# Laplace Transform of Cosine/Proof 1

## 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 $a \in \R$ is constant, and $\Re \paren s > a$.

## Proof

By definition of the Laplace Transform:

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

From Integration by Parts:

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

Here:

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

So:

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

Consider:

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

Again, using Integration by Parts:

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

Here:

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

So:

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

Substituting this into $(1)$:

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

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

 $\displaystyle \paren {1 + \frac {a^2} {s^2} } \laptrans {\cos at}$ $=$ $\displaystyle \left.{-e^{-s t} \paren {\frac 1 s \cos a t - \frac a {s^2} \sin a t} } \right\vert_{t \mathop = 0}^{t \mathop \to +\infty}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle 0 - \paren {-1 \paren {\frac 1 s \times 1 + \frac a {s^2} \times 0} }$ $\quad$ Boundedness of Real Sine and Cosine, Complex Exponential Tends to Zero $\quad$ $\displaystyle$ $=$ $\displaystyle \frac 1 s$ $\quad$ $\quad$ $\displaystyle \implies \ \$ $\displaystyle \laptrans {\cos at}$ $=$ $\displaystyle \frac 1 s \paren {1 + \frac {a^2} {s^2} }^{-1}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac 1 s \paren {\frac {s^2} {a^2 + s^2} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \frac s {s^2 + a^2}$ $\quad$ $\quad$

$\blacksquare$