# Taylor expansion of exponential of a Lie derivative

In this paper 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 which is concerned with the same paper but ask a different question, still.

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.
• It's also, I believe, the integral remainder formula for the first order Taylor polynomial of $f(x) = e^x$, evaluated at $x = tD_V$. See en.wikipedia.org/wiki/… – Deane Yang Jan 20 at 22:09
• So I figured that the integral led my onto a false lead because I thought this would be the integral formula for the remainder and I unsuccessfully tried to fit in the "false" order of $D_V$. – Jakob Möller Jan 21 at 9:25