Laplace Transform of Sine/Proof 4

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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

By definition of the Laplace transform:

$\ds \laptrans {\sin a t} = \int_0^{\to +\infty} e^{-s t} \sin a t \rd t$


From Integration by Parts:

$\ds \int f g' \rd t = f g - \int f' g \rd t$

Here:

\(\ds f\) \(=\) \(\ds \sin a t\)
\(\ds \leadsto \ \ \) \(\ds f'\) \(=\) \(\ds a \cos a t\) Derivative of $\sin a x$
\(\ds g'\) \(=\) \(\ds e^{-s t}\)
\(\ds \leadsto \ \ \) \(\ds g\) \(=\) \(\ds -\frac 1 s e^{-s t}\) Primitive of $e^{a x}$

So:

\(\text {(1)}: \quad\) \(\ds \int e^{-s t} \sin a t \rd t\) \(=\) \(\ds -\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t\)


Consider:

$\ds \int e^{-s t} \cos a t \rd t$

Again, using Integration by Parts:

$\ds \int h j\,' \rd t = h j - \int h' j \rd t$

Here:

\(\ds h\) \(=\) \(\ds \cos a t\)
\(\ds \leadsto \ \ \) \(\ds h'\) \(=\) \(\ds -a \sin a t\) Derivative of Cosine Function
\(\ds j\,'\) \(=\) \(\ds e^{-s t}\)
\(\ds \leadsto \ \ \) \(\ds j\) \(=\) \(\ds -\frac 1 s e^{-s t}\) Primitive of Exponential Function


So:

\(\ds \int e^{-s t} \cos a t \rd t\) \(=\) \(\ds -\frac 1 s e^{-st} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t\)


Substituting this into $(1)$:

\(\ds \int e^{-s t} \sin a t \rd t\) \(=\) \(\ds -\frac 1 s e^{-s t} \sin a t + \frac a s \paren {-\frac 1 s e^{-s t} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t}\)
\(\ds \) \(=\) \(\ds -\frac 1 s e^{-s t} \sin a t - \frac a {s^2} e^{-s t} \cos a t - \frac {a^2} {s^2} \int e^{-s t} \sin a t \rd t\)
\(\ds \leadsto \ \ \) \(\ds \paren {1 + \frac {a^2} {s^2} } \int e^{-s t} \sin a t \rd t\) \(=\) \(\ds -e^{-s t} \paren {\frac 1 s \sin a t + \frac a {s^2} \cos a t}\)


Evaluating at $t = 0$ and $t \to +\infty$:

\(\ds \paren {1 + \frac {a^2} {s^2} } \laptrans {\sin a t}\) \(=\) \(\ds \intlimits {-e^{-s t} \paren {\frac 1 s \sin a t + \frac a {s^2} \cos a t} } {t \mathop = 0} {t \mathop \to +\infty}\)
\(\ds \) \(=\) \(\ds 0 - \paren {-1 \paren {\frac 1 s \times 0 + \frac a {s^2} \times 1} }\) Boundedness of Real Sine and Cosine, Complex Exponential Tends to Zero
\(\ds \) \(=\) \(\ds \frac a {s^2}\)
\(\ds \leadsto \ \ \) \(\ds \laptrans {\sin a t}\) \(=\) \(\ds \frac a {s^2} \paren {1 + \frac {a^2} {s^2} }^{-1}\)
\(\ds \) \(=\) \(\ds \frac a {s^2} \paren {\frac {s^2} {a^2 + s^2} }\)
\(\ds \) \(=\) \(\ds \frac a {s^2 + a^2}\)

$\blacksquare$


Sources

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