I have a smooth function $x \rightarrow f_\epsilon (x)$ on $x\in[-1\ldots 1]$ (dependent on the continuous parameter $\epsilon$) and I want to approximate the integral $ I=\int_{-1}^1 f_\epsilon (x) dx $ Assume I know the analytic (and finite) values of the integrals $ I_0=\int_{-1}^1 f_0 (x) dx $ and $ I_i=\int_{-1}^1 f_{\epsilon,i} (x) dx $ where $f_{\epsilon,i}$ is the $i$th term in a series expansion of $f_{\epsilon}$ about $x=0$, i.e. $ f_{\epsilon}(x) = \sum_{i=1} f_{\epsilon,i}(x) $ Since $x=-1\ldots 1$, the expansion in $x$ is not necessarily a good one. But assume the $f_{\epsilon,i}$ are such that the series of $I_i$ converges (for all $\epsilon$) continuously against $I$ nevertheless. The convergence of $I$ (depending on $\epsilon$) can be slow, however. To which extent can I use the additional knowledge of $I_0$ to improve an approximation of $I$ based on a finite number of terms of it's expansion. The goal is to have a continuous approximation of $I$ as a function of $\epsilon$.
For clarity consider the example $ f_\epsilon (x) = \frac{1}{\sqrt{1+\epsilon-x}} $ where $\epsilon >0$. I get a (finite) series expansion (expanding $f_\epsilon$ about $x=0$ and integrating) $ J(\epsilon) = \frac{1}{64 (1+\epsilon)^{9/2}}(151+544\epsilon+784\epsilon^2+512\epsilon^3+128\epsilon^4) $ In this case I also know the full $I$ (for comparison): $ I(\epsilon) = 2\left[\sqrt{2+\epsilon}-\sqrt{\epsilon}\right] $ and the result for $\epsilon=0$: $ I_0 = 2\sqrt{2} $ The convergence is bad for small values of $\epsilon$ I can however construct approximations which are exact at $\epsilon =0$ and have good convergence for large $\epsilon$ $ I = g(\epsilon) I_0 (\epsilon) + (1-g(\epsilon)) J(\epsilon) $ where $0\leq g(\epsilon)\leq 1$ with $g(0)=1$ and $g(\infty)=0$ is a continuous function of $\epsilon$. However there is a lot of freedom in the choice of $g$. So I wonder if I can construct $g$ such that it yields (fast) convergence for all $\epsilon$, if I increase the number of terms in the series contribution $J$.