The following distributional limit is well-known: lim a → 0 + ( 1 x − i a − 1 x + i a ) = 2 i π δ ( x ) .

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The following distributional limit is well-known:      lim          a      →              0        +                  (          1              x        −        i        a              −          1              x        +        i        a              )    =  2  i  π  δ  (  x  )  .

barbesdestyle2k · Accepted Answer

Define       f    α    (  y  )  :=      1          (      y      −      i              )        α              −      1          (      y      +      i              )        α              ,        I    α    :=      ∫                  R                  f    α    (  y  )  d  y so distributionally      lim          a      →              0        +                  a          α      −      1            (          1              (        x        −        i        a                  )          α                      −          1              (        x        +        i        a                  )          α                      )    =      lim          a      →              0        +                  1    a        f    α        (          x      a        )    =      I    α    δ  (  x  )  .In particular,       I    α    =      ∫                  R                        (      y      +      i              )        α            −      (      y      −      i              )        α                    (              y        2            +      1              )        α              d  y is finite for   ℜ  α  ∈  (  0  ,    1  ), because its integrand is asymptotic to   2  i  α      y          α      −      1            |    y      |  . But you&#039;ve asked about what we get by multiplying this by       a          1      −      α      , which for   ℜ  α  ∈  (  0  ,    1  ) gives 0.

The following distributional limit is well-known: <munder> <mo movablelimits="true" form="pr

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