# two improper integrals

Let us consider first the improper integral

 $I(k)\;:=\;\int_{0}^{\infty}\frac{1-\cos{kx}}{x^{2}}\,dx.$

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

 $I^{\prime}(k)\;=\;\int_{0}^{\infty}\left(\frac{\partial}{\partial k}\frac{1-% \cos{kx}}{x^{2}}\right)dx\;=\;\int_{0}^{\infty}\frac{\sin{kx}}{x}\,dx\;=\;\int% _{0}^{\infty}\frac{\sin{t}}{t}\,dt$

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

 $\displaystyle\int_{0}^{\infty}\frac{1-\cos{kx}}{x^{2}}\,dx\;=\;\frac{\pi k}{2}.$ (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

 $\displaystyle\int_{0}^{\infty}\!\left(\!\frac{\sin{x}}{x}\!\right)^{2}dx\;=\;% \frac{\pi}{2}.$ (2)
Title two improper integrals TwoImproperIntegrals 2013-03-22 18:43:11 2013-03-22 18:43:11 pahio (2872) pahio (2872) 6 pahio (2872) Example msc 26A42 msc 26A06 msc 26A03