laplace-transform-L-t-sin-t-t-F-s-F-s-then-calculate-0-e-

Question Number 190420 by mnjuly1970 last updated on 02/Apr/23

laplace transform

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L_{t} {\frac{\sin (t)}{t}} = F (s)

F (s) = ?

t h e n c a l c u l a t e .

Ω = \int_{0}^{\infty} \frac{e^{- 2 t} s i n (t)}{t} d t = ?

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Commented by mahdipoor last updated on 02/Apr/23

L (\frac{s i n (t)}{t}) = L (\frac{t - \frac{t^{3}}{3!} + \frac{t^{5}}{5!} - \dots}{t}) = L (1 - \frac{t^{2}}{3!} + \frac{t^{4}}{5!} - \dots)

= L (1) - \frac{1}{3!} L (t^{2}) + \frac{1}{5!} L (t^{4}) - \dots =

\frac{1}{s} - \frac{2!}{3! s^{3}} + \frac{4!}{5! s^{5}} - \dots . = \frac{1}{s} - \frac{1}{3 s^{3}} + \frac{1}{5 s^{5}} - \dots = G (s)

G (s) = ? \Rightarrow Ω = G (2) = ?

Commented by mnjuly1970 last updated on 02/Apr/23

t h x \dots G (s) = a r c t a n (\frac{1}{s})

G (2) = a r c t a n (\frac{1}{2}) \dots

Answered by witcher3 last updated on 04/Apr/23

F (s) = Im \int_{0}^{\infty} t^{- 1} e^{- t (s - i)} dt, s > 0

F^{'} (s) = Im \int_{0}^{\infty} - e^{- ts + it} dt

= Im \frac{1}{- s + i}

= - \frac{1}{s^{2} + 1} \Rightarrow F (s) = - \arctan (s) + c

\underset{x \to \infty}{\lim F} (x) = 0 \Rightarrow c = \frac{π}{2}

Ω = - \arctan (2) + \frac{π}{2} = \arctan (\frac{1}{2})

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