Trying to solve an inertial navigation problem in terms of quaternions, I ended up with the following indefinite integral $$ I = \int e^{t \ln q} \, p \, e^{-t \ln q} dt, $$ where $q \in \mathbb H$ is a unit quaternion, $p \in \mathbb H$ is a pure imaginary quaternion, and both $q$ and $p$ are constant with respect to the integration variable $t \in \mathbb R$.

Assuming $q \neq 1$, I integrated by parts: $$ I = \int \underbrace{\mathstrut{}e^{t \ln q} p}_u \, \underbrace{\mathstrut{}e^{-t \ln q}}_{v'} \, dt = e^{t \ln q} p e^{-t \ln q} (-\ln q)^{-1} - \int e^{t \ln q} (\ln q) \, p e^{-t\ln q} (-\ln q)^{-1} \, dt, \\ I = e^{t \ln q} p \, e^{-t \ln q} (-\ln q)^{-1} + (\ln q) \, I \, (\ln q)^{-1}, $$ and got the equation $$ (\ln q) \, I - I \, (\ln q) = e^{t \ln q} \, p \, e^{-t \ln q}. $$

From this question I learned that the general solution for the $ax + xb = c$ equation is $$ x= \left( |b|^2 + 2b_0a + a^2 \right)^{-1} \left( ac +c b^* \right), $$ but it did not work for my case because of zero quaternion inversion. I also tried to expand this equation into quaternion components, but came to a system of linear equations with a singular matrix. This looks quite strange to me, since I'm pretty sure this integral should exist.

Could please someone point me to a mistake in my computations and show how to calculate this antiderivative properly? Might there be a well known solution for the problem?


Note that conjugation preserves vectors, so the integrand and hence $\mathbf{I}$ are pure imaginary.

For pure quaternions $\mathbf{a}$ and $\mathbf{b}$,

$$ \mathbf{ab}=-\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\times\mathbf{b}. $$

Thus, writing $``\ln q "=\mathbf{v}$ we have

$$ (\ln q)\mathbf{I}-\mathbf{I}(\ln q)=2\mathbf{v}\times \mathbf{I}=\exp(t\mathbf{v})\mathbf{p}\exp(-t\mathbf{v}). $$

Note that $\mathbf{v}\times\mathbf{I}$ deletes the $\mathbf{v}$-component of $\mathbf{I}$, so this equation fails to "see" this component. However, this actually isn't a problem: conjugating by $\exp(t\mathbf{v})$ fixes the $\mathbf{v}$-component of $\mathbf{p}$, call it $\mathbf{p}_{\|}$, within the integrand for $\mathbf{I}$, so we may conclude the $\mathbf{v}$-component for $\mathbf{I}$ must be $\mathbf{I}_{\|}=t\mathbf{p}_{\|}$.

Applying $\mathbf{v}\times$ to $2\mathbf{v}\times\mathbf{I}_{\perp}=\exp(t\mathbf{v})\mathbf{p}\exp(-t\mathbf{v})$ (with $v=\|\mathbf{v}\|$) yields

$$ -2v^2\mathbf{I}_{\perp}=\mathbf{v}\times\big(e^{t\mathbf{v}}\mathbf{p}e^{-t\mathbf{v}}\big) $$

from which we may conclude

$$ \mathbf{I}=t\big(\mathbf{p}\cdot\frac{\mathbf{v}}{\|\mathbf{v}\|}\big)\frac{\mathbf{v}}{\|\mathbf{v}\|}+\frac{1}{2}\big(e^{t\mathbf{v}}\mathbf{p}e^{-t\mathbf{v}}\big)\times\frac{\mathbf{v}}{\,\,\|\mathbf{v}\|^2}+\mathbf{C}. $$

Another route is to expand the exponentials $\exp(t\mathbf{v})$ using Euler's formula so you can multiply out and integrate directly. This works especially nice if, setting $\hat{\mathbf{v}}=\frac{\mathbf{v}}{\|\mathbf{v}\|}$ you write $\mathbf{p}$ with respect to an oriented orthonormal basis $\{\mathbf{u},\hat{\mathbf{v}},\mathbf{w}\}$.

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