MHB Alexander's question via email about Laplace Transforms

[Prove It]

Evaluate $\displaystyle \mathcal{L}\left\{ H\left( t - 4 \right) \sin{ \left[ 6 \left( t - 4 \right) \right] } \,\mathrm{e}^{5\,t} \right\} $

The Heaviside function suggests a second shift, but to do that, the entire function needs to be a function of $\displaystyle t - 4$.

Let $\displaystyle u = t - 4 \implies t = u + 4$, then

$\displaystyle \begin{align*} \mathrm{e}^{5\,t} &= \mathrm{e}^{5\left( u + 4 \right) } \\ &= \mathrm{e}^{5\,u + 20} \\
&= \mathrm{e}^{5\left( t - 4 \right) + 20} \\ &= \mathrm{e}^{20}\,\mathrm{e}^{5\left( t - 4 \right) } \end{align*}$

So

$\displaystyle \begin{align*} \mathcal{L}\left\{ H \left( t - 4 \right) \sin{\left[ 6\left( t - 4 \right) \right] } \,\mathrm{e}^{5\,t} \right\} &= \mathrm{e}^{20}\,\mathcal{L}\left\{ H\left( t - 4 \right) \sin{ \left[ 6\left( t - 4 \right) \right] }\, \mathrm{e}^{5\left( t - 4 \right) } \right\} \\ &= \mathrm{e}^{20}\,\mathrm{e}^{-4\,s} \,\mathcal{L} \left\{ \sin{ \left( 6\,t \right) }\, \mathrm{e}^{5\,t} \right\} \textrm{ by the second shift theorem} \\ &= \mathrm{e}^{20 - 4\,s } \left[ \frac{6}{s^2 + 6^2} \right] _{s \to s - 5} \textrm{ by the first shift theorem} \\ &= \mathrm{e}^{20 - 4\,s} \left[ \frac{6}{\left( s - 5 \right) ^2 + 36} \right] \end{align*}$

 

[benorin]

Looks correct to me.