Determine the Laplace transform given:

Question

How can you perform the following transformation? $$\begin{equation} \mathscr{L}\left\{\int_{0}^{t}\frac{e^{-3t}\sin(2t)}{t}\,dt\right\} \end{equation} $$ I'm not sure how to use the convolution.

Answer 1

HINT

$$ \mathcal{L}\left(\int_0^t f(x)dx\right)(s) = \frac{\mathcal L(f(t))(s)}{s}$$ (integration by parts) and also, $$\mathcal{L}\left(\frac{f(t)}{t}\right) = \int_s^\infty \mathcal{L}(f(t))(z)dz$$ and $$\mathcal{L}(e^{at}\sin(bt))$$ is standard.

Answer 2

If $a,b>0$ we have $$ \int_{0}^{+\infty}\frac{\sin(at)e^{-bt}}{t}\,dt=\arctan\frac{a}{b} \tag{1}$$ hence the Laplace transform of $\frac{\sin(2t)e^{-3t}}{t}$ is given by $\arctan\frac{2}{3+s}$ and the wanted Laplace transform is given by $\frac{1}{s}\arctan\frac{2}{3+s}$. $(1)$ can be derived from $$ \int_{0}^{+\infty}\cos(at)e^{-bt}\,dt \stackrel{\text{IBP}}{=}\frac{b}{a^2+b^2}\tag{0} $$ by integrating both sides with respect to $a$.