I am confused about the possible values of b in the confluent hypergeometric function of the second kind U(a,b,z). Specifically can b=0? I know that the U function can be expressed as $$U(a,b,z)=\pi \csc (\pi b)\bigg[\frac{ _1F_1(a,b,z)}{\Gamma(a-b+1)}-\frac{ z^{1-b} \ _1F_1(a-b+1,2-b,z)}{\Gamma(a)}\bigg]$$ but this page claims that $_1F_1$ is undefined for $b\le0$, but for b=0 $$_1F_1(a,0,z)=\sum\limits_{k=0}^\infty(a)_k \frac{z^k}{k!} $$ so this claim is not immediately apparent to me.
In addition to this I doubt the validity of this claim for the b=0 case since I have run across output of programs like mathematica and maple in terms of U(a,0,z) for example $$\int_0^\infty \exp(-\beta^2 u^2)\bigg[2\sqrt{b-a+u^2}-2u\bigg]du=$$ $$\frac{U\bigg(-1/2,0,(b-a)\beta^2\bigg)-1}{\beta^2}$$. I am trying to determine if I should believe this output or if it is contradicting the definition of the function.
Yes, $b=0$ is allowed. This is the logarithmic case, see e.g. http://functions.wolfram.com/HypergeometricFunctions/HypergeometricU/06/01/03/02/0001/