Can b=0 in the confluent hypergeometric function U(a,b,z)?

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.

• Undefined parameters are usually denoted by a bar, ie "-", as in ${}_{1}F_{1}(a; -; x) = \sum_{k=0}^{\infty} (a)_{k} \frac{x^{k}}{k!} = (1-x)^{-a}$. – Leucippus Jul 1 '14 at 16:13

Yes, $b=0$ is allowed. This is the logarithmic case, see e.g. http://functions.wolfram.com/HypergeometricFunctions/HypergeometricU/06/01/03/02/0001/
• So is the claim that $_1F_1$ is undefined for b=0 incorrect? – Anode Jul 1 '14 at 14:27
• @Anode: No, it is not incorrect, but $_1F_1$ is not $U.\;$ For integer $b$ and especially $b=0,-1,-2, \dots$ the $U$ function is defined by a limiting process. See e.g. N.N. Lebedev, Special Functions, Ch.9.10. – gammatester Jul 1 '14 at 14:34
• Then there are limitations on when U can be expressed in terms of $_1F_1$? – Anode Jul 1 '14 at 14:41