Confluent Hypergeometric Function of the Second Kind has the Maclaurin series of $z^{1-b}$ and asymptotic series of $\frac{1}{z^a}$, so it appeared that if $b>1$ (for $z\ll1$) and $a>0$(for $z\gg1$) will make it converge.
However, whenever I tried to put it to $z=0$, it went up to $\infty$. Yet when I plot the graph, it seemed as if the square root of the integral is converge.
How to deal with this issue, and does Confluent Hypergeometric Function of the Second Kind being square integratable or not?