I would like to see a rigorous proof that: lim R → ∞ ∫ 1 R ( 1 x 2 − 1 − 1 x ) d x 1 − x 2 / R 2 = ∫ 1 ∞ ( 1 x 2 − 1 − 1 x ) d x .

Question

I would like to see a rigorous proof that:      lim          R      →      ∞            ∫    1    R        (          1                        x          2                −        1              −          1      x        )              d      x              1      −              x        2                    /                    R        2              =      ∫    1    ∞        (          1                        x          2                −        1              −          1      x        )    d  x  .

Korotnokby · Accepted Answer

Define   f  ,  g  ,  h  :  (  2  ,  ∞  )  ×      R    →      R   by (      1    I   is the indicator function of the set I)                    f        (        R        ,        x        )                            =                                            1                              [                1                ,                R                ]                                      (            x            )                                1            −                          x              2                                      /                                      R              2                                                (                      1                                          x                2                            −              1                                −                      1            x                    )                        ,                            g        (        R        ,        x        )                            =                                            1                              [                1                ,                R                                  /                                2                ]                                      (            x            )                                1            −                          x              2                                      /                                      R              2                                                (                      1                                          x                2                            −              1                                −                      1            x                    )                        ,                            h        (        R        ,        x        )                            =                                            1                              (                R                                  /                                2                ,                R                ]                                      (            x            )                                1            −                          x              2                                      /                                      R              2                                                (                      1                                          x                2                            −              1                                −                      1            x                    )                        .            Then   f  =  g  +  h. For   R  &amp;gt;  2 and   x  ∈      R   we have  g  (  R  ,  x  )  ≤      2          3            (          1                        x          2                −        1              −          1      x        )        1          [      1      ,      ∞      )        (  x  )(the right-hand side is integrable on       R  ),      lim          R      →      ∞        g  (  R  ,  x  )  =      (          1                        x          2                −        1              −          1      x        )        1          [      1      ,      ∞      )        (  x  )and  h  (  R  ,  x  )  =                    1                  (          R                      /                    2          ,          R          ]                    (      x      )                      1        −                  x          2                          /                          R          2                    x                        x          2                −        1            (      x      +                        x          2                −        1            )        ≤            4      ⋅              1                  [          0          ,          R          ]                    (      x      )                      1        −                  x          2                          /                          R          2                    (              R        2            −      4              )                  3                      /                    2                        .Therefore, we can apply the dominated convergence theorem to the integral over g, while the integral over h goes to zero:      ∫                  R              h  (  R  ,  x  )        d    x  ≤      4          (              R        2            −      4              )                  3                      /                    2                          ∫    0    R                      d            x              1      −              x        2                    /                    R        2              =            2      π      R              (              R        2            −      4              )                  3                      /                    2                          ⟶          R      →      ∞        0    .This yields      lim          R      →      ∞            ∫                  R              f  (  R  ,  x  )        d    x  =      lim          R      →      ∞            ∫                  R              g  (  R  ,  x  )        d    x      =    DCT        ∫    1    ∞        (          1                        x          2                −        1              −          1      x        )          d    x  =  log  ⁡  (  2  )as claimed.

I would like to see a rigorous proof that: <munder> <mo movablelimits="true" form="prefix">l

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