I saw this "standard" identity in a physics paper and I was wondering how to prove it \begin{align*} \frac{d}{dx} e^{A+xB}\bigg|_{x = 0} = e^A\int_0^1 e^{A\tau}B e^{-A\tau}\,d\tau \end{align*}
I tried using Baker-Campbell-Hausdorff but I don't really know how to continue and I'm especially confused where the integral comes from.
2023 edit: I made a typo in the formula above, the integrand of the RHS should instead be $e^{-A\tau}Be^{A\tau}$, as in martini's answer.
Consider the function $$ g(s) = \exp(-sA)\frac{\partial}{\partial x}\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0} $$ We have, taking derivatives \begin{align*} g'(s) &= -A\exp(-sA)\frac{\partial}{\partial x}\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0}+ \exp(-sA)\frac{\partial}{\partial x}(A+xB)\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0}\\ &= -A\exp(-sA)\frac{\partial}{\partial x}\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0} + \exp(-sA)B\exp(sA)\\ &\qquad\qquad{} + A\exp(-sA)\frac{\partial}{\partial x}\exp\bigl(s(A+xB)\bigr)\bigr|_{x=0}\\ &= \exp(-sA)B\exp(sA) \end{align*} Now we have \begin{align*} \frac{\partial}{\partial x}\exp(A+xB)\bigr|_{x=0} &= \exp(A)g(1)\\ &= \exp(A)\int_{0}^1 g'(s)\, ds\\ &= \exp(A)\int_0^1 \exp(-sA)B\exp(sA)\, ds \end{align*}
Partial answer: consider $f(x,\tau)=e^{\tau(A+xB)}$. Then
$$\dfrac{d}{dx}e^{A+xB}=\dfrac{\partial f}{\partial x}|_{x=0,\tau=1}=\int_{\tau=0}^{\tau=1}\dfrac{\partial^2 f}{\partial x \partial \tau}|_{x=0}d\tau$$
Now $\dfrac{\partial^2 f}{\partial x \partial \tau}=\dfrac{\partial}{\partial x}[(A+xB)e^{A+xB}]$ and then we have to evaluate at $x=0$. Now we get $Be^{A\tau}+A(...)$, where $(...)$ is related to what we want to calculate... so we need hand it simbolically and pass it to the other side and take common factor. At this I do not know what to do, but it seems quite a good beginning and I hope it provides background enough for another one to fulfill the proof.
