This post is one of the posts addressing the Laplace transform of common functions using the definition. We have the following theorem:

## Theorem:

Let’s prove this result!

## 1. Apply the Laplace transform

From the definition of Laplace Transform:

\begin{aligned} & \mathscr{L}\{f(t)\} = \int_{0}^{\infty}e^{-st}f(t)dt \\ & \mathscr{L}\{f(t)\} = \int_{0}^{\infty}e^{-st}e^{at}\sin(bt)dt \\ & \mathscr{L}\{f(t)\} = \lim_{c\to\infty}\int_{0}^{c}e^{-(s-a)t}\sin(bt)dt \end{aligned}

## 2. Use integration by parts

If u=\sin(bt) and dv=e^{-(s-a)t}dt, then:

Using the formula of integration by parts:

\int udv=uv-\int vdu

we get

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

Note that

\left.\sin(bt)\cdot\left(\frac{e^{-(s-a)t}}{-(s-a)}\right)\right|_{t=0}^{t=c}=\sin(bc)\cdot\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)

because \sin(0)=0. And to the integral:

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

## 3. Use integration by parts for the second time

we will proceed in the same way we did before. Let:

so

Thus, when we integrate by parts, we have:

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

Since \cos(0)=e^{0}=1, the expression becomes:

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

## 4. Simplified Laplace integral

Let

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

then we may rewrite as follows:

\displaystyle I=\sin(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{b}{s-a}\left\{\cos(bc)\left(\frac{e^{-(s-a)c}}{-(s-a)}\right)+\frac{1}{s-a}-\frac{b}{s-a}I\right\}

which gives us:

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

## 5. Final step: calculate the limit

As 1\leq\sin(x)\leq1 and

\lim_{x\to\infty}e^{-x}=0

so, by the squeeze theorem, we may conclude that

\displaystyle \lim_{x\to\infty}\sin(x)e^{-x}=0

And in the same way

\displaystyle \lim_{x\to\infty}\cos(x)e^{-x}=0

Hence, for s-a>0\Rightarrow s>a , when we evaluate the limit of the expression above as c\to\infty , we have:

\begin{aligned} & \left[\frac{(s-a)^{2}+b^{2}}{(s-a)^{2}}\right]\lim_{c\to\infty}I=\frac{b}{(s-a)^{2}} \\ & \left[\frac{(s-a)^{2}+b^{2}}{(s-a)^{2}}\right]\mathscr{L}\{f(t)\}=\frac{b}{(s-a)^{2}} \\ & \mathscr{L}\{f(t)\}=\frac{b}{(s-a)^{2}}\cdot\frac{(s-a)^{2}}{(s-a)^{2}+b^{2}} \end{aligned}

Therefore,

\displaystyle \mathscr{L}\{f(t)\}=\frac{b}{(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.