I know how to solve basic Sturm-Liouville problems of the form $$ -(p f')' + q f = \lambda w f , $$ but the following singular SP-Problem, $$ - f''(x) + \log^2(x) f(x) = \frac{1}{x} f(x) \ , $$ with $p=1$, $q = \log^2$, $\lambda=1$ and $w=\frac 1 x$, defined on the positive half-line $\mathbb{R}_+$, with boundary values $ f(0) = f(\infty) = 0 $, seems to evade my knowledge.
I have found the solution $$ f_1(x) = c_1 e^{-x\log(x) + x}, $$ without plugging in the boundary values just yet, and I know that there's a second linear independent one which may be found by the Spectral Parameter Power Series Method as $$ f_2(x) = c_2 f_1(x) \int_0^x \frac{1}{p(t)(f_1(t))^2}\ dt $$ but - contrary to Sturm-Liouville theory as I understand it - neither one is square integrable with respect to the measure $w dx = \frac{dx}{x}$, that is, $$ \int_{0}^\infty (-f''(x) + \log^2(x) f(x)) \overline{f(x)}\ dx =\int_{0}^\infty |f(x)|^2 \frac{dx}{x} $$ diverges, no matter the constants $c_1$, $c_2$.
What's going on here - am I missing something fundamental? Aren't there any non-trivial solutions?
Note that I also realize the parallels to the LHS of a Schrödinger equation for a single particle in the potential $V = \log^2$, although the RHS of the equation is not a constant and thus cannot be interpreted as the energy of a system, but this does not contribute to my understanding.
The limit-point, limit circle classification is usually applied to intervals where one endpoint is finite and regular, and where the other is singular. Singular endpoints can include an infinite endpoint, or a singular coefficient. In your case, it appears you have two singular endpoints. If you were to look at the operator on $(0,a]$ or on $[a,\infty)$, then you would have the proper context in which to study the limit-point/limit-circle cases, but there is no guarantee of $L^2_{x}$ solutions on an interval with two singular endpoints. And, there is no guarantee of $L^2_{x}$ solutions for real eigenvalue parameters, only for complex ones. The classification of limit point and limit circle is associated with the singular endpoint, while the other endpoint is a regular one.
For example, $$ -y''=\lambda y, \;\;\; -\infty < x < \infty, $$ has no $L^2(\mathbb{R})$ solutions, regardless of the choice of $\lambda$. However, if you look at the same equation on $[0,\infty)$, there are $L^2[0,\infty)$ solutions for non-real $\lambda$, but not for real $\lambda$. The fact that there are no $L^2[0,\infty)$ solutions for real $\lambda$ tells you that the equation is in the limit point case because an equation in the limit circle case must also have two independent $L^2$ solutions for real parameters as well.