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

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

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

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$\quad + 1 \quad \ddot\smile\quad$ –  amWhy Mar 4 '13 at 0:44
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.
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 : $$\frac{d}{ds}F(s)= \frac{-1}{\pi} [\frac{\frac{10}{\pi}}{(\frac{10}{\pi})^2s^2+1}] =\frac{(\frac{1}{\pi})(\frac{-\pi}{10})}{s^2+(\frac{\pi}{10})^2}$$ $$\mathcal{L}^{-1} [\frac{d}{ds}F(s)] = -t f(t)$$ $$t f(t) = \frac{1}{\pi}sin(\frac{\pi}{10}t)$$ $$f(t)=\frac{sin(\frac{\pi t}{10})}{\pi t}$$