I'm running into some confusion in classifying singular points of the ODE: $$ y'' + p(x)y' + q(x)y = 0 $$ With $p(x) = 1/x$ and $q(x) = 1$. It's clear to me that zero is a singular point. Now, to classify this singularity we check to see whether or not $xp(x)$ and $x^2 q(x)$ are analytic at zero. This is where my confusion comes in.
So it seems to me, $xp(x) = x (1 / x)$ is not even defined at $x = 0$. But in my class notes, and in various texts, the $x$ would be canceled out. Ie it would be asserted that "$xp(x) = 1$ is analytic at zero". Could someone help me see what I'm missing?
For a reference see the first example from: https://atmos.washington.edu/~breth/classes/AM568/lect/lect15.pdf
edit: I believe this is because we have the limits in mind, but I would like verification.
I believe the authors are just using "$x\cdot p(x)$ is analytic" as shorthand for "the function $x\cdot p(x)$ (defined for $x\not = 0$) has an analytic extension to all of $\mathbb{R}$", so there is an analytic function defined on all real numbers which is equal to $x\cdot p(x)$ wherever that is defined.
It is often convenient to say functions $f$ and $g$ are `equivalent' as functions even if they have equal values everywhere except one, or finitely many, exceptions. For instance, the existence and value of integrals of such functions will be the same. If you are familiar with things like measure theory or point-set topology, there are fancier ways to say two functions are "the same" if they are equal except from a "small" set.
Alternatively, if you want a more `algebraic' way of thinking, say a function $f(x)$ has a pole of order $k$ at $x_0$ if there is a Laurent series $$f(x) = \sum_{n=-k}^\infty a_n (x-x_0)^n$$ which converges on some annulus $(a,x_0-\epsilon)\cup (x_0+\epsilon,b)$. Then a "regular single point" as described in the document is asking that each coefficient $p_k(x)$ has a pole of order (at most) $n-k$.