Suppose a function $f(x)$ that has a radius of convergence $r_0$ and we want to calculate the integral $$ \int_{-(r_0+\epsilon_1)}^{-(r_0+\epsilon_2)} f(x)\,dx \; $$ with $\epsilon_1\ll r_0$ and $\epsilon_2\ll r_0$. Now we use the Taylor expansion of $f$ (e.g. because $f(x)$ is not analytically integrable) and perform the integration of this second order approximation. I think the approximation of the integral should be much better when choosing the second order Taylor approximation of the integrand than only the zeroth order approximation. On the other side the integration goes beyond the radius of converge. Are there any arguments clarifying which approximation (0th or 2nd order) is better? Currently I would just make the vague (and not that helpful) statement that using the second order approximation is better only if $\epsilon_i$ is very small and otherwise it is better to rely on the 0th order approximation of the integrand.
The background to this question is a numerical integration where I have equidistant intervals that cannot be changed (I have to cope with the given intervals). The integrand is expanded at the center of each interval. Furthermore, I have checked that in some cases the radius of convergence of the Taylor series of the integrand is exceeded when coming close to the boundaries of the interval we are currently considering. Now I have the two options to just take the zeroth order approximation of the integrand for the whole interval or to take the second order approximation of the integrand. The following picture shows an example of the Taylor expansion (0., 1., 2., 10. and 100. order) of the integrand $1/(a+(x-b)^2)$ at the origin with a radius of convergence $r_0=3$. The dark blue curve is the exact integrand. The above described problem then occurs for example if we want to obtain the approximate integral over the interval $[-3.1,3.1]$.
