How to calculate this? #int_(1/(n+2))^(1/n)[1/x]dx#.

3 Answers
Frederico Guizini S.
May 11, 2017

See the answer below:
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Cesareo R.
May 11, 2017

For #I_n = int_(1/(n+2))^(1/n) floor(1/x) dx#

#I_n= 1/(n+1)+1/(n+2)#

Explanation:

Changing variables

#y = 1/x# we have

#int_(1/(n+2))^(1/n)floor (1/x)dx equiv int_n^(n+2) floor(y) /y^2 dy# or

#int_n^(n+1)n/y^2 dy + int_(n+1)^(n+2)(n+1)/y^2 dy= 1/(n+1)+1/(n+2)#

Cesareo R.
May 12, 2017

Now for #I_n=int_(1/(n+2))^(1/n) [1/x] dx# is

#I_n= (8 (n+1))/(4 n (n+2)+3)#

Explanation:

Changing variables

#y = 1/x# we have

#int_(1/(n+2))^(1/n) [1/x] dx equiv int_n^(n+2) [[y]] /y^2 dy# or

#int_n^(n+1/2)n/y^2 dy + int_(n+1/2)^(n+3/2)(n+1)/y^2 dy+int_(n+3/2)^(n+2)(n+2)/y^2 dy = 1/(3 + 2 n) + (4 (1 + n))/((1 + 2 n) (3 + 2 n)) + n/(n + 2 n^2) = (8 (1 + n))/(3 + 4 n (2 + n))#

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