# Laplace Transform of Sine/Proof 4

## 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 \ \$ $\ds \laptrans {\sin a t}$ $=$ $\ds \frac a {s^2} \paren {1 + \frac {a^2} {s^2} }^{-1}$ $\ds$ $=$ $\ds \frac a {s^2} \paren {\frac {s^2} {a^2 + s^2} }$ $\ds$ $=$ $\ds \frac a {s^2 + a^2}$

$\blacksquare$