# How can I find the Laplace transform of −e^(bt)sin(omega_0 t)u(−t) ?

Laplace transform of $-{e}^{bt}\mathrm{sin}\left({\omega }_{0}t\right)u\left(-t\right)$
I know that Laplace transform of
${e}^{-\alpha t}\mathrm{sin}\left({\omega }_{0}t\right)u\left(t\right)$
is $\frac{{\omega }_{0}}{\left(s+\alpha {\right)}^{2}+{\omega }_{0}^{2}}$ where $Re\left(s\right)>-Re\left(\alpha \right)$ and $u\left(t\right)={\mathbf{1}}_{t\ge 0}$. Knowing that, how can I find the Laplace transform of
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Let
$f\left(t\right)={e}^{-bt}\mathrm{sin}\left({\omega }_{0}t\right)u\left(t\right)$
Notice that
$f\left(-t\right)=-{e}^{bt}\mathrm{sin}\left({\omega }_{0}t\right)u\left(-t\right)$
Laplace transforms have a nice property that
$\mathcal{L}\left\{f\left(-t\right)\right\}\left(s\right)=\mathcal{L}\left\{f\left(t\right)\right\}\left(-s\right)$
$\mathcal{L}\left\{f\left(t\right)\right\}\left(s\right)=\frac{{\omega }_{0}}{\left(s+b{\right)}^{2}+{\omega }_{0}}$
we can use the property above to see that
$\mathcal{L}\left\{f\left(-t\right)\right\}\left(s\right)=\frac{{\omega }_{0}}{\left(-s+b{\right)}^{2}+{\omega }_{0}}=\frac{{\omega }_{0}}{\left(s-b{\right)}^{2}+{\omega }_{0}}$