By replacing$a$in L2 by$a+ib$and then by$a-ib$, and adding and subtracting the results and verify L13 and L14.

$\left(e^{-at}\sin bt\right)$

The given function is$L\left(\frac{\sin at}{t}\right)={\tan }^{-1}\left(\frac{a}{p}\right)$

A transformation of a function $f\left(x\right)$into the function $g\left(t\right)$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$F\left(s\right)$is the piecewise-continuous and exponentially-restricted real function$f\left(t\right)$.

The properties of Laplace’s transformation are as shown:

$$\begin{gathered} L\left(f\right(t\left)\right)=F\left(p\right)L\left(\frac{f\left(t\right)}{t}\right) \\ ={\int }_p^{\pi }F\left(p\right)dpL\left(f\left(t\right)\right.) \\ ={\int }_0^{\infty }{e}^{-pt}f(t)dt \\ =F(p) \end{gathered}$$

To verify the expression $\left(LI3\right)L\left(e^{-ai}\sin bt\right)=\frac{b}{{(p+a)}^2+b^2}$and$\left(L14\right)L\left(e^{-at}\cos bt\right)=\frac{p+a}{{(p+a)}^2+b^2}$used

From, property (LI2) of Laplace transforms

$L\left(e^{-w}\right)=\frac{1}{(p+a)}$

Replace by $(a+ib)$and $(a-ib)$ as,

$$\begin{gathered} L\left(e^{-(a+ib)u}\right)=\frac{1}{(p+a+ib)}\dots \left(1\right) \\ L\left(e^{-(a-ib)u}\right)=\frac{1}{(p+a-it)}\dots \left(2\right) \end{gathered}$$

Add equation (1) and (2) as,

$L\left(e^{-(a+ib))}+e^{-(m-ib)c}\right)=\frac{1}{(p+\omega +ib)}+\frac{1}{(p+a-ib)}$

Since,$e^{iux}=\cos ax+i\sin ax$

Calculate the value of.$e^{-(\alpha +\overline{b})t}$

$$\begin{gathered} e^{-(\omega +ib)t}=e^{-at}-e^{-ibe} \\ =e^{-a}(\cos bt-i\sin bt) \end{gathered}$$

Calculate the value of $e^{-(\mathbf{\mu }-\overline{\mathbf{b}})\text{s}}$.

$$\begin{gathered} e^{-(a-b)x}=e^{-at}·e^{ibt} \\ =e^{-w}(\cos bt+i\sin bx)t \end{gathered}$$

Calculate the value of$e^{-(\alpha +ib]t}+e^{-(a-a)d)}$.

$$\begin{gathered} e^{-[a+ab)}+e^{-(a-\overline{b})t}=e^{-af}(\cos bt-i\sin bt)+e^{-at}(\cos bt+i\sin bt) \\ =e^{-ai}(\cos bt-i\sin bt+\cos bt+i\sin bt)\backslash \mathrm{kern}1pt \\ =2e^{-at}\cos bt \\ L\left(e^{-(a+\dot{a})s}+e^{-(a-\tilde{b})c}\right)=\frac{1}{(p+a+\tilde{b})}+\frac{1}{(p+a-ib)} \end{gathered}$$

Divide both sides by 2 as,

$$\begin{gathered} 2L\left(e^{-at}\cos bt\right)=\frac{2jp+a)}{{(p+a)}^2+b^2} \\ L\left(e^{-at}\cos bt\right)=\frac{(p+a)}{{(p+a)}^2+b^2} \end{gathered}$$

Hence, the expression is $L\left(e^{-at}\cos bt\right)=\frac{(p+a)}{{(p+a)}^2+b^2}$verified.

Subtract equation (1) and as,

$L\left(e^{-(a+ib)1}-e^{-(\omega -ib)c}\right)=\frac{1}{(p+w+ib)}-\frac{1}{(p+a-ib)}$

The expression is given as,

$e^{iax}=\cos ax+i\sin ax$

Calculate the value of $e^{-(\mu +i\overline{b})t}$.

$$\begin{gathered} e^{-(\omega +\overline{b})}=e^{-at}-e^{-ibe} \\ =e^{-al}(\cos bt-i\sin br) \end{gathered}$$

Calculate the value of $e^{-\left[\right(u-\overline{a})r}$.

$$\begin{gathered} e^{-\{a-a)x}=e^{-at}·e^{ibt} \\ =e^{-w}(\cos bt+i\sin bx)t \end{gathered}$$

Calculate the value of $e^{-(\mu +\tilde{b})e}-e^{-\{a-at)}$.

$$\begin{gathered} e^{-(a+ibk}-e^{-(\alpha -ib)t}=e^{-at}(\cos bt-i\sin bt)-e^{-at}(\cos bt+i\sin bt) \\ =e^{-at}(\cos bt-i\sin bt-\cos bt+i\sin bt) \\ =2i{e}^{-at}\sin bt \end{gathered}$$

Calculate the Laplace transform.

$$\begin{gathered} L\left(e^{-(a+aL)d}-e^{-(a-\overline{u}]r}\right)=\frac{1}{(p+a+ib)}-\frac{1}{(p+a-ib)} \\ L\left(-2i{e}^{-at}\sin bt\right)=\frac{1}{\mid (p+a)+ib]}-\frac{1}{\left|\right(p+a)-ib|}n-2iL\left(e^{-ai}\sin br\right) \\ =\frac{(p+a)-it-(p+a)-a}{\left|{(p+a)}^2-{\left(ib\right)}^2\right|}-2iL\left(e^{-at}\sin bt\right) \\ =\frac{-2ib}{\left|{(p+a)}^2+{\left(ib\right)}^2\right|}n-2iL\left(e^{-at}\sin br\right) \end{gathered}$$

$=\frac{-2ib}{{(p+a)}^2+b^2}$

Dividing both sides by $-2i$as,

$$\begin{gathered} -2iL\left(e^{-at}\sin bt\right)=\frac{-2ib}{{(p+a)}^2+b^2} \\ L\left(e^{-at}\sin bt\right)=\frac{b}{{(p+a)}^2+b^2} \end{gathered}$$

Hence, the expression$L\left(e^{-at}\sin bt\right)=\frac{b}{{(p+a)}^2+b^2}$ is verified.