Evaluate this definite integral:

1
∫ [1/y]^-1 dy
0

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

(In reply to Numerical answer by Charlie)

your summation of 1/(n^2(n+1)) for n=1 to infinity can be evaluated analytically

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

thus we simply need to evaluate the sum 1/n^2 and 1/(n(n+1)) and subtract them

1/n^2 is known to be pi^2/6 via the reiman zeta function

1/(n(n+1))=1/n - 1/(n+1) all terms except the first cancel out thus we are left with this sum being 1

thus the desired sum is equal to pi^2/6 -1 or (p^2-6)/6

Posted by Daniel on 2008-10-15 14:34:46