I have the function $(1-z)^{-z}$, analytic except on $\mathbb{R}_{\geq 1}$
Now in the text, it says the "singular expansion" at $z=1$ is $\displaystyle \frac{1}{1-z} + \log(1-z)+O((1-z)^{1/2})$
I'm not sure what, in general, a singular expansion is... it doesn't look like a laurent series. Also, I'm unsure of how to calculate it, but I have verified through SAGE that this is the taylor expansion at 1... but I don't see how to use taylors formula to find this.
Let $f(z) = (1-z)^{-z}$. We hence get that $$\log(f(z)) = -z \log(1-z) = -\log(1-z) + (1-z) \log(1-z)$$ Hence, we have $$f(z) = e^{-\log(1-z)} \cdot e^{(1-z)\log(1-z)} = \dfrac1{1-z}\left(1+(1-z) \log(1-z) + \dfrac{(1-z)^2 \log^2(1-z)}{2!} + \cdots\right)$$