$ F(s)=\frac{\cot^{-1}(\frac{10s}{\pi})}{\pi} $ $ f(t) = ?$
How to Find Inverse Laplace Transform of $ F(s)=\frac{1}{\pi} \cot^{-1}(\frac{10s}{\pi}) $
3 Answers
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} $
Hint: $-\pi tf(t)=\mathcal{L}^{-1}\left(\frac{d}{ds}\cot^{-1}(10s/\pi)\right)$
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0$\quad + 1 \quad \ddot\smile\quad$ – 2013-03-04
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.