In physics, do we have \begin{align} \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} =(1-q^{-1}) \delta(y)? \end{align} If $y \not\to 1$, then \begin{align} \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} = 0. \end{align} If $y \to 1$, then \begin{align} & \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} \\ & =(1-q^{-1})\left(\frac{1}{1-y} + \frac{1}{1-y^{-1}}\right) \\ & = (1-q^{-1})\left(\sum_{n \geq 0} y^n + \sum_{m \leq 0} y^m\right) \\ & =\left (1-q^{-1}\right)\left(\sum_{n \in \mathbb{Z}} y^n + 1\right) \\ & = \left(1 - q^{-1}\right)\left(\delta(y)+1\right). \end{align} Where is the mistake in my computations? Thank you very much.
Edit: q is a complex number and y is a variable. This type of computations happen in the paper.
I think that maybe the following computations are correct. \begin{align} & \frac{q^{-1}y-1}{y-1}-\frac{qy^{-1}-1}{q(y^{-1}-1)} \\ & =\left(1-q^{-1}y\right)\left(\frac{1}{1-y} + \frac{y^{-1}}{1-y^{-1}}\right) \\ & = \left(1-q^{-1}y\right)\left(\sum_{n \in \mathbb{Z}} y^n\right) \\ & = \left(1-q^{-1}y\right)\delta(y) \\ & = \left(1-q^{-1} \right)\delta(y). \end{align}