Calculate Laplace transform of e^{at}cos(bt)

This post is one of the posts addressing the Laplace transform of common functions using the definition. It addresses the Laplace transform of the function $f(t)=e^{at}\cos(bt)$ . We have the following theorem:

Theorem:

Let’s prove this result!

1. Apply the Laplace transform

To prove the identity above, we will follow the same steps we did in this post. By definition, we have

$\displaystyle \mathscr{L}\{f(t)\} =\int_{0}^{\infty}e^{-st}e^{at}\cos(bt)dt$

which may be rewritten as

$\displaystyle F(s)=\lim_{c\to\infty}\int_{0}^{c}e^{-(s-a)t}\cos(bt)dt$

2. Use integration by parts

Let $u=\cos(bt)$ and $dv=e^{-(s-a)t}dt$ , so

$\displaystyle du=-b\sin(bt)dt$

as well

$\displaystyle v=\frac{e^{-(s-a)t}}{-(s-a)}$

When we integrate by parts, we get

$\begin{aligned} \int_{0}^{c}e^{-(s-a)t}\cos(bt) & =\left.\cos(bt)\left(\frac{e^{-(s-a)t}}{-(s-a)}\right)\right|_{t=0}^{t=c}-\int{0}^{c}\frac{e^{-(s-a)t}}{-(s-a)}(-b\sin(bt))dt \\& =\cos(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{1}{s-a}-\frac{b}{s-a}\int_{0}^{c}e^{-(s-a)t}\sin(bt)dt \end{aligned}$

due to the fact that $\cos(0)=e^{0}=1$ . To the last integral, we know from the proof above:

$\displaystyle \int_{0}^{c}e^{-(s-a)t}\sin(bt)dt=\sin(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{b}{s-a}\int_{0}^{c}e^{-(s-a)t}\cos(bt)dt$

3. Simplified Laplace integral

Let us denote $I$ as:

$\displaystyle I=\int_{0}^{c}e^{-(s-a)t}\cos(bt)dt$

which gives us:

$\displaystyle \int_{0}^{c}e^{-(s-a)t}\sin(bt)dt=\sin(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{b}{s-a}I$

Therefore,

$\begin{aligned} & I = \cos(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{1}{s-a}-\frac{b}{s-a}\left\{\sin(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{b}{s-a}I\right\} \\ & I = \cos(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{1}{s-a}-\frac{b}{s-a}\sin(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)-\frac{b^{2}}{(s-a)^{2}}I \end{aligned}$

i.e.,

$\begin{aligned} & I+\frac{b^{2}}{(s-a)^{2}}I = \cos(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{1}{s-a}-\frac{b}{s-a}\sin(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right) \\ & \left[1+\frac{b^{2}}{(s-a)^{2}}\right]I = \cos(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{1}{s-a}-\frac{b}{s-a}\sin(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right) \end{aligned}$

4. Final step: calculate the limit

And when we apply $\displaystyle\lim_{c\to\infty}$ in both sides of the equality above, we get:

$\displaystyle \lim_{c\to\infty}\cos(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)=\lim_{c\to\infty}\sin(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)=0$

as well

$\displaystyle\lim_{c\to\infty} I=\lim_{c\to\infty}\int_{0}^{c}e^{-(s-a)t}\cos(bt)dt=F(s)$

hence

$\begin{aligned} & \left[1+\frac{b^{2}}{(s-a)^{2}}\right]F(s)=\frac{1}{s-a} \\ & \left[\frac{(s-a)^{2}+b^{2}}{(s-a)^{2}}\right]F(s)=\frac{1}{s-a} \end{aligned}$

$\displaystyle F(s)=\frac{1}{s-a}\frac{(s-a)^{2}}{(s-a)^{2}+b^{2}}$

Therefore,

$\displaystyle F(s)=\frac{s-a}{(s-a)^{2}+b^{2}}$

This ends the proof!

We reached the end of this post, for more proof of Laplace transform of common functions, check this page.