Laplace transform of hyperbolic sine

Using the definition, find the Laplace transform of the hyperbolic sine function $f(t)=\sinh{at}$ .

1. Apply Laplace transform definition

$\displaystyle\mathscr{L}\{\sinh{at}\}=\int_{t=0}^{\infty}\sinh{at}\cdot e^{-st}dt$

The hyperbolic sine function $\sinh{at}$ can be written in term of exponential functions, i.e:

$\displaystyle\sinh{at}=\frac{1}{2}(e^{at}-e^{-at})$

Thus, the Laplace integral becomes:

$\displaystyle\mathscr{L}\{\sinh{at}\}=\int_{t=0}^{\infty}\frac{1}{2}(e^{at}-e^{-at})\cdot e^{-st}dt$

Simplifying the above equation, by multiplying from the left side in right part

$\displaystyle\mathscr{L}\{\sinh{at}\}=\frac{1}{2}\int_{t=0}^{\infty}(e^{at}\cdot e^{-st}-e^{-at}\cdot e^{-st})dt$

Using the power of exponents and difference rule of integration:

$\displaystyle\mathscr{L}\{\sinh{at}\}=\frac{1}{2}\left(\int_{t=0}^{\infty}e^{-(s-a)t}dt-\int_{t=0}^{\infty} e^{-(s+a)t}dt\right)$

Rewriting the above equation as

$\displaystyle\mathscr{L}\{\sinh{at}\}=\frac{1}{2}\left(\int_{t=0}^{\infty}e^{-(s-a)t}dt-\int_{t=0}^{\infty} e^{-(s-(a))t}dt\right)$

It is clear that this depends on the Laplace transform of exponential functions. This becomes:

$\displaystyle\mathscr{L}\{\sinh{at}\}=\frac{1}{2}\left(\mathscr{L}\{e^{at}\}-\mathscr{L}\{e^{-at}\}\right)$

By the help of formula for Laplace transform of exponential function, derived in this post:

$\displaystyle \mathscr{L}\{\sinh{at}\}=\frac{1}{2}\left(\frac{1}{s-a}-\frac{1}{s+a}\right)$

Taking the LCM on right hand side to simplify the expression:

$\displaystyle\mathscr{L}\{\sinh{at}\}=\frac{1}{2}\left(\frac{s+a-s+a}{(s-a)(s+a)}\right)$

Multiplying the terms in denominator and adding and subtracting the term in numerator, we get:

$\displaystyle\mathscr{L}\{\sinh{at}\}=\frac{1}{2}\left(\frac{2a}{s^2-a^2}\right)$

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