Question

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

7.

Short answer

The Laplace transform of above function is,

$I=\frac{1}{s}{e}^{-s}+\frac{1}{s^2}{e}^{-s}$

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$f\left(t\right)=\left\{\begin{array}{c}0,0<t<1\\ t,t>1\end{array}\right.$

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$[0,1)$ and$[1,\infty )$ Laplacian if f is$L\left\{f\left(t\right)\right\}$expressed as the sum of two integrals.

$L\left\{f\left(t\right)\right\}={\int }_0^1{e}^{-st}f\left(t\right)dt+{\int }_1^{\infty }{e}^{-st}f\left(t\right)dt$

$={\int }_0^1{e}^{-st}\left(0\right)dt+{\int }_1^{\infty }{e}^{-st}\left(t\right)dt$

$=0+{\int }_1^{\infty }t.e^{-st}dt$

Let solve first, by using integral by parts formula

$I={\int }_1^{\infty }t.e^{-st}dt$

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

Where u and v we choose according to ILATE rule;

I= Inverse

L= Logarithmic

A= Arithmetic

T= Trigonometry

E= Exponential

Asarithmeticfunctioncomesfirst,

Therefore,

$I={\int }_0^{\pi }\sin t.e^{-st}dt$

$u=e^{-st};v=t$

$=\left\{t\int e^{-st}dt-\int \left[\frac{d}{dt}t.\int e^{-st}dt\right]dt\right\}$

$=\left\{t\left(\frac{-e^{-st}}{s}\right)-\int \left[1.\left(\frac{-e^{-st}}{s}\right)\right]dt\right\}$

$I={\left\{t\left(-\frac{e^{-st}}{s}\right)-\frac{1}{s^2}{e}^{-st}\right\}}_1^{\infty }$

Simplification

$I={\left\{t\left(-\frac{e^{-st}}{s}\right)-\frac{1}{s^2}{e}^{-st}\right\}}_1^{\infty }$

$I=\left(\infty .\left(-\frac{e^{-s.\infty }}{s}\right)-\frac{1}{s^2}{e}^{-s.\infty }\right)-\left(1.\left(-\frac{e^{-s.1}}{s}\right)-\frac{1}{s^2}{e}^{-s.1}\right)$

$I=\frac{1}{s}{e}^{-s}+\frac{1}{s^2}{e}^{-s}$

Therefore the required Laplace transform of function is,

$I=\frac{1}{s}{e}^{-s}+\frac{1}{s^2}{e}^{-s}$