $t=\infty$ stays $u=\infty$. Since $f$ is periodic, $f(u+T)=f(u)$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {\int }_T^{\infty }{e}^{-st}f(t)dt}& {\displaystyle ={\int }_0^{\infty }{e}^{-s(u+T)}f(u+T)du}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-sT}{\int }_0^{\infty }{e}^{-su}f(u)du=e^{-sT}F(s)}\hfill \end{array}}$$

Substitute back, and write $F(s)$ as $ℒ\{f(t)\}$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{f(t)\}}& {\displaystyle ={\int }_0^T{e}^{-st}f(t)dt+e^{-sT}ℒ\{f(t)\}}\hfill \\ \hfill {\displaystyle ℒ\{f(t)\}(1-e^{-sT})}& {\displaystyle ={\int }_0^T{e}^{-st}f(t)dt}\hfill \\ \hfill {\displaystyle ℒ\{f(t)\}}& {\displaystyle =\frac{1}{1-e^{-sT}}{\int }_0^T{e}^{-st}f(t)dt✓}\hfill \end{array}}$$

Example

Find the Laplace transform of the given periodic function.

Triangular wave $f(t)$ with period $T=2$. On the first period it rises along $y=t$ from $(0,0)$ to $(1,1)$, then falls along $y=2-t$ to $(2,0)$. The same shape repeats: peaks at $t=1,3,\dots$ and zeros at $t=0,2,4,\dots$.

Solution

On one period,

$$f(t)=\{\begin{array}{cc}{\displaystyle t,}\hfill & {\displaystyle 0\le t<1}\hfill \\ {\displaystyle 2-t,}\hfill & {\displaystyle 1\le t<2}\hfill \end{array},f(t+2)=f(t)$$

so $f$ is periodic with period $T=2$. By the theorem,

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{f(t)\}}& {\displaystyle =\frac{1}{1-e^{-2s}}{\int }_0^2{e}^{-st}f(t)dt}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{1-e^{-2s}}[{\int }_0^1{e}^{-st}tdt+{\int }_1^2{e}^{-st}(2-t)dt]}\hfill \end{array}}$$

Integrate each by parts: $u=t$, $dv=e^{-st}dt$ for the first, and $u=2-t$, $dv=e^{-st}dt$ for the second. The two integrals are

$${\displaystyle {\int }_0^1{e}^{-st}tdt=\frac{1}{s^2}-\frac{e^{-s}}{s}-\frac{e^{-s}}{s^2},{\int }_1^2{e}^{-st}(2-t)dt=\frac{e^{-s}}{s}-\frac{e^{-s}}{s^2}+\frac{e^{-2s}}{s^2}}$$

Their sum is ${\displaystyle \frac{1}{s^2}(1-2e^{-s}+e^{-2s})}$. Factor $1-e^{-2s}=(1+e^{-s})(1-e^{-s})$ and cancel:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{f(t)\}}& {\displaystyle =\frac{1}{1-e^{-2s}}(1-2e^{-s}+e^{-2s})\frac{1}{s^2}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\frac{1}{(1+e^{-s})(1-e^{-s})}{(1-e^{-s})}^2·\frac{1}{s^2}}\hfill \end{array}}$$