two improper integrals

Let us consider first the improper integral

$$I(k):={\int }_0^{\mathrm{\infty }}\frac{1-\cos kx}{x^2}𝑑x.$$

The derivative $I^{\prime }(k)$ may be formed by differentiating under the integral sign (http://planetmath.org/DifferentiationUnderIntegralSign):

$$I^{\prime }(k)={\int }_0^{\mathrm{\infty }}\left(\frac{\partial }{\partial k}\frac{1-\cos kx}{x^2}\right)𝑑x={\int }_0^{\mathrm{\infty }}\frac{\sin kx}{x}𝑑x={\int }_0^{\mathrm{\infty }}\frac{\sin t}{t}𝑑t$$

Here, the last form has been gotten by the substitution (http://planetmath.org/ChangeOfVariableInDefiniteIntegral)  $kx=t$. But since by the parent entry (http://planetmath.org/SineIntegralInInfinity) we have

$${\int }_0^{\mathrm{\infty }}\frac{\sin t}{t}𝑑t=\frac{\pi }{2}$$

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 $I(k)$:

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{1-\cos kx}{x^2}}𝑑x={\displaystyle \frac{\pi k}{2}}.$(1)

The formula (1) gives

$$I(1)={\int }_0^{\mathrm{\infty }}\frac{1-\cos x}{x^2}𝑑x=\frac{\pi }{2}.$$

We use here the consequence formula

$$1-\cos x=\mathrm{\hspace{0.33em}2}{\sin }^2\frac{x}{2}$$

of the double angle formula  $\cos 2\alpha =1-2{\sin }^2\alpha$,  obtaining

$$\frac{\pi }{2}=\mathrm{\hspace{0.33em}2}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^2\frac{x}{2}}{x^2}𝑑x={\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}u}{u^2}𝑑u,$$

where the substitution  $\frac{x}{2}=u$  has produced the last form.  Accordingly, we can write as result the formula

${\displaystyle {\int }_0^{\mathrm{\infty }}}{\left({\displaystyle \frac{\sin x}{x}}\right)}^2𝑑x={\displaystyle \frac{\pi }{2}}.$(2)