Calculate \[ \int_0^\infty {\sin^3(x) \over x^3} \, dx .\]
Solution
Let \[ I=\int_0^\infty\frac{\sin^3x}{x^3}\,dx. \] Integrating by parts with \(dv=x^{-3}\,dx\) gives \[ I=\frac32\int_0^\infty\frac{\sin^2x\cos x}{x^2}\,dx, \] because the boundary term vanishes at both \(0\) and \(\infty\). Using \[ \cos x-\cos3x=4\sin^2x\cos x, \] we obtain \[ I=\frac38\int_0^\infty\frac{\cos x-\cos3x}{x^2}\,dx. \]
For \(0<a<b\), integration by parts together with the Dirichlet integral gives \[ \int_0^\infty\frac{\cos(ax)-\cos(bx)}{x^2}\,dx =\int_0^\infty\frac{b\sin(bx)-a\sin(ax)}x\,dx =\frac\pi2(b-a). \] Taking \(a=1\) and \(b=3\) yields \[ \int_0^\infty\frac{\cos x-\cos3x}{x^2}\,dx=\pi. \] Therefore \[ \boxed{I=\frac{3\pi}{8}}. \]