This is an old qualifying exam problem that has me stumped.
Suppose $$F(x):=\sum_{n=0}^{\infty}a_n x^n$$ converges in some neighborhood of the origin. I want to compute $$\sup\left\{\delta>0 : \text{ there exists } \epsilon>0 \text{ such that } \int_{-\epsilon}^{\epsilon}\big|F(x)\big|^{-\delta}dx<\infty \right\}$$ The integrals are to be interpreted as improper Riemann integrals if $F(0)=0$.
I'd greatly appreciate any hints or suggestions on how to get started with this.
The $F(0)\ne0$ case is easy.
When $F(0)=0$, the point is that near zero the term with the lowest degree will dominate. So let $k=\min\{n:\ a_n\ne0\}$ (i.e. $k$ is the multiplicity of $0$ as a root of $F$). Then $F^{(k)}(0) =a_k\ne0$. By writing the remainder of the Taylor polynomial we have $$ F(x)=\frac1{k!}\,F^{(k)}(\xi)\,x^k $$ for some $\xi$ between $0$ and $x$. By choosing $\epsilon$ small enough, we deduce from the continuity of $F^{(k)}$ that there exist positive constants $k_1$ and $k_2$ with $$ k_1\,|x|^k\leq|F(x)|\leq k_2\,|x|^k, \ \ \ x\in(-\epsilon,\epsilon). $$
For the improper integral $$ \int_{-\epsilon}^\epsilon|x^k|^{-\delta}\,dx $$ to be finite, we need $k\delta<1$, or $\delta<1/k$. As $$ k_2^{-\delta}\int_{\epsilon}^\epsilon|x^k|^{-\delta}\,dx\leq\int_{\epsilon}^\epsilon|F(x)|^{-\delta}\,dx\leq k_1^{-\delta}\int_{\epsilon}^\epsilon|x^k|^{-\delta}\,dx, $$ the integral in the middle will be finite precisely when $\delta<1/k$. So the desired supremum is $1/k$.