Evaluate an integral using Laplace transform

Your partial result does not hold. According to Laplace transform properties $$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx=\cal{L}\left(\frac{\sin bx}{x}\right)(a)=\int_a^{+\infty} \cal{L}\left(\sin bx\right)(p)dp=\int_a^{+\infty} \frac{b}{p^2+b^2}\, dp.$$ Can you take it form here?

the standard form of LT is $$\mathcal{L}f(t)=\int_{0}^{\infty}e^{-st}f(t)\,dt$$ So for the given integral you just have to calculate LT of $\frac{\sin bx}{x}$ then put $s=a$ these two properties of LT may be useful $$\mathcal{L} \sin (bx)=\frac{b}{b^2+s^2}$$and $$\mathcal{L}\left\{\frac{f(t)}{t}\right\}=\int_{s}^{\infty}F(s)\,ds$$