# How to expand an integral of a function as a function of integrals?

Consider an unspecified smooth norm $|\vec{y}| = f(y_1,y_2,...,y_n)$ on $\mathbb{R}^n$. Now consider the following integral $$I = \int_{-\infty}^\infty f(a_1(x),a_2(x),...,a_n(x)) dx$$ where $\{a_i(x)\}_{i=1}^n$ are smooth real functions with compact support in $[0,L]$.

In the limit $L \to 0$ we can then approximate the integral to lowest order by $I \approx f(a_1(x_0),a_2(x_0),...)L$, where $x_0 \in [0,L]$. However, we can also rewrite this approximation using $a_i(x_0) L \approx \int_{-\infty}^\infty a_i dx$ so that $$I \approx f\left(\frac{1}{L}\int_{-\infty}^\infty a_1 dx,\frac{1}{L}\int_{-\infty}^\infty a_2 dx,...\right) L = f\left(\int_{-\infty}^\infty a_1 dx,\int_{-\infty}^\infty a_2 dx,...\right)$$ where I have used the scalable property of the norm.

What is stated above is the integral up to $\mathcal{O}(L)$ order expressed in terms of simple functionals of $a_i$ and without reference to $L$. I am looking for something similar to general, $\mathcal{O}(L^n)$ order.

So my question is:

Is there a way to obtain an asymptotic expansion of $I$ for $L \to 0$, so that

1) $L$ is not stated anywhere in the expansion similarly to the expression above, and

2) the expansion is stated only in terms of the norm, its derivatives, and infinite integrals not involving the norm such as $\int_{-\infty}^\infty x a_i dx, \int_{-\infty}^\infty a_i'' dx,...$?