Laplace Transform of Cosine/Proof 1

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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

\(\ds \map {\laptrans {\cos {a t} } } s\) \(=\) \(\ds \int_0^{\to +\infty} e^{-s t} \cos {a t} \rd t\) Definition of Laplace Transform
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \int_0^L e^{-s t} \cos {a t} \rd t\) Definition of Improper Integral
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \intlimits {\frac {e^{-s t} \paren {-s \cos a t + a \sin a t} } {\paren {-s}^2 + a^2} } 0 L\) Primitive of $e^{a x} \cos b x$
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \paren {\frac {e^{-s L} \paren {-s \cos a L + a \sin a L} } {s^2 + a^2} - \frac {e^{-s \times 0} \paren {-s \, \map \cos {0 \times a} + a \, \map \sin {0 \times a} } } {s^2 + a^2} }\)
\(\ds \) \(=\) \(\ds \lim_{L \mathop \to \infty} \paren {\frac {s \, \map \cos {0 \times a} - a \, \map \sin {0 \times a} } {s^2 + a^2} - \frac {e^{-s L} \paren {-s \cos a L + a \sin a L} } {s^2 + a^2} }\)
\(\ds \) \(=\) \(\ds \frac {s \, \map \cos {0 \times a} - a \, \map \sin {0 \times a} } {s^2 + a^2} - 0\) Exponential Tends to Zero
\(\ds \) \(=\) \(\ds \frac {s \cos 0 - a \sin 0} {s^2 + a^2}\) simplifying
\(\ds \) \(=\) \(\ds \frac s {s^2 + a^2}\) Sine of Zero is Zero, Cosine of Zero is One

$\blacksquare$


Sources

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