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

Taniyah Estrada 2022-06-30 Answered
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 .
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Answers (1)

Korotnokby
Answered 2022-07-01 Author has 19 answers
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 > 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.

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