Question

The function $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}{\mathbf{te}}^{{\mathbf{t}}^{\mathbf{2}}}{\mathbf{cost}}^{\mathbf{2}}$ is not of exponential order. Nevertheless, show that the Laplace transform$\mathcal{L}\left\{2{\mathrm{te}}^{t^2}{\mathrm{cose}}^{t^2}\right\}$ exists. [Hint: Start with integration by parts.]

Short answer

$\mathcal{L}\left\{2{\mathrm{te}}^{t^2}{\mathrm{cose}}^{t^2}\right\}$ exists.

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

Using the formula for Laplace transform

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

we get

$\mathcal{\text{L}}\left\{2{\mathrm{te}}^{{\mathrm{t}}^2}{\mathrm{cose}}^{{\mathrm{t}}^2}\right\}={\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}\left(\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{sine}}^{{\mathrm{t}}^2}\right)\mathrm{dt}\left(\frac{\mathrm{d}}{\mathrm{dt}}{\mathrm{sine}}^{{\mathrm{t}}^2}=2{\mathrm{te}}^{{\mathrm{t}}^2}{\mathrm{cose}}^{{\mathrm{t}}^2}\right)$

Integration by parts

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

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

$={\left.{\mathrm{e}}^{-\mathrm{st}}{\mathrm{sine}}^{{\mathrm{t}}^2}\right|}_0^{\infty }+\mathrm{s}{\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}{\mathrm{sine}}^{{\mathrm{t}}^2}\text{d}\mathrm{t}\left(\mathrm{uv}-\int \mathrm{v}\text{d}\mathrm{u}\right)$

$=-\sin 1+\mathrm{s}{\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}{\mathrm{sine}}^{{\mathrm{t}}^2}\text{d}\mathrm{t}$

localid="1663863267458" $\stackrel{\left(1\right)}{=}s\mathcal{L}\left\{\sin e^{t^2}\right\}-\sin 1$

Since $\sin e^{t^2}$is continuous and of exponential order$\mathcal{L}\left\{\sin e^{t^2}\right\}$exists, and therefore $\mathcal{L}\left\{2t{e}^{t^2}\cos e^{t^2}\right\}$exists.