This post is about a family of integrals that are important when computing state transition matrices in certain navigation problems as well as some other applications, namely: $$ C_n(v, T) : = \int_0^{T} \int_0^{s_n} \ldots \int_0^{s_2}\exp(s_1 [v]_\times) ds_1 \; ds_2 \;\ldots \; ds_n \qquad \; n \in \{1, 2, \ldots \}. $$ where $v \in \mathbb{R}^3$ and $T \geq 0$. If we define $C_0(v, t) := \exp(t[v]_\times)$ then these integrals can also be defined by: $$ C_{n+1}(v, T) = \int_0^T C_n(v, s) \; ds. $$

Useful Series of Functions

The solution of the above integrals can be stated simply in terms of two series of functions: $$ \gamma_0(\theta) = \cos(\theta), \qquad \gamma_{n+1}(\theta) = \int_0^\theta \gamma_n(\alpha) \; d\alpha,\qquad n \in \{0, 1, 2, \ldots \} $$ and: $$ f_n(\theta) = \frac{\gamma_n(\theta)}{\theta^n}, \qquad n \in \{0, 1, 2, \ldots \}. $$ Here are the first view elements of these series:

The significance of these functions is related to the Rodrigues rotation formula via: $$ \begin{equation} \exp([\alpha u]_\times) = \gamma_0(\alpha) I + \gamma_1(\alpha) [u]_\times + \gamma_2(\alpha) uu^T ,\qquad \alpha \geq 0, \; \| u \| = 1. \label{eq:rodrigues_gamma} \end{equation} $$ and: $$ \begin{equation} \exp([v]_\times) = f_0(\lVert v \lVert) I + f_1(\lVert v\rVert) [v]_\times + f_2(\rVert|v \Vert|) vv^T, \qquad v \in \mathbb{R}^n. \label{eq:rodrigues_f} \end{equation} $$

It is important to note that each function $f_n(\cdot)$ has a finite limit an $0$, specifically: $$ f_n(0) = \frac{1}{n!} \qquad n \in \{1, 2, 3, \ldots\}. $$

Proof: We proceed by induction with the base case of $f_1(0) = \lim_{\theta \rightarrow 0}frac{\sin(\theta)}{\theta} = 1$. Now, assuming $f_n(0) = \frac{1}{n!}$, we have: $$ \begin{align} \lim_{\theta\rightarrow 0} f_{n+1}(\theta) &= \lim_{\theta\rightarrow 0} \frac{\gamma_{n+1}(\theta)}{\theta^{n+1}} \\
&= \lim_{\theta\rightarrow 0} \frac{\gamma_{n}(\theta)}{(n+1)\theta^{n}} & (\textrm{L’Hospitals rule})\\
&= \frac{1}{n+1} f_n(0) = \frac{1}{(n+1)!}. \end{align} $$

Solution

The solution of the above integrals is given by: $$ C_n(v, T) = T^n \left(f_n(T \lVert v \rVert) I + f_{n+1}(T \lVert v \rVert) [Tv]_\times + f_{n+2}(T \lVert v \rVert) (Tv)(Tv)^T\right). $$

Proof: When $\lVert v \rVert = 0$ the desired result is easy to verify as $f_n(0) = \frac{1}{n!}$, otherwise let:

• $\theta := T \lVert v \rVert$
• $u := \frac{v}{\lVert v \rVert}$

and apply the change of variables $\alpha_k = s_k \lVert v \rVert $: $$ \begin{align} C_n(v, T) &= \int_0^{T} \int_0^{s_n} \ldots \int_0^{s_2}\exp(s_1 [v]_\times) ds_1 \; ds_2 \;\ldots \; ds_n, \\
&= \frac{1}{\lVert v \rVert^n}\int_0^{\theta} \int_0^{\alpha_n} \ldots \int_0^{\alpha_2}\exp(\alpha_1 [u]_\times) d\alpha_1 \; d\alpha_2 \;\ldots \; d\alpha_n, \\
&= \frac{1}{\lVert v \rVert^n}C_n(u, \theta). \end{align} $$ Next we show that: $$ C_n(u, \theta) = \gamma_n(\theta) I + \gamma_{n+1}(\theta) [u]_\times + \gamma_{n+2}(\theta) uu^T. $$ by induction. The base case of $C_0(u, \theta)$ follows from from Equation $\eqref{eq:rodrigues_gamma}$. That: $$ C_{n+1}(u, \theta) = \int_0^\theta C_n(u, \alpha) \; d\alpha $$ then follows from the definition of the $\gamma_\cdot$ functions and linearity of integration. Finally we have: $$ \begin{align} C_n(v, T) &= \frac{1}{\lVert v \rVert^n}C_n(u, \theta) \\
&= \frac{T^n}{\theta^n}\left( \gamma_n(\theta) I + \gamma_{n+1}(\theta) [u]_\times + \gamma_{n+2}(\theta) uu^T \right) \\
&= T^n\left( f_n(\theta) I + f_{n+1}(\theta) [\theta u]_\times + f_{n+2}(\theta) (\theta u)(\theta u)^T \right). \end{align} $$ Since $\theta u = T \lVert v \rVert u = T v$ this completes the proof.

Power Series of $\gamma_n(\cdot)$ and $f_n(\cdot)$

To wrap up, I want to point out a few properties of the series of functions mentioned above.

It is worth noting that these functions have a simple relationship to the Taylor expansions for sine and cosine. In particular: $$ \begin{align} \gamma_{2m}(\alpha) &= (-1)^m\sum_{k=m}^\infty \frac{\alpha^{2k}}{2k!} (-1)^k \\
\gamma_{2m+1}(\alpha) &= (-1)^m\sum_{k=m}^\infty \frac{\alpha^{2k+1}}{(2k+1)!} (-1)^k \\
\end{align} $$ (starting from $\gamma_0(\alpha) = \cos(\alpha)$ the above can be proved by term-wise integration), and: $$ \begin{align} f_{2m}(\alpha) &= (-1)^m\sum_{k=m}^\infty \frac{\alpha^{2(k-m)}}{2k!} (-1)^k \\
f_{2m+1}(\alpha) &= (-1)^m\sum_{k=m}^\infty \frac{\alpha^{2(k-m)}}{(2k+1)!} (-1)^k. \end{align} $$

From these last two equations we see that:

• The $f_n$ functions are all even.
• $\gamma_n(\theta) + \gamma_{n+2}(\theta) = \frac{\theta^n}{n!}$ for all $n \in \{0, 1, 2, \ldots \}$.
• $f_n(\theta) + \theta^2 f_{n+2}(\theta) = \frac{1}{n!}$ for all $n \in \{0, 1, 2, \ldots \}$.

Finally, these expressions provide a method for approximating these functions near $0$.