AimettiA8J

2022-11-22

How to find the Laplace transform of $|\mathrm{sin}\left(t\right)|$?

Lukas Arias

${\mathcal{L}}_{s}\left(|\mathrm{sin}t|\right)={\int }_{0}^{\mathrm{\infty }}|\mathrm{sin}\left(t\right)|{\mathrm{e}}^{-st}\mathrm{d}t=\sum _{n=0}^{\mathrm{\infty }}{\int }_{\pi n}^{\pi \left(n+1\right)}|\mathrm{sin}\left(t\right)|{\mathrm{e}}^{-st}\mathrm{d}t=\sum _{n=0}^{\mathrm{\infty }}{\mathrm{e}}^{-s\pi n}{\int }_{0}^{\pi }\mathrm{sin}\left(t\right){\mathrm{e}}^{-st}\mathrm{d}t=\frac{1}{1-\mathrm{exp}\left(-\pi s\right)}{\int }_{0}^{\pi }\mathrm{sin}\left(t\right){\mathrm{e}}^{-st}\mathrm{d}t$
The latter integrate is easy to evaluate as follows:
${\int }_{0}^{\pi }\mathrm{sin}\left(t\right){\mathrm{e}}^{-st}\mathrm{d}t=\mathrm{\Im }{\int }_{0}^{\pi }{\mathrm{e}}^{it-st}\mathrm{d}t=\mathrm{\Im }\left(\frac{\mathrm{exp}\left(i\pi -\pi s\right)-1}{i-s}\right)=\frac{1+\mathrm{exp}\left(-\pi s\right)}{1+{s}^{2}}$
Thus
${\mathcal{L}}_{s}\left(|\mathrm{sin}t|\right)=\frac{1+\mathrm{exp}\left(-\pi s\right)}{1+{s}^{2}}\cdot \frac{1}{1-\mathrm{exp}\left(-\pi s\right)}$

Jewel Hall

Here is a short derivation. First note that $|\mathrm{sin}t|$ has a period of $\pi$. Then
$\mathcal{L}\left(|\mathrm{sin}t|\right)$
$=\frac{1}{1-{e}^{-\pi s}}\phantom{\rule{1em}{0ex}}{\int }_{0}^{\pi }{e}^{-st}|\mathrm{sin}t|\phantom{\rule{thinmathspace}{0ex}}dt$
$=\frac{1}{1-{e}^{-\pi s}}\phantom{\rule{1em}{0ex}}{\int }_{0}^{\pi }{e}^{-st}\mathrm{sin}t\phantom{\rule{thinmathspace}{0ex}}dt$
$=\frac{1}{1-{e}^{-\pi s}}\phantom{\rule{1em}{0ex}}{\left(-{e}^{-st}\frac{s\phantom{\rule{thinmathspace}{0ex}}\mathrm{sin}t+\mathrm{cos}t}{{s}^{2}+1}\phantom{\rule{1em}{0ex}}|}_{0}^{\pi }\right)$
$=\frac{1}{1-{e}^{-\pi s}}\phantom{\rule{1em}{0ex}}\left(\frac{{e}^{-\pi s}}{{s}^{2}+1}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}+\frac{1}{{s}^{2}+1}\right)$
$=\frac{1+{e}^{-\pi s}}{1-{e}^{-\pi s}}\phantom{\rule{1em}{0ex}}\cdot \frac{1}{{s}^{2}+1}$

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