Solving Langevin equation In a past exam paper that I am looking at, there is the following question:

Given that the displacement, $\mathbf{x}$, of a particle in $3$-dimensional Brownian motion is given by: $$m\ddot{\mathbf{x}}=-6\pi \mu a\dot{\mathbf{x}}+\mathbf{F}(t)$$
where $m$ is the mass of the particle and $\mathbf{F}(t)$ is a stochastic force. Show that the expectation of squared displacment satisfies:
$$\langle \mathbf{x}\cdot\mathbf{x}\rangle=\frac{k_{B}T}{\pi \mu a}\left(t-t_{0}\left(1-e^{-t/t_{0}}\right)\right)$$
Where $t_{0} = m/6\pi \mu a$. [Use the equipartition theorem to set $\langle\dot{\mathbf{x}}\cdot\dot{\mathbf{x}} \rangle=3k_{B}T/m$].

This is a classic Langevin equation, and so I began by using an integrating factor $e^{\gamma t}$, where:
$$\gamma \equiv \frac{6\pi \mu a}{m}$$
This gives me:
$$\mathbf{v}(t)e^{\gamma t}-\mathbf{v}_{0}=\frac{1}{m}\int_{0}^{t}\mathbf{F}(t^{\prime})e^{\gamma t^{\prime}}\:\mathrm{d}t^{\prime}$$
We know that $\mathbf{x}(t) = \int_{0}^{t}\mathbf{v}(t^{\prime})\:\mathrm{d}t^{\prime}$. We therefore can find:
$$\begin{align}\mathbf{x}(t) &= \int_{0}^{t}\left[\mathbf{v}_{0}e^{-\gamma t^{\prime}}+\frac{e^{-\gamma t^{\prime}}}{m}\int_{0}^{t^{\prime}}\mathbf{F}(t^{\prime\prime})e^{\gamma(t^{\prime\prime})}\mathrm{d}t^{\prime\prime}\right]\mathrm{d}t^{\prime} \\ &= \frac{\mathbf{v}_{0}}{\gamma}\left(1-e^{-\gamma t}\right)-\frac{1}{\gamma m}\int_{0}^{t}\mathbf{F}(t^{\prime\prime})\left(e^{\gamma(t^{\prime\prime}-t)}-1\right)\:\mathrm{d}t^{\prime\prime}\end{align}$$
Squaring:
$$\mathbf{x}(t)\cdot\mathbf{x}(t)=\frac{1}{\gamma^{2}}(1-e^{-\gamma t})^{2} + \frac{2v_{0}}{m\gamma^{2}}\int_{0}^{t}\mathbf{v}_{0}\cdot\mathbf{F}(t^{\prime\prime})(1-e^{\gamma(t^{\prime\prime}-t)})\mathrm{d}t^{\prime\prime}+\frac{1}{m^{2}\gamma^{2}}\left(\int_{0}^{t}\mathbf{F}(t^{\prime\prime})\left(1-e^{\gamma(t^{\prime\prime}-t)}\right)\right)^{2}$$
But taking the expectation of this doesn't yield the correct result? What is the correct way to approach this problem?
 A: Taking the expectation of $\mathbf{x}\cdot\mathbf{x}$, and noting that $\langle \mathbf{F}(t)\rangle = 0$ and $\langle \mathbf{F}(t)\mathbf{F}(t^{\prime})\rangle = \Gamma\delta(t-t^{\prime})$ we get:
$$\langle \mathbf{x}\cdot\mathbf{x}\rangle = \left(v_{0}^{2}-\frac{\Gamma}{2\gamma m^{2}}\right)\left(\frac{1-e^{-\gamma t}}{\gamma}\right)^{2}+\frac{\Gamma}{\gamma^{2}m^{2}}\left(t-\frac{1-e^{-\gamma t}}{\gamma}\right)$$
However, we are told that $\langle \dot{\mathbf{x}}\cdot\dot{\mathbf{x}}\rangle=3k_{B}T/m$. We can find:
$$\langle \dot{\mathbf{x}}\cdot\dot{\mathbf{x}}\rangle = v_{0}^{2}e^{-2\gamma t}+2\frac{v_{0}e^{-\gamma t}}{m}\int_{0}^{t}\langle \mathbf{F}(t^{\prime})\rangle e^{\gamma(t^{\prime}-t)}\:\mathrm{d}t^{\prime}+\frac{1}{m^{2}}\int_{0}^{t}\int_{0}^{t}\langle\mathbf{F}(t^{\prime})\cdot\mathbf{F}(t^{\prime\prime})\rangle e^{\gamma(t^{\prime}-t)}e^{\gamma(t^{\prime\prime}-t)}\:\mathrm{d}t^{\prime}\:\mathrm{d}t^{\prime\prime}$$
Simplifying, we get:
$$\langle \dot{\mathbf{x}}\cdot\dot{\mathbf{x}}\rangle = v_{0}^{2}e^{-2\gamma t} + \frac{\Gamma}{2\gamma m^{2}}(1-e^{-2\gamma t})$$
But we know that $\langle \dot{\mathbf{x}}\cdot\dot{\mathbf{x}}\rangle$ has no time dependence, so $v_{0}^{2} = \frac{\Gamma}{2\gamma m^{2}}$, and $\frac{\Gamma}{2\gamma m^{2}} = \frac{3k_{B}T}{m}$. Therefore:
$$\langle \mathbf{x}\cdot\mathbf{x}\rangle = \frac{\Gamma}{\gamma^{2}m^{2}}\left(t-\frac{1-e^{-\gamma t}}{\gamma}\right) = \frac{k_{B}T}{\pi \mu a}\left(t-t_{0}\left(1-e^{-t/t_{0}}\right)\right)$$
As required.
