= Solution
For the chosen <Fourier transform> convention, <Parseval's identity> is $\int|f|^2dx=(2\pi)^{-1}\int|\widetilde f|^2dk$. For smooth rapidly decaying functions its bilinear form follows by inserting the inverse transform of $g$ into $\int f\overline g\,dx$ and interchanging the integrals:
$$
\int f(x)\overline{g(x)}dx=\frac1{2\pi}\int\widetilde f(k)\overline{\widetilde g(k)}dk.
$$
Take $g=f$. Approximation by such functions extends the identity to <L2 space>, which includes the compact triangular function here.
For that triangle, $\int|f|^2dx=2\int_0^1(1-x)^2dx=2/3$, while $\widetilde f(k)=[\sin(k/2)/(k/2)]^2$. Thus
$$
\frac23=\frac1{2\pi}\int_{-\infty}^{\infty}\left(\frac{\sin(k/2)}{k/2}\right)^4dk
=\frac2\pi\int_0^\infty\left(\frac{\sin u}u\right)^4du,
$$
giving
$$
\boxed{\int_0^\infty\frac{\sin^4u}{u^4}du=\frac\pi3.}
$$
The continuous limiting integrand at zero is one. This is the <sinc fourth-power integral>.
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