$\partial_{\mu} \exp[X] = \int_{0}^{1} d \alpha \exp[\alpha X] \partial_{\mu}{X} \exp[(1-\alpha)X]$

Question

Suppose $X$ is a matrix, I want to prove \begin{align} \partial_{\mu} \exp[X] = \int_{0}^{1} d \alpha \exp[\alpha X] \partial_{\mu}{X} \exp[(1-\alpha)X] \end{align} One naive guess is using the BCH formula and switching the order of $\partial_{\mu} X \exp[(1-\alpha)X]$, but I am not sure how $\alpha$ integrations are done and eventually disappear.

Answer 1

By writing

$$\exp[X] = \lim_{N\to\infty}\left(1+\frac{X}{N}\right)^N \;\; ,$$

we can express $\partial_{\mu} \exp[X]$ as follows

$$ \partial_{\mu} \exp[X] = \lim_{N\to\infty}\sum_{k=1}^{N}\left(1+\frac{X}{N}\right)^{k-1}\frac{\partial_{\mu}X}{N}\left(1+\frac{X}{N}\right)^{N-k} \;\; .$$

Note that this has to be further formally justified because I switched the limit and the derivative. I'll leave that aside. Then, rearranging a bit we can write

$$ \partial_{\mu} \exp[X] = \lim_{N\to\infty}\sum_{k=1}^{N}\left[\left(1+\frac{X}{N}\right)^N\right]^{\frac{k-1}{N}}\partial_{\mu}X\left[\left(1+\frac{X}{N}\right)^N\right]^{1-\frac{k}{N}}{\frac{1}{N}} \;\; .$$

In the limit, this becomes an integral, $k/N \to \alpha$, $1/N \to d\alpha$. Thus,

\begin{align} \partial_{\mu} \exp[X] = \int_{0}^{1} d \alpha \exp[\alpha X] \partial_{\mu}{X} \exp[(1-\alpha)X] \;\; . \end{align}

You can find a slightly different approach in Feynman.