(0)

\begin{align*} \int\frac{\sin^{3}z}{z^{2}}dz & =-\frac{\sin^{3}z}{z}+3\int\frac{\sin^{2}z\cos z}{z}dz\\ & =-\frac{\sin^{3}z}{z}+3\int\frac{\cos z-\cos^{3}z}{z}dz\\ & =-\frac{\sin^{3}z}{z}+3\int\frac{\cos z-\frac{1}{4}\left(\cos3z+3\cos z\right)}{z}dz\\ & =-\frac{\sin^{3}z}{z}+\frac{3}{4}\int\frac{\cos z-\cos3z}{z}dz\\ & =-\frac{\sin^{3}z}{z}+\frac{3}{4}\left(\int\frac{\cos z}{z}dz-\int\frac{\cos3z}{z}dz\right)\\ & =-\frac{\sin^{3}z}{z}+\frac{3}{4}\left(-\int_{z}^{\infty}\frac{\cos t}{t}dt+\int_{z}^{\infty}\frac{\cos3t}{t}dt\right)+C\\ & =-\frac{\sin^{3}z}{z}+\frac{3}{4}\left(-\int_{z}^{\infty}\frac{\cos t}{t}dt+\int_{3z}^{\infty}\frac{\cos t}{t}dt\right)+C\\ & =-\frac{\sin^{3}z}{z}+\frac{3}{4}\left(\Ci\left(z\right)-\Ci\left(3z\right)\right)+C \end{align*} \begin{align*} \int_{0}^{\infty}\frac{\sin^{3}z}{z^{2}}dz & =\left[-\frac{\sin^{3}z}{z}+\frac{3}{4}\left(\Ci\left(z\right)-\Ci\left(3z\right)\right)\right]_{0}^{\infty}\\ & =-\frac{3}{4}\lim_{z\rightarrow+0}\left(\Ci\left(z\right)-\Ci\left(3z\right)\right)\\ & =\frac{3}{4}\lim_{z\rightarrow+0}\left(\Ci\left(3z\right)-\Ci\left(z\right)\right)\\ & =\frac{3}{4}\log3 \end{align*}

(0)-2

\begin{align*} \int_{0}^{\infty}\frac{\sin^{3}x}{x^{2}}dx & =\int_{0}^{\infty}\sin^{3}x\mathcal{L}_{t}\left[tU\left(t\right)\right]\left(x\right)dx\\ & =\int_{0}^{\infty}\sin^{3}x\left(\int_{0}^{\infty}te^{-xt}dt\right)dx\\ & =\int_{0}^{\infty}tdt\int_{0}^{\infty}\sin^{3}xe^{-xt}dx\\ & =\frac{1}{4}\int_{0}^{\infty}tdt\int_{0}^{\infty}\left(3\sin x-\sin3x\right)e^{-xt}dx\\ & =\frac{1}{4}\int_{0}^{\infty}tdt\mathcal{L}_{t}\left[\left(3\sin x-\sin3x\right)U\left(x\right)\right]\left(t\right)\\ & =\frac{1}{4}\int_{0}^{\infty}t\left(3\frac{1}{t^{2}+1}-\frac{3}{t^{2}+3^{2}}\right)dt\\ & =\frac{3}{4}\int_{0}^{\infty}\left(\frac{t}{t^{2}+1}-\frac{t}{t^{2}+3^{2}}\right)dt\\ & =\frac{3}{4}\left[\frac{1}{2}\log\left(t^{2}+1\right)-\frac{1}{2}\log\left(t^{2}+3^{2}\right)\right]_{0}^{\infty}\\ & =\frac{3}{4\cdot2}\log3^{2}\\ & =\frac{3}{4}\log3 \end{align*}