Laplace transform of $\sin(at)$ by definition

Question

I want to find Laplace transform of $\sin(at)$ by definition $$\displaystyle \mathcal L \left\{{\sin at}\right\} = \int_0^{\to +\infty}e^{-st}\sin at \, \mathrm dt$$

After taking integration by parts twice, I reached the following at last $$\displaystyle \mathcal L \left\{ {\sin at}\right\}=\displaystyle \left.{-e^{-st} {\frac {s^2}{s^2+a^2} }\left({\frac 1 s \sin at + \frac a {s^2} \cos at}\right)}\right\vert_{t \mathop = 0}^{t \mathop \to +\infty}$$

I can' t do the rest. What is the limit of $\sin at$ or $\cos at$ while $t\rightarrow \infty$

For the rest, Can anybody explain it step by step pls.?

Answer 1

You are forgetting to evaluate $-e^{-st}$ as $t\to\infty$:

$$\mathcal{L}\left\{\sin at\right\}=\frac{s^2}{s^2+a^2} \left[-\color{red}{e^{-st}}\left(\frac{1}{s}\sin at + \frac{a}{s^2}\cos at\right)\right]_{t= 0}^{t=\infty}$$

What happens to the expression in square brackets when $t\to\infty$?

Answer 2

$$\displaystyle \mathcal L \left\{{\sin at}\right\} = \int_0^{\to +\infty}e^{-st}\sin at \, \mathrm dt$$

$$I=\int_0^{\to +\infty}e^{-st}\sin at =\frac 1 {2i}\int_0^{\to +\infty}e^{-st}(e^{iat}-e^{-iat}) dt=\frac 1 {2i}\int_0^{\to +\infty}(e^{-t(-ia+s)}-e^{-t(ia+s}) dt$$ $$2iI=\left|\frac {e^{-t(-ia+s)}}{(ia-s)}\right|_0^{\infty}+\left|\frac {e^{-t(ia+s)}}{(ia+s)}\right|_0^{\infty}$$ $$2iI=\frac {1}{(s-ia)}- \frac {1}{(s+ia)}$$ $$I=\frac a {s^2+a^2}$$