regular singular point

A singular point mild enough that xp(x) and x^2 q(x) are still analytic there, so a Frobenius series still works.

A singular point $x_0$ (where $p$ or $q$ fails to be analytic) is regular if $(x-x_0)p(x)$ and $(x-x_0)^2q(x)$ are both analytic at $x_0$ -- exactly the condition the indicial equation's coefficients $p_0,q_0$ are extracted from, and the condition guaranteeing at least one Frobenius series solution exists. A singular point that fails this test is irregular, and the Frobenius method does not apply there.