= Solution
Centre the <infinite square well> at zero. Its <ground state> <wavefunction> is $\psi(x)=\sqrt{2/L}\cos(\pi x/L)$ inside the well and zero outside. The <momentum representation in quantum mechanics> gives the <wavevector> amplitude
$$
\phi(k)=\frac1{\sqrt{2\pi}}\int\psi(x)e^{-ikx}dx
=2\sqrt{\pi L}\,\frac{\cos(kL/2)}{\pi^2-(kL)^2}.
$$
Consequently the unnormalised <wavevector distribution of an infinite-well ground state> is
$$
\boxed{p(k)\propto\frac{\cos^2(kL/2)}{[\pi^2-(kL)^2]^2}.}
$$
The apparent singularities at $k=\pm\pi/L$ are removable. The normalized <probability density> is $4\pi L$ times the boxed expression and has limit $L/(4\pi)$ there. Translating the well changes only the <Fourier transform>'s phase. The <wavevector> is continuously distributed; a box <energy eigenstate> is not a free <momentum eigenstate>.
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