Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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

share|cite|improve this answer
$\quad + 1 \quad \ddot\smile\quad$ – 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.

share|cite|improve this answer
Thank you so much. – Amir Alizadeh Dec 21 '12 at 22:34
up vote 0 down vote accepted

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 : $$ \frac{d}{ds}F(s)= \frac{-1}{\pi} [\frac{\frac{10}{\pi}}{(\frac{10}{\pi})^2s^2+1}] =\frac{(\frac{1}{\pi})(\frac{-\pi}{10})}{s^2+(\frac{\pi}{10})^2} $$ $$ \mathcal{L}^{-1} [\frac{d}{ds}F(s)] = -t f(t)$$ $$ t f(t) = \frac{1}{\pi}sin(\frac{\pi}{10}t) $$ $$ f(t)=\frac{sin(\frac{\pi t}{10})}{\pi t} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.