# Determine the Laplace transform of the following functions. \frac{2}{t}\sin 3at where a is any positive constant and s>a

Determine the Laplace transform of the following functions.
$\frac{2}{t}\mathrm{sin}3at$ where a is any positive constant and s>a
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Step 1
Given:
The function is $\frac{2}{t}\mathrm{sin}\left(3at\right)$, where a is any positive constant and s>a
Step 2
Let the function
$f\left(t\right)=\frac{2}{t}\mathrm{sin}\left(3at\right)$
if
We know that
$L\left[\mathrm{sin}\left(at\right)\right]=\frac{a}{{s}^{2}+{a}^{2}}$
Then
$L\left[\mathrm{sin}\left(3at\right)\right]=\frac{3a}{{s}^{2}+{\left(3a\right)}^{2}}$
$⇒L\left[\mathrm{sin}\left(3at\right)\right]=\frac{3a}{{s}^{2}+9{a}^{2}}$
Step 3
Then
$L\left[\frac{2}{t}\mathrm{sin}\left(3at\right)\right]=2L\left[\frac{1}{t}\mathrm{sin}\left(3at\right)\right]$
Since $L\left[\frac{1}{t}f\left(t\right)\right]={\int }_{s}^{\mathrm{\infty }}F\left(s\right)ds$
$2L\left[\frac{1}{t}\mathrm{sin}\left(3at\right)\right]=2{\int }_{0}^{\mathrm{\infty }}\frac{3a}{{s}^{2}+9{a}^{2}}ds$
$L\left[\frac{2}{t}\mathrm{sin}\left(3at\right)\right]=6a{\int }_{0}^{\mathrm{\infty }}\frac{1}{{s}^{2}+{\left(3a\right)}^{2}}ds$
$L\left[\frac{2}{t}\mathrm{sin}\left(3at\right)\right]=6a{\left[\frac{1}{3a}{\mathrm{tan}}^{-1}\left(\frac{1}{3a}\right)\right]}_{0}^{\mathrm{\infty }}$
$L\left[\frac{2}{t}\mathrm{sin}\left(3at\right)\right]=2\left[\frac{\pi }{2}-0\right]$
Therefore, $L\left[\frac{2}{t}\mathrm{sin}\left(3at\right)\right]=\pi$
Step 4