Evaluate \[ \int_0^\infty \frac{1-\cos(ax)}{x^2}\,dx \] for \(a\in\mathbb R\).
Solution
1. For \(a\neq0\),\[ \int_0^\infty \frac{1-\cos(ax)}{x^2}\,dx = \abs{a} \int_0^\infty\frac{1-\cos t}{t^2}\,dt. \]
The integrand depends on \(a\) only through \(\abs{a}\), since cosine is even. For \(a\neq0\), set \[ t=\abs{a}x. \] Then \(dx=dt/\abs{a}\) and \(x^2=t^2/\abs{a}^2\), which gives the displayed identity.
2. One has\[ \int_0^\infty\frac{1-\cos t}{t^2}\,dt = \int_0^\infty\frac{\sin t}{t}\,dt. \]
For \(0<\varepsilon<R\), integration by parts gives \[ \int_\varepsilon^R\frac{1-\cos t}{t^2}\,dt = \left[-\frac{1-\cos t}{t}\right]_{\varepsilon}^{R} +\int_\varepsilon^R\frac{\sin t}{t}\,dt. \] As \(\varepsilon\downarrow0\), \[ \frac{1-\cos\varepsilon}{\varepsilon}\longrightarrow0, \] and as \(R\to\infty\), \[ \frac{1-\cos R}{R}\longrightarrow0. \] The integral of \(\sin t/t\) converges by Dirichlet’s test, so passage to the improper limits yields the claim.
3. For every \(s>0\),\[ F(s)\coloneqq\int_0^\infty e^{-st}\frac{\sin t}{t}\,dt = \frac\pi2-\arctan s. \]
Fix \(s_0>0\). For \(s\geq s_0/2\), the derivative of the integrand with respect to \(s\) has absolute value at most \(e^{-s_0t/2}\), which is integrable on \([0,\infty)\). Thus differentiation under the integral sign is justified at \(s_0\). Since \(s_0\) was arbitrary, \[ F'(s) = -\int_0^\infty e^{-st}\sin t\,dt = -\frac1{1+s^2}. \] Also \(F(s)\to0\) as \(s\to\infty\): after the substitution \(u=st\), \[ F(s) = \int_0^\infty e^{-u}\frac{\sin(u/s)}u\,du, \] and \[ \left|\frac{\sin(u/s)}u\right|\leq\frac1s, \] so \(\abs{F(s)}\leq1/s\).
Therefore \[ F(s) = -\int_s^\infty F'(v)\,dv = \int_s^\infty\frac{dv}{1+v^2} = \frac\pi2-\arctan s. \]
4. The Dirichlet integral is\[ \int_0^\infty\frac{\sin t}{t}\,dt = \frac\pi2. \]
Fix \(R>0\). On \([0,R]\), dominated convergence gives \[ \lim_{s\downarrow0} \int_0^R e^{-st}\frac{\sin t}{t}\,dt = \int_0^R\frac{\sin t}{t}\,dt. \] For every \(s\geq0\), the function \[ t\longmapsto\frac{e^{-st}}t \] is positive and decreasing on \([R,\infty)\). Dirichlet’s estimate therefore gives, uniformly in \(s\geq0\), \[ \left| \int_R^\infty e^{-st}\frac{\sin t}{t}\,dt \right| \leq \frac{2}{R}. \] Thus the tails are uniformly negligible as \(R\to\infty\), and consequently \[ \int_0^\infty\frac{\sin t}{t}\,dt = \lim_{s\downarrow0}F(s). \] Step 3 gives \[ \lim_{s\downarrow0}F(s)=\frac\pi2. \]
5. For every \(a\in\RR\),\[ \boxed{ \int_0^\infty \frac{1-\cos(ax)}{x^2}\,dx = \frac\pi2\abs{a} }. \]
If \(a=0\), the integrand vanishes identically. If \(a\neq0\), combine steps 1, 2, and 4.
6. Q.E.D.
Step 5 is the required evaluation.