Inversely Integrating An Inverse (Posted on 2008-10-15)

	Submitted by K Sengupta
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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)

	Subject	Author	Date
	re: Numerical answer	Daniel	2008-10-15 14:34:46
	re: Numerical answer -- then Google search	Charlie	2008-10-15 13:11:25
	Numerical answer	Charlie	2008-10-15 12:28:38