Laplace Transform of Sine/Proof 1

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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

\(\ds \map {\laptrans {\sin {a t} } } s\) \(=\) \(\ds \int_0^{\to +\infty} e^{-s t} \sin {a t} \rd t\) Definition of Laplace Transform
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \int_0^L e^{-s t} \sin {a t} \rd t\) Definition of Improper Integral
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \intlimits {\frac {e^{-s t} \paren {-s \sin a t - a \cos a t} } {\paren {-s}^2 + a^2} } 0 L\) Primitive of $e^{a x} \sin b x$
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \paren {\dfrac {e^{-s L} \paren {-s \sin a L - a \cos a L} } {s^2 + a^2} - \dfrac {e^{-s \times 0} \paren {-s \, \map \sin {0 \times a} - a \, \map \cos {0 \times a} } } {s^2 + a^2} }\)
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \paren {\dfrac {s \sin 0 + a \cos 0} {s^2 + a^2} - \dfrac {e^{-s L} \paren {s \sin a L + a \cos a L} } {s^2 + a^2} }\) Exponential of Zero and rearranging
\(\ds \) \(=\) \(\ds \dfrac {s \sin 0 + a \cos 0} {s^2 + a^2} - 0\) Exponential Tends to Zero
\(\ds \) \(=\) \(\ds \frac a {s^2 + a^2}\) Sine of Zero is Zero, Cosine of Zero is One

$\blacksquare$


Sources

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