= Solution
Put $F(r)=r^4u^3K(r)$ and retain the scalar trace correlation $C(r)$. For radial functions the <Laplacian> is $\nabla^2C=r^{-2}(r^2C')'$. Multiplying the trace <Kármán-Howarth equation> by the radial volume element and integrating gives
$$
\dot L=4\pi\left[\frac{F'}r\right]_0^\infty
+8\pi\nu[r^2C']_0^\infty.
$$
Regularity at the origin removes the lower endpoints. The stated triple-correlation tail gives $F=u^3(a+b/r+\cdots)$ and hence $F'/r\to0$. Rapid enough decay of the trace correlation removes the viscous endpoint. Therefore
$$
\boxed{\dot L=0.}
$$
For the <Loitsyansky integral>, multiplying by $-4\pi r^4$ gives the nonlinear contribution
$$
\dot I_{\rm nl}=-4\pi\int_0^\infty r^2\left(\frac{F'}r\right)'dr
=-4\pi[rF']_0^\infty+8\pi[F]_0^\infty
=8\pi F(\infty).
$$
The given tail removes $rF'$ but permits a nonzero limit of $F$. The viscous contribution is
$$
\begin{aligned}
\dot I_\nu
&=-8\pi\nu\int_0^\infty r^2(r^2C')'dr\\
&=-8\pi\nu[r^4C']_0^\infty+16\pi\nu[r^3C]_0^\infty
-48\pi\nu\int_0^\infty r^2C(r)dr\\
&=-12\nu L.
\end{aligned}
$$
Here the stronger weighted endpoints vanish under the assumed sufficiently rapid correlation decay. Combining the terms yields
$$
\boxed{\frac{dI}{dt}=8\pi[r^4u^3K]_\infty-12\nu L.}
$$
The conserved <Saffman integral> measures the density of large-volume <momentum> fluctuations established in part (i). Internal interactions redistribute <momentum> without changing that leading extensive <variance> when forcing and far-field <momentum> fluxes are absent. This is a statistical expression of <linear momentum> conservation, not the assertion that kinetic energy is conserved during viscous decay.
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