# Laplace Transform of Sine

## 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

 $\ds \laptrans {\sin a t}$ $=$ $\ds \laptrans {\frac {e^{i a t} - e^{-i a t} } {2 i} }$ Sine Exponential Formulation $\ds$ $=$ $\ds \frac 1 {2 i} \paren {\laptrans {e^{i a t} } - \laptrans {e^{-i a t} } }$ 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

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

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

$\blacksquare$