rhenan5v

2022-10-26

How to Find Inverse Laplace Transform of $F\left(s\right)=\frac{1}{\pi }{\mathrm{cot}}^{-1}\left(\frac{10s}{\pi }\right)$

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Expert

Hint:
$-\pi tf\left(t\right)={\mathcal{L}}^{-1}\left(\frac{d}{ds}{\mathrm{cot}}^{-1}\left(10s/\pi \right)\right)$

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Paloma Sanford

Expert

We know :
$co{t}^{-1}\left(x\right)=\frac{\pi }{2}-ta{n}^{-1}\left(x\right)$
So :
$F\left(s\right)=\frac{co{t}^{-1}\left(\frac{10s}{\pi }\right)}{\pi }=\frac{1}{\pi }\left(\frac{\pi }{2}-ta{n}^{-1}\left(\frac{10s}{\pi }\right)\right)$
Also we know :
$\mathcal{L}\left[-tf\left(t\right)\right]=\frac{d}{ds}F\left(s\right)$
So :
$\frac{d}{ds}F\left(s\right)=\frac{-1}{\pi }\left[\frac{\frac{10}{\pi }}{\left(\frac{10}{\pi }{\right)}^{2}{s}^{2}+1}\right]=\frac{\left(\frac{1}{\pi }\right)\left(\frac{-\pi }{10}\right)}{{s}^{2}+\left(\frac{\pi }{10}{\right)}^{2}}$
${\mathcal{L}}^{-1}\left[\frac{d}{ds}F\left(s\right)\right]=-tf\left(t\right)$
$tf\left(t\right)=\frac{1}{\pi }sin\left(\frac{\pi }{10}t\right)$
$f\left(t\right)=\frac{sin\left(\frac{\pi t}{10}\right)}{\pi t}$

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