I'm having a little difficulty getting the right answer for this. So far I've got
$$F(k) = \frac{1}{2\pi}\int_{-\infty}^0 (1+x)e^{-ikx}dx+ \frac{1}{2\pi}\int_0^{\infty}(1-x)e^{-ikx}dx\\ =\frac{1}{2\pi}\bigg[ -\frac{1}{k^2}e^{ik} -\frac{1}{k^2}e^{-ik}+\frac{2}{k^2} \bigg]\\ =\frac{1}{2\pi}\bigg( \frac{e^{ik / 2}}{ik} - \frac{e^{-ik/2}}{ik} \bigg)^2\\ =\frac{1}{\pi}\frac{\sin^2({k/2})}{k^2}$$
And the question wants to to evaluate using Parseval's theorem $$\int_0^\infty\frac{\sin^4({k/2})}{k^4} $$ For which I get $$\frac{\pi}{2}\int_0^\infty(1-x)^2dx = \frac{\pi}{6} $$ The actual answer is supposed to be $\frac{\pi}{24}$ so I'm not sure where I went wrong here. Any help is appreciated.
You lost a factor of two when squaring $e^{it}-e^{-it}=2i\sin t$. Fixing this gives $$F(k)=\frac2\pi\frac{\sin^2(k/2)}{k^2}. $$ The lost factor of $2$ becomes a factor of $4$ when $F$ gets squared. Parseval's theorem then yields $$ \int_0^\infty\frac{\sin^4({k/2})}{k^4}=\frac{\pi}{8}\int_0^\infty(1-x)^2dx = \frac{\pi}{24}. $$