By integrating the appropriate formula with respect to, verify L19.

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

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

A transformation of a function f(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)

The properties of Laplace’s transformation are as shown:

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

Consider the function:$f\left(t\right)=\sin t$

Now, simplify${\int }_p^{\infty }F\left(p\right)dp$ as follows:

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

Using property of Laplace transformation, we get

role="math" localid="1664283228365" $$\begin{gathered} {\int }_p^{\infty }F\left(p\right)dp=L\left(\frac{f\left(t\right)}{t}\right)L\left(\frac{f\left(t\right)}{t}\right) \\ ={\int }_p^{\infty }F\left(p\right)dp \end{gathered}$$

Therefore,

$$\begin{gathered} L\left(\sin at\right)=\frac{a}{p^2+a^2}L\left(\frac{\sin at}{t}\right) \\ ={\int }_p^{\infty }\frac{a}{p^2+a^2}dp \\ =\frac{\pi }{2}-{\tan }^{-1}\frac{p}{a} \\ ={\tan }^{-1}\frac{p}{a} \end{gathered}$$