= Solution
Write $K=k_{\max}r$. Using the transform in part (a), the <band-limited linear density correlation> is
$$
\xi(r)=\frac{A}{2\pi^2r}\int_0^{k_{\max}}k^2\sin(kr)\,dk.
$$
Two integrations by parts give the antiderivative $-k^2\cos(kr)/r+2k\sin(kr)/r^2+2\cos(kr)/r^3$. Subtract its value at zero to obtain, for $r>0$,
$$
\boxed{\xi(r)=\frac{A}{2\pi^2r^4}\left[-K^2\cos K+2K\sin K+2(\cos K-1)\right].}
$$
The origin is a removable limit. The bracket starts with $K^4/4$, so
$$
\boxed{\xi(0)=\frac{Ak_{\max}^4}{8\pi^2}.}
$$
This is also the density variance obtained directly by setting $\sin(kr)/(kr)=1$ in the integral. The sharp spectral edge produces oscillatory correlations, with leading large-$r$ term $-Ak_{\max}^2\cos(k_{\max}r)/(2\pi^2r^2)$. A negative correlation at some separation is compatible with a nonnegative power spectrum.
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