Laplace Transform of Hyperbolic Cosine

Theorem

Let $\cosh t$ be the hyperbolic cosine, where $t$ is real.

Let $\laptrans f$ denote the Laplace transform of the real function $f$.

Then:

$\laptrans {\cosh a t} = \dfrac s {s^2 - a^2}$

where $a \in \R_{>0}$ is constant, and $\map \Re s > a$.

Proof 1

 $\ds \laptrans {\cosh {a t} }$ $=$ $\ds \int_0^{\to +\infty} e^{-s t} \cosh {a t} \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \intlimits {\frac {e^{-s t} \paren {-s \cosh a t + a \sinh a t} } {\paren {-s}^2 - a^2} } {t \mathop = 0} {t \mathop \to +\infty}$ Primitive of $e^{a x} \cosh b x$ $\ds$ $=$ $\ds 0 - \frac {-s \, \map \cosh {0 \times a} + a \, \map \sinh {0 \times a} } {s^2 - a^2}$ Exponential Tends to Zero, Exponential of Zero $\ds$ $=$ $\ds \frac {s \cosh 0 - a \sinh 0} {s^2 - a^2}$ simplifying $\ds$ $=$ $\ds \frac s {s^2 - a^2}$ Hyperbolic Sine of Zero is Zero, Hyperbolic Cosine of Zero is One

Proof 2

 $\ds \laptrans {\cosh at}$ $=$ $\ds \laptrans {\frac {e^{at} + e^{-at} } 2}$ Definition of Hyperbolic Cosine $\ds$ $=$ $\ds \frac 1 2 \paren {\laptrans {e^{at} } + \laptrans {e^{-at} } }$ Linear Combination of Laplace Transforms $\ds$ $=$ $\ds \frac 1 2 \paren {\frac 1 {s - a} + \frac 1 {s + a} }$ Laplace Transform of Exponential $\ds$ $=$ $\ds \frac 1 2 \paren {\frac {s + a + s - a} {\paren {s - a} \paren {s + a} } }$ $\ds$ $=$ $\ds \frac s {s^2 - a^2}$

$\blacksquare$

Proof 3

 $\ds \laptrans {\sinh a t}$ $=$ $\ds \laptrans {\frac {e^{a t} + e^{-a t} } 2}$ Definition of Hyperbolic Cosine $\ds$ $=$ $\ds \int_0^{\to +\infty} e^{-s t} \paren {\frac {e^{a t} + e^{-a t} } 2} \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \dfrac 1 2 \int_0^{\to +\infty} e^{-s t} e^{a t} \rd t + \dfrac 1 2 \int_0^{\to +\infty} e^{-s t} e^{-a t} \rd t$ Linear Combination of Laplace Transforms $\ds$ $=$ $\ds \dfrac 1 2 \laptrans {e^{a t} } + \dfrac 1 2 \laptrans {e^{-a t} }$ Definition of Laplace Transform $\ds$ $=$ $\ds \frac 1 2 \paren {\frac 1 {s - a} + \frac 1 {s + a} }$ Laplace Transform of Exponential $\ds$ $=$ $\ds \frac 1 2 \paren {\frac {s + a + s - a} {\paren {s - a} \paren {s + a} } }$ $\ds$ $=$ $\ds \frac s {s^2 - a^2}$

$\blacksquare$