Calculate \[ \int_0^\infty {\cos(x) - \cos(4x) \over x^2} \, dx .\]
Solution
For \(a>0\), integration by parts gives \[ \int_0^\infty\frac{1-\cos(ax)}{x^2}\,dx =a\int_0^\infty\frac{\sin(ax)}x\,dx. \] The boundary term vanishes at both endpoints. By the standard indented upper-half-plane contour for \(e^{iaz}/z\), \[ \int_0^\infty\frac{\sin(ax)}x\,dx=\frac\pi2. \] Hence \[ \int_0^\infty\frac{1-\cos(ax)}{x^2}\,dx=\frac{\pi a}{2}. \] Subtracting the cases \(a=1\) and \(a=4\) in the appropriate order, \[ \int_0^\infty\frac{\cos x-\cos4x}{x^2}\,dx =\frac{4\pi}{2}-\frac{\pi}{2} =\boxed{\frac{3\pi}{2}}. \]