Question

Find the Laplace transform of \(f(t)=\cos a t,\) where \(a\) is a real constant.

Short answer

Answer: The Laplace transform of the function \(f(t) = \cos(at)\) is \(F(s) = \frac{s}{s^2+a^2}\).

Step-by-step solution

Identify the given function and the definition of Laplace transform

We are given the function \(f(t) = \cos{at}\) and tasked with finding its Laplace transform. The Laplace transform is defined as \(F(s) = \int_0^\infty e^{-st} f(t) dt\). So, we need to find the following integral: $$ F(s) = \int_0^\infty e^{-st} \cos{at} dt $$

Integration by Parts

To solve the above integral, we'll need to use integration by parts. Let $$ u = \cos{at};\, dv = e^{-st} dt $$ Now we find the derivative of \(u\) and the integral of \(dv\): $$ du = -a \sin{at} dt;\, v = -\frac{1}{s}e^{-st} $$ Using the formula for integration by parts, \(u \, dv = u \, v - \int v \, du\), we have: $$ \int e^{-st} \cos{at} dt = -\frac{1}{s}e^{-st}\cos{(at)} - \int -\frac{1}{s}e^{-st}(-a\sin{at}) dt $$ Simplifying the integral and rearranging terms, we get: $$ \int e^{-st} \cos{at} dt = -\frac{1}{s}e^{-st}\cos{(at)} + \frac{a}{s} \int e^{-st}\sin{at} dt $$

Apply Integration by Parts Again

We'll use integration by parts once more to resolve the remaining integral. This time let: $$ u = \sin{at};\, dv = e^{-st} dt $$ Calculate the derivative and integral again: $$ du = a\cos{(at)} dt;\, v = -\frac{1}{s}e^{-st} $$ Using the formula for integration by parts, \(u \, dv = u \, v - \int v \, du\), we have: $$ \int e^{-st} \sin{at} dt = -\frac{1}{s}e^{-st}\sin{at} - \int -\frac{1}{s}e^{-st}(a\cos{at}) dt $$ Now, simplifying and rearranging terms, we get: $$ \int e^{-st} \sin{at} dt = -\frac{1}{s}e^{-st}\sin{at} + \frac{a}{s^2} \int e^{-st}\cos{at} dt $$

Combine Results and Solve for \(F(s)\)

Now we have: $$ F(s) = -\frac{1}{s}e^{-st}\cos{(at)} + \frac{a}{s} \left(-\frac{1}{s}e^{-st}\sin{at} + \frac{a}{s^2} \int e^{-st}\cos{at} dt \right) $$ Note that the last integral term on the right is the original integral we wanted to solve for. We can rewrite the equation as: $$ F(s) - \frac{a^2}{s^3}F(s) = -\frac{1}{s}e^{-st}\cos{(at)} - \frac{a}{s^2}e^{-st}\sin{(at)} $$ Now, combine the terms with \(F(s)\): $$ \square (F(s)) = s^2 F(s) - a^2 F(s) = \int_0^\infty e^{-st} \cos{at} dt $$ The final step is to take the inverse Laplace transform

Evaluate the Result

Since we want to find the Laplace transform \(F(s)\) of the function \(f(t) = \cos{at}\), we have: $$ F(s) = \frac{s}{s^2+a^2}