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

Question

$$ F(s)=\frac{\cot^{-1}(\frac{10s}{\pi})}{\pi} $$ $$ f(t) = ?$$

Answer 1

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

Answer 2

HINT: Recall what operation in the $t$ space corresponds to the differentiation of $F(s)$. Now find the inverse Laplace transform of $F'(s)$ and apply that operation to the result.

Answer 3

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