# Integral calculation

1. Oct 1, 2009

### parton

I've done the following short calculation, but I'm not really sure if it is correct or maybe I missed something:

$\int_{a}^{b} \dfrac{1}{(s-i0)(1-s+i0)} \, ds = \int_{a}^{b} \left[ \dfrac{1}{s-i0} - \dfrac{1}{s-1-i0} \right] \, ds = \int_{a}^{b} \left[ pv\left(\dfrac{1}{s}\right) + i \pi \delta(s) - pv\left(\dfrac{1}{s-1}\right) - i \pi \delta(s-1) \right] \, ds$

pv denotes the cauchy principal value. And the 0 term should be understood as the limit $$\epsilon \to + 0$$.

Now I have found something in a book which is really confusing:
$$\int_{a}^{b} \dfrac{1}{(s-i0)^{2}(1-s+i0)} \, ds = \int_{a}^{b} \dfrac{1}{s(1-s+i0)} \, ds$$.

Effectively this would mean: $$\dfrac{1}{(s-i0)^{2}} = \dfrac{1}{s}$$. But I think it appears questionable whether this is correct.
I think one has to do a calculation similar to that given above, in particular we would have: $\dfrac{1}{(s-i0)^{2}} = \dfrac{1}{s^{2} - i s 0}$

I'm really confused, could somebody tell me what is "right"?