Matrix exponential integral over external parameter

Question

Given a matrix $A$, consider the matrix exponential $e^{Ax}$, where $A$ does not depend on the parameter $x$. How can we evaluate an expression of the following form?

\begin{eqnarray} \int_{a}^{b} dx e^{Ax}=? \end{eqnarray}

Using the series expansion I get:

\begin{eqnarray} \sum_{n=0}^{\infty}\int_{a}^{b}dx\frac{(Ax)^{n}}{n!}=A^{-1}\sum_{n=0}^{+\infty}\frac{(Ax)^{n+1}}{(n+1)!}\bigg|_{a}^{b}=A^{-1}\left(e^{Ax}-1\right)\bigg|_{a}^{b} \end{eqnarray}

Simplifying further gives:

\begin{eqnarray} A^{-1}\left(e^{Ab}-e^{Aa}\right). \end{eqnarray}

Your guess is not correct. That's why you will not obtain it by any correct mean. — KBS, Sep 22 '22 at 15:10
In that case, I am not even sure if one can define the exponential operator (?). But formally, the above series expansion only makes sense if $A$ is an invertible matrix. — Zarathustra, Sep 22 '22 at 15:32
For the general case, you can try to use something like the Jordan decomposition. — Rodrigo de Azevedo, Sep 22 '22 at 15:35
Now that you have gotten a good answer, it's worth noting that that for specific matrices you can often achieve other closed forms that do not depend on the pseudoinverse. These can be more or less useful depending on the context in which you encounter them. — Christian E. Ramirez, Sep 22 '22 at 17:38
@Zarathustra That's not correct, the series expansion makes sense for any matrix. — ViktorStein, Sep 22 '22 at 19:43

greg · Answer 1 · 2022-09-23T07:04:19.183

$ \def\p{\partial} \def\A{A^D} \def\E{e^{tA}} \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\ndx#1{\op{index}\LR{#1}} \def\rnk#1{\op{rank}\LR{#1}} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} \def\fracLR#1#2{\LR{\frac{#1}{#2}}} $Your calculation is okay, except that you failed to consider singular matrices, which may not even be a concern for your intended purposes.

If singular matrices are a concern, then I would recommend consulting
$\;$ Matrix Mathematics: Theory, Facts, and Formulas $\,$ by Bernstein

The full solution involves the Drazin inverse $\A$ $$\eqalign{ \int \E\:dt &= \A\E + \LR{I-A\A}\sum_{j=1}^k \fracLR{t^j}{j!}A^{j-1} +C \\ }$$ where $C$ is a constant of integration, $A^0=I,\,$ and $k$ is the smallest integer such that $$\rnk{A^k} = \rnk{A^{k+1}} \qiq k=\ndx{A}$$

If $A$ is invertible, then $\A$ is the standard matrix inverse and the second term disappears.

The reason that the sum cuts off at $j=k$ is due to the following property of the Drazin inverse $$\eqalign{ A\A A^{j-1} &\ne A^{j-1} \qquad ({\rm for}\:j\le k) \\ A\A A^k &= A^k \\ }$$

The Drazin inverse is a tool which is more theoretical than practical, so here's a nice way to evaluate the integral without calculating the Drazin inverse (which is notoriously difficult).

Better yet, if you have access to a matrix library which provides the phi-function (e.g. ExpoKit) then you can use this exact result $$\eqalign{ \int e^{At}dt = t\cdot\phi(At)\;+\; C \\ }$$

Matrix exponential integral over external parameter

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