Question

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

Short answer

The Laplace transform of above function is,

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

Step-by-step solution

Define Laplace transform:

Let f be a function define for$t⩾0$.Then the integral role="math" localid="1663930571148" $\mathbf{L}\left\{\mathbf{f}\left(\mathbf{t}\right)\right\}\mathbf{=}{\int }_{\mathbf{0}}^{\infty }{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{f}\left(\mathbf{t}\right)\mathbf{dt}$ is said to be Laplace transform of f provide that integral converges.

Solve for the Laplace transform as:

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

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.

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

Since, the integral is:

$I={\int }_0^{\infty }t.\cos t.e^{-st}dt$

Letsolve it first, by using integral by parts formula;

So formula is $L\left\{\cos t\right\}=\frac{s}{1+s^2}$

Therefore,

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

Simplify the integral to determine the Laplace transform:

Simplification;

$I={\int }_0^{\infty }t.\cos t.e^{-st}dt=L\left\{t.\cos t\right\}$

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

Therefore the required Laplace transform of function is,

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