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Inversely Integrating An Inverse (Posted on 2008-10-15) Difficulty: 2 of 5
Evaluate this definite integral:

1
∫ [1/y]-1 dy
0

Note: [x] is the greatest integer ≤ x.

  Submitted by K Sengupta    
Rating: 3.0000 (1 votes)
Solution: (Hide)
Substituting y = 1/z, we have: dy = -(z)-2.dz

Thus, the given integral:

= Integr.(1 to infinity) (z)-2*[z]-1 dz

Now, for any positive integer n, we must have:

Integr.(n to n+1) z-2*[z]-1 dz

= n-1*Integr.(n to n+1) z-2 dz

= n-1( n-1 - (n+1)-1)

= n-2 - ( n-1*(n+1)-1)

= n-2 - ( n-1 - (n+1)-1)

= f(n) (say)

Thus, the required integral:

= Sum (n = 1 to infinity) f(n)

= Sum(n = 1 to infinity) n-2   - 1

= pi 2/6 - 1
(Since, we know that: Sum(n = 1 to infinity) n-2 = pi2/6, by way of Basel problem).

= 0.644934 (correct to six places)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: Numerical answerDaniel2008-10-15 14:34:46
Solutionre: Numerical answer -- then Google searchCharlie2008-10-15 13:11:25
SolutionNumerical answerCharlie2008-10-15 12:28:38
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