Using Identity to Handle Term: \dfrac{1}{(x-i0)(1-x+i0)}

[parton]

There is the identity

$$\dfrac{1}{1-x+i0} = PV \dfrac{1}{x} - i \pi \delta(1-x)$$

PV corresponds to Cauchy principal value.

But how can I handle a term like

$$\dfrac{1}{(x-i0)(1-x+i0)}$$

and how can I use the identity above? I tried several things such as writing

$$\dfrac{1}{(x-i0)(1-x+i0)} = \dfrac{1}{x(1-x) + i (x-1) 0} = PV \dfrac{1}{x(1-x)} - i \pi \delta(x(1-x)) \text{signum}(x-1)$$

But I don't know to handle this term correctly. Could anyone help me please?

 

[kurt.physics]

The term can be written as \dfrac{1}{(x-i0)(1-x+i0)} = \dfrac{1}{(x-i0)(1-x)} + \dfrac{i0}{(x-i0)(1-x)} Using the identity above, we have \dfrac{1}{(x-i0)(1-x+i0)} = PV \dfrac{1}{(x-i0)(1-x)} - i \pi \delta(1-(x-i0)) \text{signum}(x-1) + \dfrac{i0}{(x-i0)(1-x)} This simplifies to \dfrac{1}{(x-i0)(1-x+i0)} = PV \dfrac{1}{(x-i0)(1-x)} - i \pi \delta(1-x) \text{signum}(x-1) where the Cauchy principal value is taken around the point $x = 1$.