Firstly $1-\cos(x)=2\big(\frac{1}{2}(1-\cos(x))\big)=2\sin^2(\frac{x}{2})$.
Also $$\sin^2(\frac{x}{2})=\Big(\frac{e^{\frac{1}{2}ix}-e^{-\frac{1}{2}ix}}{2i}\Big)^2=\frac{1}{4}(2-e^{ix}-e^{-ix})$$.
So the integral becomes: $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{(2-e^{ix}-e^{-ix})}{x^2}e^{ikx}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2e^{ikx}-e^{ix(k+1)}-e^{ix(k-1)}}{x^2}dx$$.
We can now consider separate cases of $k$.
When $k=0$, we get $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{2-e^{ix}-e^{-ix}}{x^2}dx$$.
Now I am stuck. The problem now is that $\int_{-\infty}^{\infty}\frac{2}{x^2}dx$ is a divergent integral. And the same problem arises when $|k|=1$. Have I made an error somewhere?
Also the pole at $x=0$ is not simple - how could we use contour integration for a non-simple pole that lies on the real line (I've not seen this before)?
You may use contour integration in the complex plane to evaluate integral. You have written integral in the form $$I(k)=\frac{1}{2}\int_{-\infty}^{\infty}\frac{2e^{ikx}-e^{ix(k+1)}-e^{ix(k-1)}}{x^2}dx$$ Now, you should consider 4 cases
a) $k>1$
b) $k<-1$
c) $k\in[0,1]$
d) $k\in[-1,0]$
We go to the complex plane and use the contour (case a)
This includes $I(k)$, integral along a big half-circle of radius $R\to\infty$ counter clockwise and a small half-circle of radius $r\to 0$ around $x=0$ clockwise.
Integral around this contour $\oint{f}(x)dx=0$, because there are not poles inside the contour. On the other hand $\oint{f}(x)dx=I(k)+\int_R+\int_r=2\pi{i}Res{f}(x)\Rightarrow I(k)=-\int_R-\int_r$, if $Res{f}(x)=0$
Integral along a big half-circle $\to0$ at $R\to\infty$ due to Jordan lemma. $I(k)=-\int_r =-\frac{1}{2}\int_r\frac{2e^{ikx}-e^{ix(k+1)}-e^{ix(k-1)}}{x^2}dx\Rightarrow$ $${I(k)}=-\frac{1}{2}\lim_{r\to0}\int_{\pi}^0\frac{2 \exp(ikr e^{it}) -\exp(i(k+1)re^{it})-\exp(i(k-1)re^{it})}{r^2e^{2it}}ire^{it}dt$$
Expanding exponent into a series $\,\exp(ire^{it})=1+ire^{it}-\frac{1}{2}r^2e^{2it}-\frac{i}{3!}r^3e^{3it}+...$ $${I(k)}=\frac{1}{2}\lim_{r\to0}\int_0^{\pi}\frac{2 ikr e^{it} -i(k+1)re^{it}-i(k-1)re^{it}}{r^2e^{2it}}ire^{it}dt=\frac{\pi{i}}{2}\Bigl(2ik-i(k+1)-i(k-1)\Bigr)=0$$
Exactly in the same way $I(k)=0$ in the case b) (the only point - we have to close the contour in the lower half-plane and integrate around the point $x=0$ in the positive direction - counter clockwise)
In the case c) we go clockwise around $x=0$, but close the contour in the upper half-plane for first two terms and in the lower half-plane - for the third term of the integrand. In this case for the third term we will also get a residual at $x=0$ (the pole is inside the closed contour).
$${I(k)}=\frac{1}{2}\lim_{r\to0}\int_0^{\pi}\frac{2 ikr e^{it} -i(k+1)re^{it}-i(k-1)re^{it}}{r^2e^{2it}}ire^{it}dt-\frac{1}{2}2\pi{i} Res\Bigl(-\frac{e^{ix(k-1)}}{x^2}\Bigr)|_{x=0}=\frac{\pi{i}}{2}\Bigl(2ik-i(k+1)-i(k-1)\Bigr)+\pi{i}i(k-1)=\pi(1-k)$$
In the same way we evaluate the integral in the case d)
$I(k)=0$ for $k<-1$ and $k>1$
$I(k)=\pi(1-k)$ for $k\in[0,1]$
$I(k)=\pi(1+k)$ for $k\in[-1,0]$