Collin's question via email about a Laplace Transform

[Prove It]

Find $\displaystyle \begin{align*} F\left( s \right) \end{align*}$ if $\displaystyle \begin{align*} f\left( t \right) = \mathrm{H}\,\left( t - 4 \right) \, \sin{ \left[ 3\,\left( t - 4 \right) \right] } \, \mathrm{e}^{6\,t} \end{align*}$

As the Heaviside function is a function of t - 4, that means all other terms must also be functions of t - 4. The sine function is, but the exponential isn't. However with a little manipulation, we get

$\displaystyle \begin{align*} f\left( t\right) &= \mathrm{H}\,\left( t - 4 \right) \,\sin{ \left[ 3\,\left( t - 4 \right) \right] } \,\mathrm{e}^{6\,\left( t - 4 \right) + 24} \\ &= \mathrm{H}\,\left( t - 4 \right) \,\sin{ \left[ 3\,\left( t - 4 \right) \right] } \,\mathrm{e}^{6\,\left( t - 4 \right) } \,\mathrm{e}^{24} \\ \\ F\left( s \right) &= \mathcal{L}\,\left\{ \mathrm{H}\,\left( t - 4 \right) \,\sin{ \left[ 3\,\left( t - 4 \right) \right] } \,\mathrm{e}^{6\,\left( t - 4 \right) } \,\mathrm{e}^{24} \right\} \\ &= \mathrm{e}^{24}\,\mathcal{L}\,\left\{ \mathrm{H}\,\left( t - 4 \right) \,\sin{ \left[ 3\,\left( t - 4 \right) \right] } \,\mathrm{e}^{6\,\left( t - 4 \right) } \right\} \\ &= \mathrm{e}^{24}\,\mathrm{e}^{-4\,s}\,\mathcal{L}\,\left\{ \sin{ \left( 3\,t \right) } \,\mathrm{e}^{6\,t} \right\} \\ &= \mathrm{e}^{24 - 4\,s}\,\mathcal{L}\,\left\{ \sin{ \left( 3\,t \right) } \right\} _{s \to s - 6} \\ &= \mathrm{e}^{24 - 4\,s} \,\left[ \frac{3}{s^2 + 3^2} \right] _{s \to s - 6} \\ &= \mathrm{e}^{24 - 4\,s} \, \left[ \frac{3}{\left( s - 6 \right) ^2 + 9} \right] \end{align*}$

 

[benorin]

Look correct to me.