I'm looking to integrate angular velocity and acceleration into a quaternion. After researching it, I came across this site, which shows that integrating angular velocity alone gives the following: $$ q_{t} = exp(\frac{1}{2}\omega t)q_{0} $$ where $\omega$ is the vector of angular velocities. Numerical integration does give an answer close to this as well.
Finding a reference for how to do this with angular acceleration was a bit more difficult, but I was able to find this paper which gives the following $$ q_{t} = exp(\frac{1}{2}\omega t +\frac{1}{2}\dot{\omega}t^{2})q_{0} $$ where $\dot{\omega}$ is the vector of angular accelerations. I thought this was interesting, since given the resemblance to the kinematic equations, I would have thought the answer would have been this $$ q_{t} = exp(\frac{1}{2}\omega t +\frac{1}{4}\dot{\omega}t^{2})q_{0} $$
And indeed, numerical integration gives an answer closer to the latter. For the tests I was using the difference was minor, so it could also just be that I made a mistake in the code I wrote for the numerical integration.
Can anyone confirm which of these is correct, or if it's a completely different answer. Appreciate it.
Let's compare some series expansions, Taylor and exponential. The DE is $$ \dot q(t)=\frac12\omega(t) q(t). $$ Its derivative is $$ \ddot q = \frac12\dot\omega q+\frac12 \omega\dot q=\left(\frac12 \dot\omega+\frac14\omega^2\right)q $$ This gives a quadratic Taylor polynomial of $$ q(t+\Delta t)=\left(1+\frac12\omega(t)\Delta t+\frac12\left(\frac12 \dot\omega(t)+\frac14\omega(t)^2\right)\Delta t^2\right)q(t)+O(\Delta t^3) $$ One the other hand we get the exponential $$ \exp(a\Delta t+b\Delta t^2)=1+(a\Delta t+b\Delta t^2)+\frac12(a\Delta t+b\Delta t^2)^2+O(\Delta t^3) $$ So yes, it has to be $a=\frac12\omega(t)$ and $b=\frac14\dot\omega(t)$, the non-commutativity of the quaternions does not play a role up to this point.
You would have to look into the sources cited in the cited paper to see if there are non-trivial reasons for the cited formula or if this is a simple typo, as this formula was not further used in the exposed theory.