Laplace Transform of Cosine/Proof 4

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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:

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


From Integration by Parts:

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

Here:

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

So:

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


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:

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

So:

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

Substituting this into $(1)$:

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


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

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

$\blacksquare$