So, suppose we want to evaluate the integral $$\int_{a}^{b+\epsilon c}f(x)\, dx$$ where $f:\mathbb{R}\to\mathbb{R}$ is assumed to be smooth and regular in the integration region $[a,b+\epsilon c]\subset\mathbb{R}$, and $\epsilon\ll 1$ (and also assuming that $\epsilon c\ll 1$). Supposing $f$ is a known function, is it possible to expand this integral perturbatively in powers of $\epsilon$?
That is, is an expansion of the form $$\int_{a}^{b+\epsilon c}f(x)\, dx=\int_{a}^{b}f(x)\, dx+\epsilon(...)+\epsilon^2(...)+...$$ possible? I don't think I've ever seen perturbation theory used in this way, but I'm curious if it can be done.
We can set $F:\Bbb R\to \Bbb R$ as $$F(x) = \int_a^xf(t)\text{d}t.$$ Then, we have $$\int_a^{b + \epsilon c}f(x)\text{d}x = \int_a^{b }f(x)\text{d}x + \int_b^{b + \epsilon c}f(x)\text{d}x.$$
What you wish is an expansion for the second integral. We note that \begin{align} \int_b^{b + \epsilon c}f(x)\text{d}x &= F(b + \epsilon c) - F(b). \end{align}
We expand $F(b + \epsilon c)$ as \begin{align} F(b + \epsilon c) = F(b) + \dfrac{F'(b)}{1!}(\epsilon c) + \dfrac{F''(b)}{2!}(\epsilon c)^2 + \cdots + \dfrac{F^{(n)}(b)}{n!}(\epsilon c)^n + \dfrac{F^{(n+1)}(\xi)}{(n + 1)!}(\epsilon c)^{n+1} && (*) \end{align} for some $\xi \in (b, b+\epsilon c)$.
If you have the additional information that $f$ is analytic, then you can write the above an infinite series like you seemed to have wanted.
Moreover, you can note that $(*)$ can be simplified in terms of $f$ using $F' = f$. Thus, you end up getting everything in terms of $f$. To summarise, you get
\begin{align} &\int_a^{b + \epsilon c}f(x)\text{d}x\\ &= \int_a^{b }f(x)\text{d}x + \dfrac{f(b)}{1!}(\epsilon c) + \dfrac{f'(b)}{2!}(\epsilon c)^2 + \cdots \end{align}