Quaternion integration

If the angular velocity is changing continuously, the following holds true

$ q(t)=q(0)\exp\left({\int_{0}^{t}\frac{q_\omega(\tau)}{2}\ d\tau}\right) \tag 1$

Specifications and Data

1. $q(t),q(0)$ represents quaternions
2. $q_\omega(\tau)$ represents the quaternion representation of angular velocity at $\tau$. It implies if $\omega(\tau) \in R^3 $ is the angular velocity,then $q_\omega(\tau)=(0,\omega(\tau))$ at $\tau$
3. Exponent of a quaternion $J=( p,v)$ can be defined as \begin{eqnarray} e^{J}=e^{p}\left(cos|v| ,\frac{v}{|v|}sin|v| \right) \end{eqnarray}$v$ is a vector. If you are given a vector, make it as a quaternion with $p=0$

Question

How do we prove equation $ q(t)=q(0)\exp\left({\int_{0}^{t}\frac{q_\omega(\tau)}{2}\ d\tau}\right) $ precisely?

Solution 1:

http://www.cim.mcgill.ca/~mpersson/docs/quat_calc_notes.pdf gives what I think is essentially the same result, $$ q(t) = q(t_0) \exp\left(\frac 12 \int_{t_0}^t \mathbf{\omega}(\tau) d\tau\right), $$ but this uses the following definition of the derivative of a quaternion as a function of $t$: $$ \frac d{dt} \mathbf q(t) = \lim_{\Delta t\rightarrow 0} \frac 1{\Delta t} 2\log(\mathbf q(t)^{-1}\mathbf q(t + \Delta t)) $$

Differential calculus over quaternions is also discussed in http://web.mit.edu/2.998/www/QuaternionReport1.pdf and http://arxiv.org/pdf/math/0310362.pdf, but those treatments seemed less helpful.