# Laplace Transform of Cosine/Proof 4

## Theorem

Let $\cos$ be the real cosine function.

Let $\laptrans f$ denote the Laplace transform of the real function $f$.

Then:

$\laptrans {\cos a t} = \dfrac s {s^2 + a^2}$

where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.

## Proof

By definition of the Laplace Transform:

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

From Integration by Parts:

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

Here:

 $\ds f$ $=$ $\ds \cos at$ $\ds \leadsto \ \$ $\ds f'$ $=$ $\ds -a \sin a t$ Derivative of Cosine Function $\ds g'$ $=$ $\ds e^{-s t}$ $\ds \leadsto \ \$ $\ds g$ $=$ $\ds -\frac 1 s e^{-s t}$ Primitive of Exponential Function

So:

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

Consider:

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

Again, using Integration by Parts:

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

Here:

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

So:

 $\ds \int e^{-s t} \sin a t \rd t$ $=$ $\ds -\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t$

Substituting this into $(1)$:

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

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

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

$\blacksquare$