 Solution

Fora finite point $z_0$ is a regular singular point when $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ extend holomorphically to $z_0$. A singularity at infinity is classified by applying this criterion after the change of variable $w=1/z$.

For the displayed Bessel differential equation,At $z=0$, the functions $zP(z)=1$ and $z^{2}Q(z)=z^2-1/4$ are holomorphic, so zero is regular singular. To inspect infinity, write $Y(w)=y(1/w)$. The equation becomesHere $w^{2}Q(w)=w^{-2}-1/4$ is not holomorphic at zero, so infinity is an irregular singular point. Thus

Set $y=z^{-1/2}f$. Direct differentiation makes every lower-order term cancel:Hence $f^{\prime \prime }+f=0$, and on any domain carrying a branch of $\sqrt{z}$ two linearly independent solutions areNear the regular singular point, these behave as $z^{-1/2}$ and $z^{1/2}$, the two Frobenius exponents. At infinity their combinations are $z^{-1/2}{e}^{\pm iz}$; after $w=1/z$ these contain $e^{\pm i/w}$, the exponential behavior characteristic of an irregular singular point.