Question

Use part (c) of Theorem 7.1.1 to show that

$\mathcal{L}\left\{e^{(a+\mathrm{ib})t}\right\}\mathbf{=}\frac{\mathbf{s}\mathbf{-}\mathbf{a}\mathbf{+}\mathbf{ib}}{{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}\mathbf{,}$

where a and b are real and ${\mathbf{i}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}$Show how Euler's formula (page 136) can then be used to deduce the results

$$\begin{gathered} \mathcal{L}\left\{e^{\mathrm{at}}\mathrm{cosbt}\right\}\mathbf{=}\frac{\mathbf{s}\mathbf{-}\mathbf{a}}{{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}} \\ \mathcal{L}\left\{e^{\mathrm{at}}\mathrm{sinbt}\right\}\mathbf{=}\frac{\mathbf{b}}{{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}\mathbf{.} \end{gathered}$$

Short answer

Therefore Euler’s formula used to deduce the results

$$\begin{gathered} \mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{cosbt}\right\}=\frac{\mathrm{s}-\mathrm{a}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2} \\ \mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{sinbt}\right\}=\frac{\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2} \end{gathered}$$

Step-by-step solution

Definition of Laplace Transform

Let 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 f, provided that the integral converges.

Use Laplace Transform

From part (c) of theorem 7.1 .1, we have

$\text{L}\left\{{\mathrm{e}}^{\mathrm{at}}\right\}=\frac{1}{\mathrm{s}-\mathrm{a}}$

So,

$$\begin{gathered} \mathcal{\text{L}}\left\{{\mathrm{e}}^{(\mathrm{a}+\mathrm{ib})\mathrm{t}}\right\}=\frac{1}{\mathrm{s}-(\mathrm{a}+\mathrm{ib})} \\ =\frac{1}{\mathrm{s}-\mathrm{a}-\mathrm{ib}}·\frac{\mathrm{s}-\mathrm{a}+\mathrm{ib}}{\mathrm{s}-\mathrm{a}+\mathrm{ib}} \\ =\frac{\mathrm{s}-\mathrm{a}+\mathrm{ib}}{{(\mathrm{s}-\mathrm{a})}^2-{\left(\mathrm{ib}\right)}^2}\left({\mathrm{i}}^2=-1\right) \\ =\frac{\mathrm{s}-\mathrm{a}+\mathrm{ib}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2}............\left(1\right) \end{gathered}$$

Euler's formula

${\mathrm{e}}^{\mathrm{i\theta }}=\mathrm{cos\theta }+\mathrm{isin\theta }$----------- (2)

$$\begin{gathered} \mathcal{L}\left\{e^{(a+ib)t}\right\}=\mathcal{L}\left\{e^{at}{e}^{ibt}\right\} \\ \stackrel{\left(2\right)}{=}\mathcal{L}\left\{e^{at}\left(\mathrm{cosbt}+\mathrm{isinbt}\right)\right\} \\ =\mathcal{L}\left\{e^{at}\cos bt+e^{at}i\sin bt\right\}(\mathcal{L}\{\alpha f\left(t\right)+\beta g\left(t\right)\}=\alpha \mathcal{L}\{f\left(t\right)\}+\beta \mathcal{L}\{g\left(t\right)\left\}\right) \\ =\mathcal{L}\left\{e^{at}\cos bt\right\}+i\mathcal{L}\left\{e^{at}\sin bt\right\}...........\left(3\right) \end{gathered}$$

Based on (2) and (3), we have.

role="math" localid="1663850436371" $$\begin{gathered} \mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{cosbt}\right\}+\mathrm{i}\mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{sinbt}\right\}=\frac{\mathrm{s}-\mathrm{a}+\mathrm{ib}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2} \\ =\frac{\mathrm{s}-\mathrm{a}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2}+\mathrm{i}\frac{\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2} \end{gathered}$$

Comparing the real and imaginary parts, we get

$\mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{cosbt}\right\}=\frac{\mathrm{s}-\mathrm{a}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2}\text{and}\mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{sinbt}\right\}=\frac{\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2+{\mathrm{b}}^2}$