Deborah Proctor

Deborah Proctor

Answered

2022-10-18

If:
λ = 0 1 d x 1 + x 3 ;
Then evaluate:
p = lim n ( r = 1 n ( n 3 + r 3 ) n 3 n ) 1 / n ;

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Answer & Explanation

Pradellalo

Pradellalo

Expert

2022-10-19Added 16 answers

First, to evaluate λ, use partial fractions:
1 1 + x 3 = 1 3 ( 1 1 + x x 2 x 2 x + 1 )
The second piece in the parentheses is more amenable to integration when expressed as follows:
x 2 x 2 x + 1 = x 1 / 2 ( x 1 / 2 ) 2 + 3 / 4 3 / 2 ( x 1 / 2 ) 2 + 3 / 4
Now, let
A = 0 1 d x 1 + x = log 2
B = 0 1 d x x 1 / 2 ( x 1 / 2 ) 2 + 3 / 4 = [ log [ ( x 1 2 ) 2 + 3 4 ] ] 0 1 = 0
C = 0 1 d x 3 / 2 ( x 1 / 2 ) 2 + 3 / 4 = 3 [ arctan [ 2 3 ( x 1 2 ) ] ] 0 1 = π 3
Therefore
λ = 1 3 ( A B + C ) = 1 3 log 2 + π 3 3
Now for the limit, which is evaluated by taking logs of both sides:
log p = lim n 1 n r = 0 n log ( 1 + r 3 n 3 )
This is a Riemann sum, in the sense that
lim n 1 n r = 0 n f ( r n ) = 0 1 d x f ( x )
Therefore
log p = 0 1 d x log ( 1 + x 3 ) = [ x log ( 1 + x 3 ) ] 0 1 0 1 d x 3 x 3 1 + x 3 = log 2 3 0 1 d x ( 1 1 1 + x 3 ) = log 2 3 + 3 λ = 2 log 2 3 + π 3 p = 4 e ( 3 π 3 )

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