Laplace Transform of Sine/Proof 5

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Theorem

Let $\sin$ denote the real sine function.

Let $\laptrans f$ denote the Laplace transform of a real function $f$.


Then:

$\laptrans {\sin at} = \dfrac a {s^2 + a^2}$

where $a \in \R_{>0}$ is constant, and $\map \Re s > 0$.


Proof

From Laplace Transform of Second Derivative:

$(1): \quad \laptrans {\map {f} t} = s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0$

under suitable conditions.


Then:

\(\ds \map f t\) \(=\) \(\ds \sin a t\)
\(\ds \leadsto \ \ \) \(\ds \map {f'} t\) \(=\) \(\ds a \cos a t\)
\(\ds \map {f} t\) \(=\) \(\ds -a^2 \sin a t\)
\(\ds \map f 0\) \(=\) \(\ds 0\)
\(\ds \map {f'} 0\) \(=\) \(\ds a\)
\(\ds \leadsto \ \ \) \(\ds \laptrans {-a^2 \sin a t}\) \(=\) \(\ds s^2 \laptrans {\sin a t} - s \times 0 - a\) from $(1)$, substituting for $\map f t$, $\map {f'} 0$ and $\map f 0$
\(\ds \leadsto \ \ \) \(\ds -a^2 \laptrans {\sin a t}\) \(=\) \(\ds s^2 \laptrans {\sin a t} - a\)
\(\ds \leadsto \ \ \) \(\ds \laptrans {\sin a t}\) \(=\) \(\ds \dfrac a {s^2 + a^2}\) rearranging

$\blacksquare$


Sources

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