Question

Show that${\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}{\mathit{f}}_{\mathbf{n}}\left(t\right)\mathit{d}\mathit{t}\mathbf{=}\mathbf{1}$for the functions${\mathit{f}}_{\mathbf{n}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$in Figures 11.3 and 11.4.

Short answer

The given function is verified

Step-by-step solution

Given information

The given expressions are$\int f_x\left(t\right)dt=1$.

Definition of Laplace Transformation

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

Verify the given function

Show that

$\int f_n\left(t\right)dt=1$

Where

$f_n\left(t\right)=n{e}^{-n\left(i-r_o\right)}t>I_0$

$f_n\left(f\right)=0$

Otherwise

${\int }_{f_n}\left(t\right)dt={\int }_{i}n{e}^{-n\left(t-n_0\right)}dt$

Solve further

$$\begin{gathered} {\int }_0{f}_n\left(t\right)dt=n{e}^{n{h}_n}\left(\frac{e^{-\mathrm{\pi }}}{-\mathrm{\pi }}\right) \\ \int f_n\left(t\right)dt=e^{n+0-n+0} \\ \int f_n\left(t\right)dt=1 \end{gathered}$$

Thus, verified.