Free parameters in logistic regression When applying logistic regression, one is essentially applying the following function $1/(1 + e^{\beta x})$ to provide a decision boundary, where $\beta$ are a set of parameters that are learned by the algorithm, and $x$ is an input feature vector. This appears to be the general framework provided by widely available packages such as Python's sklearn. 

This is a very basic question, and can be manually implemented by normalization of the features, but shouldn't a more accurate decision boundary be given by:  $1/(1 + e^{\beta (x - \alpha)})$, where $\alpha$ is an offset? Of course an individual can manually subtract a pre-specified $\alpha$ from the features ahead of time and achieve the same result, but wouldn't it be best for the logistic regression algorithm to simply let $\alpha$ be a free parameter that is trained, like $\beta$? Is there a reason this is not routinely done?
You get the same effect from including a bias term, i.e. $\frac{1}{1+\exp(-\beta x + bias)}$, and the defaults for most software is to include such bias.

To see why, you can do the math, expanding $\beta(x-a)$ and seeing that you can express the sum of those differences as a single variable.