Integral of $\exp(A t)$ with $A$ antisymetric $3\times 3$ matrix

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

$ \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
$$\eqalign{ \int R\,dt &= It - K\fracLR{\cos(\a t)}{\a} + K^2\LR{t-\frac{\sin(\a t)}{\a}} \;+\; const \\ }$$