Question

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t\). The Laplace Transform of \(f(t)\) is defined by \(F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t\) if the improper integral exists. Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function. $$ f(t)=\cosh a t $$

Short answer

The Laplace Transform of \(\cosh a t\) is \(s/(s^2 - a^2)\) for \(s > a\).

Step-by-step solution

Understanding the definition of the Laplace Transform

The Laplace Transform of a function \(f(t)\) is given by \(F(s) = \int_{0}^{\infty} e^{-st}f(t) dt\), where \(s\) is a complex number frequency parameter. Following this definition, we can write the Laplace Transform of \(f(t) = \cosh a t\) as \(F(s) = \int_{0}^{\infty} e^{-st}\cosh a t dt\).

Expressing the \(\cosh\) function in exponential form

The hyperbolic cosine function \(\cosh(x)\) can be expressed in exponential form as \(\cosh(x) = (e^{x} + e^{-x})/2\). So we can express \(f(t) = \cosh a t\) in exponential form as \(f(t) = (e^{at} + e^{-at})/2\). Substituting \(f(t)\) back into the Laplace transform gives \(F(s) = \int_{0}^{\infty} e^{-st}(e^{at} + e^{-at})/2 dt\). Dividing through by 2 gives \(F(s) = \frac{1}{2} \int_{0}^{\infty} e^{-st} (e^{at} + e^{-at}) dt\) or \(F(s) = \frac{1}{2} \int_{0}^{\infty} (e^{-st+at} + e^{-st-at}) dt\).

Solving the Integral

The integral of \(e^x\) is \(e^x\), so we just have to integrate \(e^{(-s+a)t}\) and \(e^{(-s-a)t}\) separately. The antiderivatives are: \(-\frac{1}{s-a}e^{(a-s)t}\) and \(-\frac{1}{s+a}e^{-(s+a)t}\). Evaluate these at the limits from 0 to infinity, then subtract the results. After simplification we finally get \(F(s) = \frac{s}{s^2-a^2}\), assuming \(s > a\).