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Take $x(0)=0$. Integrating the velocity solution givesThe initial/noise cross term vanishes under the independence assumption. Evaluating the remaining integral yields the general integrated Ornstein-Uhlenbeck displacementUse the thermal initial velocity and the fluctuation-dissipation relation for a Langevin particle. With diffusion coefficient $D=k_{B}T/\zeta$, this simplifies toFor $t\ll \tau$, expansion gives $⟨x^2⟩=(k_{B}T/m)t^2+O(t^3)$: displacement is initially $v_{0}t$, so the particle is ballistic. For $t\gg \tau$, $⟨x^2⟩=2Dt-2D\tau +o(1)$, hence the leading behavior is one-dimensional diffusion with $D=k_{B}T/(6\pi \mu a)$.

The ballistic conclusion depends on the initial state. For a particle prepared exactly at rest, $v_0=0$, the first expression gives $⟨x^2⟩=A{t}^3/(3m^2)+O(t^4)$ at very short times. A thermal initial velocity, or a nonzero initial velocity variance, is required for the stated $t^2$ mean-square law. It should not be replaced silently by a zero-velocity initial condition.

Set $\zeta =\gamma /m>0$ and $\mathbf{A}={\mathbf{A}}^{\prime }/m$. Multiplying the Langevin equation by $e^{\zeta t}$ and integrating givesThe expected value of $\mathbf{U}$ is zero. For the printed dot-product force covariance, write $C_A(s-s^{\prime })=\varphi (|s-s^{\prime }|)/m^2$, so $\int C_A(y)dy=\tau$. Without any white-noise approximation the trace variance isThe intended short-correlation approximation replaces $C_A(y)$ by $\tau \delta (y)$, where $\delta$ is the Dirac delta. It requires the force correlation time to be much shorter than ${\zeta }^{-1}$ and the observation time. ThenAt thermal equilibrium, the equipartition theorem gives $m⟨|\mathbf{u}{|}^2⟩/2=3k_{B}T/2$. Taking the long-time limit determines the vector noise normalization in inertial Langevin dynamics:This factor six corresponds to the dot-product covariance across three components. If $\varphi$ were defined for one component instead, its integrated strength would be $2\gamma k_{B}T$ before dividing by $m^2$, and the acceleration strength would be $2\zeta k_{B}T/m$.

Finite rapid correlation is not literally a Dirac delta. For a stationary force and local drag, its exact limiting velocity trace isThis follows by splitting the double convolution into $s\ge s^{\prime }$ and $s^{\prime }\ge s$. Thus equipartition fixes the weighted integral, not just $\tau$, unless the short-correlation approximation is made. For example, $C_A(y)=\tau e^{-|y|/t_c}/(2t_c)$ gives $\tau /[2\zeta (1+\zeta t_c)]$. The remaining explicit formulae use the intended continuous-time Gaussian white noise limit.

Assume isotropic jointly Gaussian noise, as appropriate to an isotropic bath. The component covariance isThe scalar moment identities in the question must now be applied to a fixed Cartesian component $X=U_i$, not to the magnitude of the vector. Its variance isIn the $2n$-fold integral for $⟨X^{2n}⟩$, the Isserlis theorem replaces the force product by a sum of complete pairings. Each pairing yields $n$ identical two-time integrals, hence $(s_t^2{)}^n$. Pair the first labelled factor in $2n-1$ ways, then the first unpaired factor in $2n-3$ ways, and continue. The Gaussian even-moment pairing count isOdd force products vanish, soFor completeness, summing the scalar moment-generating function givesThe same Gaussian process argument for every linear combination of components gives the vector characteristic function $⟨e^{i\mathbf{k}\cdot \mathbf{U}}⟩=e^{-s_t^2|\mathbf{k}{|}^2/2}$. Inverting this Fourier transform yields the normalized multivariate Gaussian distributionThe initial condition is a velocity Dirac delta as $t\downarrow 0$, and the long-time limit is the thermal velocity density. The converted TeX's exponent is damaged; the PDF has the squared norm displayed here. The scalar pairing formula cannot be interpreted as a vector-norm formula: the radial moments of an isotropic Gaussian vector instead give $⟨|\mathbf{U}{|}^{2n}⟩=(2n+1)!!(s_t^2{)}^n$ in three dimensions.

Integrate the velocity once more. Interchanging the time integrals gives the integrated Ornstein-Uhlenbeck displacementFor the stated fixed initial position and velocity,Let $\mathbf{R}=\mathbf{r}-⟨\mathbf{r}⟩$ and $D=k_{B}T/\gamma$. Its component covariance comes from integrating the squared noise kernel:Therefore the total centered variance isThe mean-square displacement from ${\mathbf{r}}_0$ additionally contains $|{\mathbf{u}}_0{|}^2(1-e^{-\zeta t}{)}^2/{\zeta }^2$. Confusing this deterministic drift contribution with centered variance changes the short-time conclusion.

For times longer than the force correlation time but $\zeta t\ll 1$, the mean motion is ${\mathbf{r}}_0+{\mathbf{u}}_{0}t+O(t^2)$ and the centered trace is $2\zeta (k_{B}T/m)t^3+O(t^4)$. Thus a nonzero fixed initial velocity produces a leading ballistic mean-square displacement $|{\mathbf{u}}_0{|}^2{t}^2$; conditioned on ${\mathbf{u}}_0=0$, the white-noise displacement variance instead starts as $t^3$. If one averages over an independent equilibrium initial velocity, its covariance adds $3(k_{B}T/m)(1-e^{-\zeta t}{)}^2/{\zeta }^2$, givingThis is the usual thermal ballistic regime, before velocity memory has relaxed. At $\zeta t\gg 1$, the conditional mean approaches ${\mathbf{r}}_0+{\mathbf{u}}_0/\zeta$, while the centered trace is $6Dt-9D/\zeta +O(e^{-\zeta t})$. Its leading behavior is diffusion with $⟨|\mathbf{R}{|}^2⟩\sim 6Dt$ and $D=k_{B}T/\gamma$, as successive displacement increments lose inertial memory. At times below a finite force correlation time the white-noise short-time expansion is not applicable; for a regular force covariance and fixed initial velocity the leading noise-induced displacement variance is of order $t^4$.