Laplace Transform of Sine

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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 1

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

$\blacksquare$


Proof 2

\(\ds \laptrans {e^{i a t} }\) \(=\) \(\ds \frac 1 {s - i a}\) Laplace Transform of Exponential
\(\ds \) \(=\) \(\ds \frac {s + i a} {s^2 + a^2}\) multiplying top and bottom by $s + i a$

Also:

\(\ds \laptrans {e^{i a t} }\) \(=\) \(\ds \laptrans {\cos a t + i \sin a t}\) Euler's Formula
\(\ds \) \(=\) \(\ds \laptrans {\cos a t} + i \laptrans {\sin a t}\) Linear Combination of Laplace Transforms

So:

\(\ds \laptrans {\sin a t}\) \(=\) \(\ds \map \Im {\laptrans {e^{i a t} } }\)
\(\ds \) \(=\) \(\ds \map \Im {\frac {s + i a} {s^2 + a^2} }\)
\(\ds \) \(=\) \(\ds \frac a {s^2 + a^2}\)

$\blacksquare$


Proof 3

\(\ds \laptrans {\sin a t}\) \(=\) \(\ds \laptrans {\frac {e^{i a t} - e^{-i a t} } {2 i} }\) Euler's Sine Identity
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \paren {\laptrans {e^{i a t} } - \laptrans {e^{-i a t} } }\) Linear Combination of Laplace Transforms
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \paren {\frac 1 {s - i a} - \frac 1 {s + i a} }\) Laplace Transform of Exponential
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \paren {\frac {s + i a - s + i a} {s^2 + a^2} }\) simplifying
\(\ds \) \(=\) \(\ds \frac 1 {2 i} \paren {\frac {2 i a} {s^2 + a^2} }\) simplifying
\(\ds \) \(=\) \(\ds \frac a {s^2 + a^2}\) simplifying

$\blacksquare$


Proof 4

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
\(\ds \) \(=\) \(\ds \frac a {s^2}\)
\(\ds \leadsto \ \ \) \(\ds \laptrans {\sin a t}\) \(=\) \(\ds \frac a {s^2} \paren {1 + \frac {a^2} {s^2} }^{-1}\)
\(\ds \) \(=\) \(\ds \frac a {s^2} \paren {\frac {s^2} {a^2 + s^2} }\)
\(\ds \) \(=\) \(\ds \frac a {s^2 + a^2}\)

$\blacksquare$


Proof 5

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

$\blacksquare$


Also see


Sources

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