jhenezhubby01ff

2022-10-05

I want to find Laplace transform of $\mathrm{sin}\left(at\right)$ by definition
$\mathcal{L}\left\{\mathrm{sin}at\right\}={\int }_{0}^{\to +\mathrm{\infty }}{e}^{-st}\mathrm{sin}at\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$
After taking integration by parts twice, I reached the following at last
$\mathcal{L}\left\{\mathrm{sin}at\right\}={-{e}^{-st}\frac{{s}^{2}}{{s}^{2}+{a}^{2}}\left(\frac{1}{s}\mathrm{sin}at+\frac{a}{{s}^{2}}\mathrm{cos}at\right)|}_{t=0}^{t\to +\mathrm{\infty }}$
I can' t do the rest.

Jase Powell

You are forgetting to evaluate $-{e}^{-st}$ as $t\to \mathrm{\infty }$:
$\mathcal{L}\left\{\mathrm{sin}at\right\}=\frac{{s}^{2}}{{s}^{2}+{a}^{2}}{\left[-{e}^{-st}\left(\frac{1}{s}\mathrm{sin}at+\frac{a}{{s}^{2}}\mathrm{cos}at\right)\right]}_{t=0}^{t=\mathrm{\infty }}$
What happens to the expression in square brackets when $t\to \mathrm{\infty }$?

beninar6u

$\mathcal{L}\left\{\mathrm{sin}at\right\}={\int }_{0}^{\to +\mathrm{\infty }}{e}^{-st}\mathrm{sin}at\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}t$
$I={\int }_{0}^{\to +\mathrm{\infty }}{e}^{-st}\mathrm{sin}at=\frac{1}{2i}{\int }_{0}^{\to +\mathrm{\infty }}{e}^{-st}\left({e}^{iat}-{e}^{-iat}\right)dt=\frac{1}{2i}{\int }_{0}^{\to +\mathrm{\infty }}\left({e}^{-t\left(-ia+s\right)}-{e}^{-t\left(ia+s}\right)dt$
$2iI={|\frac{{e}^{-t\left(-ia+s\right)}}{\left(ia-s\right)}|}_{0}^{\mathrm{\infty }}+{|\frac{{e}^{-t\left(ia+s\right)}}{\left(ia+s\right)}|}_{0}^{\mathrm{\infty }}$
$2iI=\frac{1}{\left(s-ia\right)}-\frac{1}{\left(s+ia\right)}$
$I=\frac{a}{{s}^{2}+{a}^{2}}$

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