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Integral calculation

  1. Oct 1, 2009 #1
    I've done the following short calculation, but I'm not really sure if it is correct or maybe I missed something:

    [itex] \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 [/itex]

    pv denotes the cauchy principal value. And the 0 term should be understood as the limit [tex] \epsilon \to + 0 [/tex].

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

    Effectively this would mean: [tex] \dfrac{1}{(s-i0)^{2}} = \dfrac{1}{s} [/tex]. 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: [itex] \dfrac{1}{(s-i0)^{2}} = \dfrac{1}{s^{2} - i s 0} [/itex]

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