Question

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

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

Short answer

The Laplace transform of above function is,

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

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$f\left(t\right)={\{}_{0,t⩾\pi }^{\sin t,0⩽t<\pi }$

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

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

$={\int }_0^{\pi }{e}^{-st}\left(sint\right)dt+{\int }_{\pi }^{\infty }{e}^{-st}\left(0\right)dt$

$I={\int }_0^{\pi }\sin t.e^{-st}dt+0$

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 u and v we choose according to ILATE rule;

I= Inverse

L= Logarithmic

A= Arithmetic

T= Trigonometry

E= Exponential

Asarithmeticfunctioncomesfirst,

Therefore,

$$\begin{gathered} I={\int }_0^{\pi }\sin t.e^{-st}dt \\ u=e^{-st};v=\sin t \\ I=e^{-st}\int \sin tdt-\int \left\{\frac{d}{dt}{e}^{-st}\int \sin tdt\right\}dt \\ ={\left\{-e^{-st}.\cos t\right\}}_0^{\pi }-\int \left\{s.e^{-st}.\cos t\right\}dt \\ ={\left\{-e^{-st}.\cos t\right\}}_0^{\pi }-s\int \left\{e^{-st}.\cos t\right\}dt \\ I={\left\{-e^{-st}.\cos t\right\}}_0^{\pi }-s.I_1 \end{gathered}$$

Againweuseintegralbypartsformulatosolveit,

$$\begin{gathered} I_1=\int e^{-st}\cos tdt \\ u=e^{-st};v=\cos t \\ =e^{-st}\int \cos tdt-\int \left\{\frac{d}{dt}{e}^{-st}\int \cos tdt\right\}dt \\ ={\left\{e^{-st}.\sin t\right\}}_0^{\pi }-{\int }_0^{\pi }-s.e^{-st}.\sin tdt \\ ={\left\{e^{-st}.\sin t\right\}}_0^{\pi }+s{\int }_0^{\pi }{e}^{-st}.\sin tdt \\ {I}_1={\left\{e^{-st}.\sin t\right\}}_0^{\pi }+s.I \end{gathered}$$

Simplification

Wecombineand for further simplification to get required integral;

$$\begin{gathered} I={\left\{-e^{-st}.\cos t\right\}}_0^{\pi }-\left\{{\left(s.e^{-st}\sin t\right)}_0^{\pi }\right\}-s^{2}I \\ I+s^{2}I={\left\{-e^{-st}.\cos t\right\}}_0^{\pi }-{\left\{s.e^{-st}\sin t\right\}}_0^{\pi } \\ \left(1+s^2\right)I={\left\{-e^{-st}.\cos t\right\}}_0^{\pi }-{\left\{s.e^{-st}\sin t\right\}}_0^{\pi } \\ I=\frac{1}{\left(1+s^2\right)}{\left\{-e^{-st}.\left(\cos t+s.\sin t\right)\right\}}_0^{\pi } \\ I=\frac{1}{\left(1+s^2\right)}\left\{\left[-e^{-s.\pi }.\left(\cos \pi +s.\sin \pi \right)\right]-\left[-e^{-s.0}.\left(\cos 0+s.\sin 0\right)\right]\right\} \\ I=\frac{1}{\left(1+s^2\right)}\left\{\left[-e^{-s.\pi }.\left(-1+s.0\right)\right]-\left[-e^0.\left(1+s.0\right)\right]\right\} \\ I=\frac{e^{-s.\pi }+1}{\left(s^2+1\right)} \end{gathered}$$

Therefore the required Laplace transform of function is,

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