= Solution
Let $\sigma_v^2=\langle v(0)^2\rangle$, assumed finite. The displacement is $\Delta X(t)=\int_0^t v(s)\,ds$. By <stationarity>, its two-time covariance is $\langle v(s)v(r)\rangle=\sigma_v^2\rho(s-r)$, with $\rho$ even for a real scalar process. Integrating over the two triangles of the square gives the <Taylor turbulent dispersion> identity
$$
\boxed{\langle\Delta X(t)^2\rangle=2\sigma_v^2\int_0^t(t-s)\rho(s)\,ds.}
$$
A sufficient condition for ordinary long-time variance growth is $\int_0^\infty|\rho(s)|\,ds<\infty$ with $\int_0^\infty\rho(s)\,ds>0$. The <dominated convergence theorem> then gives
$$
\frac{\langle\Delta X(t)^2\rangle}{2t}\longrightarrow D,
\qquad \boxed{D=\sigma_v^2\int_0^\infty\rho(s)\,ds.}
$$
More generally the necessary asymptotic condition for a finite positive variance-based <diffusion coefficient> is convergence of $\sigma_v^2\int_0^t(1-s/t)\rho(s)\,ds$ to that coefficient. Absolute integrability is sufficient, not necessary. A zero integral does not give nondegenerate ordinary <diffusion>. Moreover a second-moment calculation alone does not establish a <normal distribution> or an entire diffusive scaling limit; those require additional probabilistic assumptions.
For the specified positive <autocorrelation>, direct integration gives, when $\alpha\ne1,2$,
$$
\langle\Delta X^2\rangle=2\sigma_v^2\left[\frac{(1+t)^{2-\alpha}-1}{(1-\alpha)(2-\alpha)}-\frac t{1-\alpha}\right].
$$
Thus the <power-law velocity-correlation dispersion> has the following regimes. If $\alpha>1$, the <Lagrangian integral time> is $1/(\alpha-1)$ and \b[ordinary variance growth holds], with $D=\sigma_v^2/(\alpha-1)$. At $\alpha=2$ the exact expression is $2\sigma_v^2[t-\log(1+t)]$. If $\alpha=1$,
$$
\langle\Delta X^2\rangle=2\sigma_v^2[(1+t)\log(1+t)-t]\sim2\sigma_v^2t\log t.
$$
For $0<\alpha<1$,
$$
\boxed{\langle\Delta X^2\rangle\sim\frac{2\sigma_v^2}{(1-\alpha)(2-\alpha)}t^{2-\alpha}.}
$$
The last two regimes spread faster than ordinary <diffusion> and have no finite long-time <diffusion coefficient>. Their slowly decaying velocity memory is responsible. At short times all these correlations instead give the ballistic law $\langle\Delta X^2\rangle\sim\sigma_v^2t^2$.
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