Finding $\int_{-\infty}^\infty |f\ast f'|^2(x)\,dx$ using Plancherel’s theorem

Question

Suppose $G(\mathbb R)\ni f(x),\mathcal{F}[f](\omega)=\frac{1}{1+|w|^3}$ find the value of $$\int_{-\infty}^\infty |f\ast f'|^2(x)\,dx$$

I thought using Plancherel’s theorem \begin{align} \int_{-\infty}^\infty |f\ast f'|^2(x)\,dx &=2\pi\int_{-\infty}^\infty|\mathcal{F}(f\ast f')|\,d\omega\\[2ex] &=2\pi\int_{-\infty}^\infty(\frac{1}{1+|w|^3}\cdot \frac{i\omega}{1+|\omega|^3})^2\,d\omega\\[2ex] &=-4\pi\int_0^\infty(\frac{\omega}{(1+\omega^3)^2})^2\,d\omega \end{align} but here I'm stuck because I can't find any Fourier series so that I'd be able to apply Parseval's identity or any other ways to simplify this integral. How can I simplify the last integral?

Answer 1

Hint: Substitute $u=\omega^3$.