Laplace Transform of Cosine

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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 1

\(\ds \map {\laptrans {\cos {a t} } } s\) \(=\) \(\ds \int_0^{\to +\infty} e^{-s t} \cos {a t} \rd t\) 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

$\blacksquare$


Proof 2

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

Also:

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

So:

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

$\blacksquare$


Proof 3

\(\ds \laptrans {\cos a t}\) \(=\) \(\ds \laptrans {\frac {e^{i a t} + e^{-i a t} } 2}\) Euler's Cosine Identity
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\laptrans {e^{i a t} } + \laptrans {e^{-i a t} } }\) Linear Combination of Laplace Transforms
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac 1 {s - i a} + \frac 1 {s + i a} }\) Laplace Transform of Exponential
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\frac {s + i a + s - i a} {s^2 + a^2} }\) simplifying
\(\ds \) \(=\) \(\ds \frac s {s^2 + a^2}\) simplifying

$\blacksquare$


Proof 4

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
\(\ds \) \(=\) \(\ds \frac 1 s\)
\(\ds \leadsto \ \ \) \(\ds \laptrans {\cos at}\) \(=\) \(\ds \frac 1 s \paren {1 + \frac {a^2} {s^2} }^{-1}\)
\(\ds \) \(=\) \(\ds \frac 1 s \paren {\frac {s^2} {a^2 + s^2} }\)
\(\ds \) \(=\) \(\ds \frac s {s^2 + a^2}\)

$\blacksquare$


Also see


Sources