Question

In Problems 1–18 use Definition 7.1.1 to find $L\left\{f\left(t\right)\right\}.$

3. $f\left(t\right)={\{}_{1,t⩾1}^{t,0⩽t<1}$

Short answer

The Laplace transform of above function is,

$L\left\{f\left(t\right)\right\}=\frac{1}{s^2}\left(1-e^{-s}\right)$

Step-by-step solution

Definition 7.1.1 Laplace transform

Let f be a function define for$t⩾0$.Then the integral

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

Is said to be Laplace transform of f provide that integral converges.

Applying the definition

Consider the function$f\left(t\right)={\{}_{1,t⩾1}^{t,0⩽t<1}$

The objective is to find$L\left\{f\left(t\right)\right\}$ using the definition.

Note that, the function f is defined for$t⩾0$.

From the definition,

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

Since f is defined in two pieces role="math" localid="1663942519342" $[0,1)$and $[1,\infty )$Laplacian if f is $L\left\{f\left(t\right)\right\}$expressed as the sum of two integrals.

$$\begin{gathered} L\left\{f\left(t\right)\right\}={\int }_0^1{e}^{-st}f\left(t\right)dt+{\int }_1^{\infty }{e}^{-st}f\left(t\right)dt \\ ={\int }_0^1{e}^{-st}\left(t\right)dt+{\int }_1^{\infty }{e}^{-st}\left(1\right)dt \\ ={\int }_0^{1}t.e^{-st}dt+{\int }_1^{\infty }{e}^{-st}dt \\ I=I_1+I_2 \end{gathered}$$

Let solve first, by using integral by parts formula

So formula is $I=\int u.vdt=u\int vdt-\int \left[\frac{d}{dt}u.\int vdt\right]dt$

Where and we choose according to ILATE rule;

I=Inverse

L=Logarithmic

A=Arithmetic

T=Trigonometry

E=Exponential

Asarithmeticfunctioncomesfirst,

Therefore,

$$\begin{gathered} I_1={\int }_0^{1}t.e^{-st}dt \\ u=t;v=e^{-st} \\ {I}_1=t\int e^{-st}dt-\int \left[\frac{d}{dt}t.\int e^{-st}dt\right]dt \\ =t\left(-\frac{e^{-st}}{s}\right)dt-\int \left[\left(1\right).\left(-\frac{e^{-st}}{s}\right)dt\right]dt \\ =t\left(-\frac{e^{-st}}{s}\right)dt+\frac{1}{s}\int e^{-st}dt\backslash \mathrm{hfill} \\ =t\left(-\frac{e^{-st}}{s}\right)dt+\frac{1}{s}\left(-\frac{e^{-st}}{s}\right) \\ {I}_1=t\left(-\frac{e^{-st}}{s}\right)dt-\frac{1}{s^2}{e}^{-st} \\ {I}_1={\left\{t\left(-\frac{e^{-st}}{s}\right)-\frac{1}{s^2}{e}^{-st}\right\}}_0^1 \end{gathered}$$

Similarly;

$$\begin{gathered} I_2={\int }_1^{\infty }{e}^{-st}dt \\ ={\left\{-\frac{e^{-st}}{s}\right\}}_1^{\infty } \end{gathered}$$

Simplification

Wecombineand to get required integral;

$$\begin{gathered} I={\left\{t\left(-\frac{e^{-st}}{s}\right)-\frac{1}{s^2}{e}^{-st}\right\}}_0^1+{\left\{-\frac{e^{-st}}{s}\right\}}_1^{\infty } \\ I=\left\{\left(1.\left(-\frac{e^{-s.1}}{s}\right)-\frac{1}{s^2}{e}^{-s.1}\right)-\left(0.\left(-\frac{e^{-s.0}}{s}\right)-\frac{1}{s^2}{e}^{-s.0}\right)\right\}+\left\{-\frac{e^{-s.\infty }}{s}+\frac{e^{-s.1}}{s}\right\} \\ I=-\frac{e^{-s}}{s}-\frac{1}{s^2}{e}^{-s}-0+\frac{1}{s^2}{e}^0-0+\frac{e^{-s}}{s} \\ I=-\frac{1}{s^2}{e}^{-s}+\frac{1}{s^2} \\ I=\frac{1}{s^2}\left(1-e^{-s}\right) \end{gathered}$$

Therefore the required Laplace transform of function is,

$L\left\{f\left(t\right)\right\}=\frac{1}{s^2}\left(1-e^{-s}\right)$