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
Best Answer
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.