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I apologise if this is trivial but I want to know how should I treat integrals having matrices as arguments.

For example consider the integral $$\int_a^b e^{-As}ds$$, $A$ is some $n \times n$ invertible matrix, and $e^{-As}$ is an exponential matrix, how should I compute this?

What I would do is treat it as a usual integral so $$\int_a^b e^{-As}ds =[-e^{-Ax}A^{-1} ]_a^b$$ but here another question rises, is it $=[-e^{-Ax}A^{-1} ]_a^b$ or $[-A^{-1}e^{-Ax}]_a^b$?

Since $e^{-Ax}$ and $A^{-1}$ don't necessarily commute

Can you explain this?

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    $\begingroup$ Why do you think $e^{-Ax}$ and $A^{-1}$ don't commute? Do you have examples? $\endgroup$ Apr 7, 2021 at 23:00
  • $\begingroup$ I think the integral is done componentwise. That is how it is usually done for integrating vector valued functions of a real variable. The fundamental theorem of calculus applies componentwise. Since the total derivative also happens to occur componentwise, your equation is correct if you can show $D(-e^{-Ax}A^{-1})(x) = e^{-Ax}$. $\endgroup$
    – Mason
    Apr 8, 2021 at 19:48

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By definition of the exponential of a matrix, $$e^{-Ax}=\sum_0^\infty\frac{(-1)^n}{n!}A^nx^n\tag1$$ It turns out that this can be integrated term-by-term, and it gives exactly the answer you have guessed.

It should be apparent from equation $(1)$ that $e^{-Ax}$ does, in fact, commute with $A$ and $A^{-1}$, so either way you write it is fine.

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Actually, they do necessarily commute, which is nice!

The exponential here is $$e^{-As}=\sum_{n=0}^\infty \frac{(-As)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^ns^nA^n}{n!}$$ and so, upon distributing, $$e^{-As}A^{-1}=\sum_{n=0}^\infty \frac{(-1)^ns^nA^{n-1}}{n!}=A^{-1}e^{-As}$$ You can be more confident in your integral by distributing the integration through the summation and calculating as usual (we handwave some steps here): $$\begin{align} \int_a^b e^{-As}\ ds &= \int_a^b \sum_{n=0}^\infty \frac{(-1)^ns^nA^n}{n!}\ ds \\ &=\sum_{n=0}^\infty \left(\frac{(-1)^nA^n}{n!}\int_a^b s^n\ ds\right) \\ &=\sum_{n=0}^\infty \left(\frac{(-1)^nA^n}{n!}\left[\frac{s^{n+1}}{n+1}\right]_{s=a}^b\right) \\ &=\left[\sum_{n=0}^\infty \left(\frac{(-1)^nA^n}{n!}\frac{s^{n+1}}{n+1}\right)\right]_{s=a}^b \\ &=\left[\sum_{n=0}^\infty \frac{(-1)^nA^ns^{n+1}}{(n+1)!}\right]_{s=a}^b \\ &=\left[\sum_{n=1}^\infty \frac{(-1)^{n-1}A^{n-1}s^{n}}{n!}\right]_{s=a}^b\\ &=\left[(-1)^{-1}A^{-1}+\sum_{n=1}^\infty \frac{(-1)^{n-1}A^{n-1}s^{n}}{n!}\right]_{s=a}^b \\ &=\left[\sum_{n=0}^\infty \frac{(-1)^{n-1}A^{n-1}s^{n}}{n!}\right]_{s=a}^b \\ &=\left[-A^{-1}\sum_{n=0}^\infty \frac{(-1)^{n}A^{n}s^{n}}{n!}\right]_{s=a}^b \\ &=\left[-A^{-1}e^{-As}\right]_{s=a}^b \end{align} $$

Note that we were able to introduce the constant term $(-1)^{-1}A^{-1}$ for the usual integration reasons: we're taking a difference. Note also the steps which might not always be justified in different settings or with different sums if we were being more rigorous!

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