Question

Use Definition 7.1.1 to find $\mathbf{L}\left\{\mathbf{f}\left(\mathbf{t}\right)\right\}\mathbf{,}$if$\mathbf{f}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{cost}\mathbf{?}$

Short answer

The Laplace transform of above function is,

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

Step-by-step solution

Define Laplace transform:

Let f be a function define for$t⩾0$ .Then the integralrole="math" localid="1663931991633" $\mathbf{L}\left\{\mathbf{f}\left(\mathbf{t}\right)\right\}\mathbf{=}{\int }_{\mathbf{0}}^{\infty }{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{f}\left(\mathbf{t}\right)\mathbf{dt}$ is said to be Laplace transform of f provide that integral converges.

Determine the Laplace transform

Consider the function $f\left(t\right)=e^t\cos t$

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

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

Since,

$I={\int }_0^{\infty }\cos t.e^{\left(1-s\right)t}dt$

Letsolve it first, by using integral by parts formula;

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

Here, u and v we choose according to ILATE rule;

I= Inverse

L= Logarithmic

A= Arithmetic

T= Trigonometry

E= Exponential

Asarithmeticfunctioncomesfirst,

Therefore,

$I={\int }_0^{\infty }\cos t.e^{\left(1-s\right)t}dt$

$u=\cos t;v=e^{\left(1-s\right)t}$

Solve for the integral as:

$$\begin{gathered} =\cos t\int e^{\left(1-s\right)t}dt-\int \left[\frac{d}{dt}\cos t.\int e^{\left(1-s\right)t}dt\right]dt \\ ={\left\{\cos t\left(\frac{e^{\left(1-s\right)t}}{1-s}\right)\right\}}_0^{\infty }+\frac{1}{1-s}\int \sin t.e^{\left(1-s\right)t}dt \\ ={\left\{\cos t\left(\frac{e^{\left(1-s\right)t}}{1-s}\right)\right\}}_0^{\infty }+\frac{1}{1-s}{\left\{\sin t\int e^{\left(1-s\right)t}dt-\int \left[\frac{d}{dt}\sin t.\int e^{\left(1-s\right)t}dt\right]\right\}}_0^{\infty } \\ ={\left\{\cos t\left(\frac{e^{\left(1-s\right)t}}{1-s}\right)\right\}}_0^{\infty }+\frac{1}{1-s}{\left\{\sin t\int e^{\left(1-s\right)t}dt-\int \left[\cos t\left(\frac{e^{\left(1-s\right)t}}{1-s}\right)\right]\right\}}_0^{\infty } \end{gathered}$$

Solve further as:

$\mathit{I}\mathbf{=}{\left\{\cos t\left(\frac{e^{\left(1-s\right)t}}{1-s}\right)\right\}}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{+}\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{s}}{\left\{\sin t\int e^{\left(1-s\right)t}dt-\frac{1}{1-s}\int cost.e^{\left(1-s\right)t}dt\right\}}_{\mathbf{0}}^{\mathbf{\infty }}$

Simplify the integral to determine the Laplace transform:

Solve as:

$I={\left\{\cos t\left(\frac{e^{\left(1-s\right)t}}{1-s}\right)\right\}}_0^{\infty }+\frac{1}{1-s}{\left\{\sin t\int e^{\left(1-s\right)t}dt-\frac{1}{1-s}\int cost.e^{\left(1-s\right)t}dt\right\}}_0^{\infty }$

$I+\frac{1}{{\left(1-s\right)}^2}I={\left\{\cos t\left(\frac{e^{\left(1-s\right)t}}{1-s}\right)\right\}}_0^{\infty }+\frac{1}{1-s}{\left\{\sin t\left(\frac{e^{\left(1-s\right)t}}{1-s}\right)\right\}}_0^{\infty }$

$I=\frac{{\left(1-s\right)}^2}{{\left(1-s\right)}^2+1}\left\{\cos \infty .\left(\frac{e^{\left(1-s\right).\infty }}{1-s}\right)-\cos 0.\left(\frac{e^{\left(1-s\right).0}}{1-s}\right)\right\}+\frac{1}{1-s}\left\{\sin \infty .\left(\frac{e^{\left(1-s\right).\infty }}{1-s}\right)-\sin 0.\left(\frac{e^{\left(1-s\right).0}}{1-s}\right)\right\}$

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

Solve further as:

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

Therefore the required Laplace transform of function is,

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