Question

Show that the function$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}{\mathbf{t}}^{\mathbf{2}}$ does not possess a Laplace transform. [Hint: Write$\mathcal{L}\left\{1/t^2\right\}$ as two improper integrals:$\mathcal{L}\left\{1/t^2\right\}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{st}}}{{\mathbf{t}}^{\mathbf{2}}}\mathbf{dt}\mathbf{+}{\mathbf{\int }}_{\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{st}}}{{\mathbf{t}}^{\mathbf{2}}}\mathbf{dt}\mathbf{=}{\mathbf{I}}_{\mathbf{1}}\mathbf{+}{\mathbf{I}}_{\mathbf{2}}\mathbf{.}$

Show that${\mathbf{I}}_{\mathbf{1}}$ diverges.]

Short answer

The definition of the Laplace transform for$\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{t}}^2}$ is not satisfied, the function does not have the Laplace transform.

Step-by-step solution

Definition of Laplace Transform

Let$\mathbf{f}$ be a function defined for $\mathbf{t}\mathbf{⩾}\mathbf{0}$. Then the integral

$\mathcal{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{x}}{\mathbf{e}}^{\mathbf{-}\mathbf{xt}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

is said to be the Laplace transform of $\mathbf{f}$, provided that the integral converges.

Use Laplace Transform

The formula for Laplace transform

$\mathcal{L}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}={\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}\mathrm{f}\left(\mathrm{t}\right)\text{d}\mathrm{t}$

We have that$\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{t}}^2}$ , so use the above formula.

$\mathcal{L}\left\{\frac{1}{{\mathrm{t}}^2}\right\}={\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}\frac{1}{{\mathrm{t}}^2}\text{d}\mathrm{t}$

The function $\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{t}}^2}$is undefined at $\mathrm{t}=0$. The endpoint $\mathrm{t}=\infty$is an undefined point for all functions.

We will split up the integral into two integrals.

$\mathcal{L}\left\{\frac{1}{{\mathrm{t}}^2}\right\}={\int }_0^1{\mathrm{e}}^{-\mathrm{st}}\frac{1}{{\mathrm{t}}^2}\text{d}\mathrm{t}+{\int }_1^{\infty }{\mathrm{e}}^{-\mathrm{st}}\frac{1}{{\mathrm{t}}^2}\text{d}\mathrm{t}$

${\mathrm{I}}_1={\int }_0^1{\mathrm{e}}^{-\mathrm{st}}\frac{1}{{\mathrm{t}}^2}\text{d}\mathrm{t}$

Integration by parts

$\mathrm{u}={\mathrm{e}}^{-\mathrm{st}}\text{d}\mathrm{u}=-{\mathrm{se}}^{-\mathrm{st}}\text{d}\mathrm{t}$

$\text{d}\mathrm{v}=\frac{1}{{\mathrm{t}}^2}\text{d}\mathrm{t}\mathrm{v}=-\frac{1}{\mathrm{t}}$

Let's consider the integral

$=-{\left.\frac{1}{\mathrm{t}}{\mathrm{e}}^{-\mathrm{st}}\right|}_0^1-\mathrm{s}{\int }_0^1{\mathrm{e}}^{-\mathrm{st}}\frac{1}{\mathrm{t}}\text{d}\mathrm{t}\left(\mathrm{uv}-\int \mathrm{v}\text{d}\mathrm{u}\right)$

$=-{\mathrm{e}}^{-\mathrm{s}}+\infty -\mathrm{s}{\int }_0^1{\mathrm{e}}^{-\mathrm{st}}\frac{1}{\mathrm{t}}\text{d}\mathrm{t}$

Therefore, we see the rest of the integral is divergent. Hence, we do not need to evaluate the rest of the problem.

Since, the integral is divergent, the original integral is also divergent. We do not need to evaluate the integral ${\mathrm{I}}_2$. So, since the definition of the Laplace transform for$\mathrm{f}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{t}}^2}$ is not satisfied, the function does not have the Laplace transform.