Evaluate \(\textstyle \int _ { 0 } ^ { \infty } { \frac { \cos x - \cos 4 x } { x ^ { 2 } } } d x\)
Solution
Set \[ F(z)=\frac{e^{iz}-e^{4iz}}{z^2}. \] For \(0<\varepsilon<R\), use the contour consisting of \[ [-R,-\varepsilon], \] the upper semicircle of radius \(\varepsilon\) traversed clockwise from \(-\varepsilon\) to \(\varepsilon\), \[ [\varepsilon,R], \] and the upper semicircle of radius \(R\) traversed counterclockwise.
1. Near \(z=0\),\[ F(z) = -\frac{3i}{z} +\frac{15}{2} +O(z). \]
The exponential expansions give \[ e^{iz} = 1+iz-\frac{z^2}{2}+O(z^3) \] and \[ e^{4iz} = 1+4iz-8z^2+O(z^3). \] Subtracting, \[ e^{iz}-e^{4iz} = -3iz +\frac{15}{2}z^2 +O(z^3). \] Divide by \(z^2\).
2. The integral over the small indented semicircle tends to\[ -3\pi \]as \(\varepsilon\to0\).
Write \[ F(z)=-\frac{3i}{z}+H(z), \] where step 1 shows that \(H\) is bounded near \(0\). On the small semicircle, parametrize \[ z=\varepsilon e^{i\theta}, \qquad \pi\geq\theta\geq0. \] Then \[ \int -\frac{3i}{z}\,dz = \int_\pi^0 (-3i)i\,d\theta = -3\pi. \] The integral of \(H\) tends to zero because \(H\) is bounded while the arc length is \(\pi\varepsilon\).
3. The integral over the large upper semicircle tends to zero as\(R\to\infty\).
For \(z\) in the upper half-plane, \[ \abs{e^{iz}}\leq1 \qquad\text{and}\qquad \abs{e^{4iz}}\leq1. \] Hence on \(\abs{z}=R\), \[ \abs{F(z)} \leq \frac2{R^2}. \] The large arc has length \(\pi R\), so the ML estimate gives \[ \abs{ \int_{\text{large arc}}F(z)\,dz } \leq \frac{2\pi}{R} \longrightarrow0. \]
4. The symmetric indented real-axis integrals satisfy\[ \lim_{\substack{R\to\infty\\ \varepsilon\to0^+}} \left( \int_{-R}^{-\varepsilon} \frac{e^{ix}-e^{4ix}}{x^2}\,dx + \int_{\varepsilon}^{R} \frac{e^{ix}-e^{4ix}}{x^2}\,dx \right) = 3\pi. \]
The indented contour contains no singularities of \(F\), so Cauchy’s theorem gives total contour integral zero. Therefore \[ \begin{aligned} &\int_{-R}^{-\varepsilon}F(x)\,dx +\int_{\varepsilon}^{R}F(x)\,dx\\ &\qquad= -\int_{\text{small arc}}F(z)\,dz -\int_{\text{large arc}}F(z)\,dz. \end{aligned} \] Let \(R\to\infty\) and \(\varepsilon\to0^+\). Steps 2 and 3 give the displayed limit \(3\pi\).
The real part of the integrand extends continuously across \(x=0\), because \[ \cos x-\cos4x=O(x^2). \] Hence \(\int_{-\infty}^{\infty}(\cos x-\cos4x)x^{-2}\,dx\) converges as an ordinary improper integral and equals the real part of the limit \(3\pi\).
5. The whole-line real integral is\[ \int_{-\infty}^{\infty} \frac{\cos x-\cos4x}{x^2}\,dx = 3\pi. \]
Take real parts in step 4.
6. The requested value is\[ \boxed{ \int_0^{\infty} \frac{\cos x-\cos4x}{x^2}\,dx = \frac{3\pi}{2}. } \]
The integrand in step 5 is even, so its whole-line integral is twice the integral over \([0,\infty)\).
7. Q.E.D.
Step 6 is the requested evaluation.