Laplace Transform of Hyperbolic Sine/Proof 1

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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

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

$\blacksquare$