Let us consider first the improper integral $$I(k) \;:=\; \int_0^\infty\frac{1-\cos{kx}}{x^2}\,dx.$$ The derivative $I'(k)$ may be formed by differentiating under the integral sign: $$I'(k) \;=\; \int_0^\infty\left(\frac{\partial}{\partial k}\frac{1-\cos{kx}}{x^2}\right) dx \;=\; \int_0^\infty\frac{\sin{kx}}{x}\,dx \;=\; \int_0^\infty\frac{\sin{t}}{t}\,dt$$ Here, the last form has been gotten by the substitution $kx = t$ . But since by the parent entry we have $$\int_0^\infty\frac{\sin{t}}{t}\,dt \;=\; \frac{\pi}{2}$$ and since $I(0) = 0$ , we can write $$I(k) \;=\; \int_0^k\frac{\pi}{2}\, dk \;=\; \frac{\pi k}{2}.$$ Thus we have evaluated the integral $I(k)$ :

(1)

The formula (1) gives $$I(1) \;=\; \int_0^\infty\frac{1-\cos{x}}{x^2}\,dx \;=\; \frac{\pi}{2}.$$ We use here the consequence formula $$1-\cos{x} \;=\; 2\sin^2{\frac{x}{2}}$$ of the double angle formula $\cos{2\alpha} = 1-2\sin^2{\alpha}$ , obtaining $$\frac{\pi}{2} \;=\; 2\int_0^\infty\frac{\sin^2\frac{x}{2}}{x^2}\,dx \;=\; \int_0^\infty\frac{\sin^2{u}}{u^2}\, du,$$ where the substitution $\frac{x}{2} = u$ has produced the last form. Accordingly, we can write as result the formula

(2)