Integral of $\exp(A t)$ with $A$ antisymetric $3\times 3$ matrix I'm working on some robot pose and velocity interpolation (in 3D).
In my calculations, I end up with an integral I don't know how to compute (other than by numerical integration).
How can I compute : $\int_{t_0}^{t_1}\exp(A t) dt$ knowing that $A$ is a $3\times 3$ anti-symetric matrix.
More precisely $$A=\begin{bmatrix}
0 & -w_z & w_y\\
w_z & 0 & -w_x\\
-w_y & w_x & 0
\end{bmatrix}$$
(from a frame transformation point of view, it's the adjoint of the rotation vector $w$, but this shouldn't matter for the problem at hand)
If $A$ had been invertible, it would have been easy : $A^{-1}\exp(At_1) - A^{-1}\exp(At_0)$. But an anti-symetric matrix is not invertible, so $A^{-1}$ does not exist.
I also tried to express it as series, but get also stuck because $A$ is not invertible:
$$\int \exp(At) dt = \int \sum_{n=0}^{\infty}\frac{(A t)^n}{n!} dt
= \sum_{n=0}^{\infty}\int \frac{(A t)^n}{n!} dt = 
\sum_{n=0}^{\infty}\frac{A^n}{n!}\int t^n dt \\ =
\sum_{n=0}^{\infty}\frac{A^n}{n!} \frac{t^{n+1}}{n+1}=
\sum_{n=0}^{\infty}\frac{A^n   t ^{n+1}}{(n+1)!} = $$
(wrong because $A$ is non invertible)
$$ = A^{-1}\sum_{n=0}^{\infty}\frac{A^{n+1}   t ^{n+1}}{(n+1)!}=
 A^{-1}\sum_{n=1}^{\infty}\frac{A^n   t^n}{n!} =
A^{-1}(\exp(A t)-\operatorname{Id})
$$
Do you have any idea how to compute this integral? Ideally in closed form, or numerically in a more precise/rapid way than numerical integration?
Thanks a lot in advance
 A: Note that for your precise matrix, $A^3=-(w_x^2+w_y^2+w_z^2)A$, so $A^{n-1}=-\frac{A^{n+1}}{w_x^2+w_y^2+w_z^2}$ for any $n\geq 2$. In other words, your sum is
\begin{equation}
\begin{split}
\sum_{n=1}^\infty\frac{A^{n-1}t^n}{n!}&=It+\sum_{n=2}^\infty\frac{A^{n-1}t^n}{n!}\\
&=It-\frac{1}{w_x^2+w_y^2+w_z^2}\sum_{n=2}^\infty\frac{A^{n+1}t^n}{n!}\\
&=It-\frac{A}{w_x^2+w_y^2+w_z^2}\left[\exp(At)-At-I\right]
\end{split}
\end{equation}
from which you can compute your integral.
A: $
\def\a{\alpha}\def\b{\beta}\def\t{\theta}
\def\o{{\tt1}}\def\p{\partial}
\def\LR#1{\left(#1\right)}
\def\BR#1{\Big(#1\Big)}
\def\op#1{\operatorname{#1}}
\def\trace#1{\op{Tr}\LR{#1}}
\def\qiq{\quad\implies\quad}
\def\c#1{\color{red}{#1}}
\def\CLR#1{\c{\LR{#1}}}
\def\fracLR#1#2{\LR{\frac{#1}{#2}}}
\def\A{\fracLR{A}{\a}}
$For typing convenience, define the variables
$$\eqalign{
\a &= \sqrt{w^2_x+w^2_y+w^2_z}\qquad K &= \A \qquad \t=\a t \\
}$$
Apply the Rodrigues rotation formula to obtain
$$\eqalign{
R &= \exp(At) \;=\; \exp(K\t) \\
 &= I + K\sin(\a t) + K^2\BR{\o-\cos(\a t)} \\
}$$
$K$ is independent of $t$ and can be pulled out of the integral, while the integrals of the remaining scalar functions are elementary
$$\eqalign{
\int R\,dt
 &= It - K\fracLR{\cos(\a t)}{\a} + K^2\LR{t-\frac{\sin(\a t)}{\a}} \;+\; const \\
}$$
A: You can always find the eigenvectors and eigenvalues of your matrix. Once that has been done, you get that $D = P*A*P^{-1}$, where $D$ is your diagonal matrix, and $P$ is the change of basis matrix.
Now, $e^{At} = e^{P^{-1}DPt} = P^{-1}e^{Dt}P$, and $e^{D}$ is easy to calculate since that is just the diagonal matrix where in each diagonal entry you obtain $e^{d_i}$ where $d_i$ is the $i$-th entry of $D$.
I hope this approach helps.
