Hint: by definition L(f)(s)=∫0∞f(t)e−stdt=∫0π0⋅e−stdt+∫π∞sin⁡te−stdt=∫π∞sin⁡te−stdt Using L(f)(s):=∫0∞f(t)e−stdt=∫0π0⋅e−stdt+∫π∞sin⁡te−stdt=∫π∞sin⁡te−stdt you can arrive at the answer.

Key to "Id" - Plainmath

Ronnie Baur

Answered question

2021-11-23

Answer & Explanation

Tionant

Beginner2021-11-24Added 17 answers

Hint: by definition
$L (f) (s) = \int_{0}^{\infty} f (t) e^{- s t} d t = \int_{0}^{π} 0 \cdot e^{- s t} d t + \int_{π}^{\infty} \sin t e^{- s t} d t = \int_{π}^{\infty} \sin t e^{- s t} d t$
Using $L (f) (s) := \int_{0}^{\infty} f (t) e^{- s t} d t = \int_{0}^{π} 0 \cdot e^{- s t} d t + \int_{π}^{\infty} \sin t e^{- s t} d t = \int_{π}^{\infty} \sin t e^{- s t} d t$ you can arrive at the answer.

breisgaoyz

Beginner2021-11-25Added 14 answers

You can also use the shifting property of the Laplace transform. Your function is
$f (t) = \sin (t) u (t - π) = - \sin (t - π) u (t - π) = g (t - π)$
with $g (t) = - \sin (t) u (t)$ . The transform of $g (t)$ is a standard result that can be found in any Laplace transform table:
$G (s) = - \frac{1}{s^{2} + 1}$
and by the shifting property
$F (s) = e^{- π s} G (s) = - \frac{e^{- π s}}{s^{2} + 1}$

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