Find the Laplace transform of the hyperbolic cosine function f(t)=\cosh{at} using the definition.


1. Apply Laplace transform definition

By the definition of Laplace transform, we have:

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

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

\displaystyle\cosh{at}=\frac{1}{2}(e^{at}+e^{-at})

Thus, the Laplace integral becomes:

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

2. Use Laplace transforms table

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

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


Finally, we get the result of Laplace transform of hyperbolic cosine function:

\displaystyle \mathscr{L}\{\cosh{at}\}=\frac{s}{s^2-a^2}


That’s all!

We reached the end of this post, If you would like to practice, here is the example of the Laplace transform of the hyperbolic sine function.