In [this paper](https://www.ams.org/mcom/2008-77-264/S0025-5718-08-02101-7/) on page 8 the author claims that the Taylor expansion for the expression $e^{tD\_V}$ where $D\_V$ is the Lie derivative with respect to a vector field $V$ (defined by $(D\_VG)(x) = \frac{d}{dt}|\_{t=0}V(\phi^t\_V(x))$ and $\phi^t\_V(x)$ is the flow of the differential equation $\dot{\psi} = V(\psi)$ looks like this


$$e^{tD\_V} = I +tD\_V +t^2\int\_0^1(1-\theta)e^{\theta t D\_V}D\_V^2 d\theta $$


I can't wrap my head around the remainder term. Shouldn't it be of third oder in $D\_V$ because the integrand involves the third derivative of the exponential? I was thinking that a change of variables is performed but it didn't work. A related question might be [this](https://mathoverflow.net/questions/309170/exponential-map-lie-derivative-in-variation-for-constant-formula-for-ode) which is concerned with the same paper but ask a different question, still.



Taylor expansion of exponential of a Lie derivative
Maybe I'm missing something, but it seems to me you're overthinking this. Since
$$
\int\_{0}^{1} d\theta (1-\theta ) \theta^{n} = \frac{1}{n+1} -\frac{1}{n+2} = \frac{n!}{(n+2)!}
$$
the third term in your expression is
$$
\sum\_{n=0}^{\infty } \frac{t^{n+2} D\_V^{n+2} }{n!} \frac{n!}{(n+2)!}
= \sum\_{n=2}^{\infty } \frac{t^n D\_V^n }{n!}
$$
exactly as it should be.


