Proving that two integrals are proportional to each other (Fourier Analysis)

Proceed from what you have obtained:

It remains to show that $$ \int_{\mathbb R} \frac{|e^{2\pi i h\xi}+e^{-2\pi i h\xi}-2|^2}{|h|^3}dh = C \xi^2\qquad (*) $$

Note that $$ e^{2\pi i h\xi}+e^{-2\pi i h\xi}-2 = -4\sin^2(\pi h\xi) $$ and thus we are looking at $$ \int_{\mathbb R} \frac{16\sin^4(\pi h\xi)}{|h|^3}dh. $$

By change of variable $x = h\xi$, we see that $$ \int_{\mathbb R} \frac{16\sin^4(\pi h\xi)}{|h|^3}dh = 16\xi^2 \int_{\mathbb R} \frac{\sin^4(\pi x)}{|x|^3}dx. $$ (Check this for $\xi > 0$ and $\xi < 0$ separately!)

The integral $$ \int_{\mathbb R} \frac{\sin^4(\pi x)}{|x|^3}dx $$ is clearly convergent, call it $K$. It follows that $C = 16K$ in $(*)$.