Find the principal value of $\int_{-\infty}^{\infty}\frac{1-\cos x}{x^2}\,\mathrm dx$

How to find the Cauchy principal value of the following integral

$$\int_{-\infty}^{\infty}\frac{1-\cos x}{x^2}\,\mathrm dx$$

How to start this problem?

Consider following parametric integral for $\alpha \ge0$ $$I(\alpha )=\int_{-\infty}^{\infty}\frac{1-\cos \alpha x}{x^2}\,\mathrm dx$$ We have $I(0)=0$ $$I'(\alpha )=\int_{-\infty}^{\infty}\frac{\sin \alpha x}{x}\,\mathrm dx=\pi$$ $$I(\alpha )=\pi \alpha +c$$ Then $I(0)=\pi \cdot0+c=0 \implies I(a)=\pi a$ $$I(\alpha )=\int_{-\infty}^{\infty}\frac{1-\cos \alpha x}{x^2}\,\mathrm dx=\pi \alpha $$ $$I(1)=\pi$$

$$\large\int_{-\infty}^{\infty}\frac{1-\cos x}{x^2}\,\mathrm dx=\pi $$

\begin{align} \int_{-\infty}^{\infty}\frac{1-\cos x}{x^2}\,\mathrm dx&=\int_{-\infty}^{\infty}\frac{2\sin^2 \left(\frac x2\right)}{x^2}\,\mathrm dx\tag1\\ &=\int_{-\infty}^{\infty}\frac{\sin^2 y}{y^2}\,\mathrm dy\tag2\\ &=-\left.\frac{\sin^2 y}{y}\;\right|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\frac{2\sin y\cos y}{y}\,\mathrm dy\tag3\\ &=0+\int_{-\infty}^{\infty}\frac{\sin 2y}{y}\,\mathrm dy\tag4\\ &=\int_{-\infty}^{\infty}\frac{\sin z}{z}\,\mathrm dz\tag5\\ &=\pi\tag6 \end{align}

Explanation :

$(1)\;$ Use identity $\;\displaystyle2\sin^2 \left(\frac x2\right)=1-\cos x$

$(2)\;$ Use substitution $\;\displaystyle y=\frac{x}{2}$

$(3)\;$ Apply integration by parts by taking $\;\displaystyle u=\sin^2y$ and use the fact that $\;\displaystyle 0\le\sin^2 y\le1$

$(4)\;$ Use identity $\;\displaystyle \sin 2 y=2\sin y\cos y$

$(5)\;$ Use substitution $\;\displaystyle z=2y$

$(6)\;$ $\displaystyle \int_{-\infty}^{\infty}\frac{\sin z}{z}\,\mathrm dz=2\int_{0}^{\infty}\frac{\sin z}{z}\,\mathrm dz$