Laplace poles

The relations used throughout this solution can be derived using Feynman's trick of differentiating under the integral sign or by simply referring to a table of Laplace transforms.

Consider:

$\mathcal{L}\left[\sin{t}\right] =\int_{0}^{\infty}\mathrm{e}^{-st} \sin{t} \ dt= \frac{1}{1+s^2}$

Therefore:

$\mathcal{L}\left[t\sin{t}\right] =-\frac{d}{ds}\left(\int_{0}^{\infty}\mathrm{e}^{-st} \sin{t} \ dt\right) =-\frac{d}{ds}\left(\frac{1}{1+s^2}\right)$ $\implies \mathcal{L}\left[t\sin{t}\right] = -\int_{0}^{\infty}\frac{\partial}{\partial s}\left(\mathrm{e}^{-st} \sin{t}\right) \ dt =\int_{0}^{\infty}\mathrm{e}^{-st} t\sin{t} \ dt= \frac{2s}{(1+s^2)^2}$

$\implies \mathcal{L}\left[t\sin{t}\right] = \frac{2s}{(1+s^2)^2} = \frac{2s}{(s+i)^2(s-i)^2}$

Therefore: $s = \pm \ i$ are the required poles, which are $\boxed{2}$ in number, where $i = \sqrt{-1}$ .