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. $$ f(t)=\cos a t $$

Short answer

The Laplace Transform of \(f(t) = \cos(at)\) is \(\frac{s}{s^2 + a^2}\).

Step-by-step solution

Define the Function

First, identify the function for which the Laplace transformation needs to be found. In this case, the function is \(f(t) = \cos(at)\).

Write Down the Laplace Transform

The Laplace Transform formula is \(F(s) = \int_{0}^{+\infty}e^{-st}f(t)dt\). Now, the goal is to calculate this integral with \(f(t)\) being \(\cos(at)\). So the Laplace Transform becomes \(F(s) = \int_{0}^{+\infty}e^{-st}\cos(at)dt\).

Solve the Integral

To solve the integral, we use integration by parts, where integration by parts formula is \(\int udv = uv - \int vdu\). Select \(u = \cos(at)\) and \(dv = e^{-st}dt\). Then \(du = -a\sin(at)dt\) and \(v = -1/s e^{-st}\). Apply integration by parts to get \(\int_{0}^{+\infty}e^{-st}\cos(at)dt = [e^{-st}\cos(at)/s]_0^{+\infty} + a/s \int_{0}^{+\infty}e^{-st}\sin(at)dt\) and solve further into \(-1/ s \cos(a/s)e^{-∞} + 1 /s \cos(0) - a/s \int_0^{+\infty}e^{-st}\sin(at)dt\). Now, the integral in the latter term needs to be integrated by parts again, repeating the process until it forms a rational function.

Simplify the Result

After a successful second integration by parts, the result would be a simple rational function. In our case, the Laplace Transform of \(\cos(at)\) concludes to \(\frac{s}{s^2 + a^2}\).