# Laplace Transform of Hyperbolic Sine

## Theorem

Let $\sinh t$ be the hyperbolic sine, where $t$ is real.

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

Then:

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

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

## Proof 1

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

$\blacksquare$

## Proof 2

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

$\blacksquare$

## Proof 3

 $\ds \laptrans {\sinh a t}$ $=$ $\ds \laptrans {\frac {e^{a t} - e^{-a t} } 2}$ Definition of Hyperbolic Sine $\ds$ $=$ $\ds \int_0^{\to +\infty} e^{-s t} \paren {\frac {e^{a t} - e^{-a t} } 2} \rd t$ 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 $\ds$ $=$ $\ds \frac 1 2 \paren {\frac {s + a - s + a} {\paren {s - a} \paren {s + a} } }$ $\ds$ $=$ $\ds \frac a {s^2 - a^2}$

$\blacksquare$