Okay, with \(\displaystyle f(t)= \frac{1}{t} sin(\alpha t)\), the Laplace transform is
\(\displaystyle \int_0^\infty \frac{1}{t} sin(\alpha t) e^{-st} dt\)
That looks pretty difficult to me! I might be inclined to try writing \(\displaystyle sin(\alpha t)\) as \(\displaystyle \frac{e^{i\alpha t}- e^{-i\alpha t}}{2i}\) and combine that with the \(\displaystyle e^{-st}\).