Laplace Transform of Cosine/Proof 2

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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

\(\displaystyle \laptrans {e^{i a t} }\) \(=\) \(\displaystyle \frac 1 {s - i a}\) $\quad$ Laplace Transform of Exponential $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {s + i a} {s^2 + a^2}\) $\quad$ multiply top and bottom by $s + i a$ $\quad$

Also:

\(\displaystyle \laptrans {e^{i a t} }\) \(=\) \(\displaystyle \laptrans {\cos a t + i \sin a t}\) $\quad$ Euler's Formula $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \laptrans {\cos a t} + i \laptrans {\sin a t}\) $\quad$ Linear Combination of Laplace Transforms $\quad$

So:

\(\displaystyle \laptrans {\cos a t}\) \(=\) \(\displaystyle \map \Re {\laptrans {e^{i a t} } }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \map \Re {\frac {s + i a} {s^2 + a^2} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac s {s^2 + a^2}\) $\quad$ $\quad$

$\blacksquare$


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