Laplace Transform of Sine/Proof 5

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 $\map \Re s > 0$.

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

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

$\blacksquare$

1965: Murray R. Spiegel: Theory and Problems of Laplace Transforms ... (previous) ... (next): Chapter $1$: The Laplace Transform: Solved Problems: Laplace Transform of Derivative: $16$