Wyatt Weeks

2022-10-11

Compute a laplace transform $\mathcal{L}\left[{e}^{-2x}\mathrm{sin}x\right]$
So I want to do ${\int }_{0}^{\mathrm{\infty }}{e}^{-2x}\mathrm{sin}x{e}^{-px}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(p+2\right)x}\mathrm{sin}x\phantom{\rule{thinmathspace}{0ex}}dx$
But due to the way $e$ and $\mathrm{sin}x$ integrates I dont know where to go from here?

Bridget Acevedo

Hint.
One may write
${\int }_{0}^{\mathrm{\infty }}{e}^{-\left(p+2\right)x}\mathrm{sin}xdx=\text{Im}{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(p+2-i\right)x}\phantom{\rule{mediummathspace}{0ex}}dx=\text{Im}\phantom{\rule{mediummathspace}{0ex}}\frac{1}{p+2-i}=\frac{1}{\left(p+2{\right)}^{2}+1},\phantom{\rule{1em}{0ex}}p>-2.$

Maverick Avery

We have
$J={\int }_{0}^{\mathrm{\infty }}{e}^{-2x}\mathrm{sin}x{e}^{-px}\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(p+2\right)x}\mathrm{sin}x\phantom{\rule{thinmathspace}{0ex}}dx=-{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(p+2\right)x}\phantom{\rule{thinmathspace}{0ex}}d\left(\mathrm{cos}x\right)=\phantom{\rule{0ex}{0ex}}=-\left({{e}^{-\left(p+2\right)x\mathrm{cos}x}|}_{0}^{\mathrm{\infty }}+\left(p+2\right){\int }_{0}^{\mathrm{\infty }}{e}^{-\left(p+2\right)x}\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}dx\right)=\phantom{\rule{0ex}{0ex}}=1-\left(p+2\right){\int }_{0}^{\mathrm{\infty }}{e}^{-\left(p+2\right)x}\mathrm{cos}x\phantom{\rule{thinmathspace}{0ex}}dx=1-\left(p+2\right)\left({{e}^{-\left(p+2\right)x\mathrm{sin}x}|}_{0}^{\mathrm{\infty }}+\left(p+2\right)\underset{J}{\underset{⏟}{{\int }_{0}^{\mathrm{\infty }}{e}^{-\left(p+2\right)x}\mathrm{sin}x\phantom{\rule{thinmathspace}{0ex}}dx}}\right)$
$J=1-\left(p+2{\right)}^{2}J⇒\phantom{\rule{1em}{0ex}}J=\frac{1}{1+\left(p+2{\right)}^{2}}$