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NightBreeze
Newbie

Posts: 26
 Odd integration problem   « on: Jun 21st, 2008, 11:17am » Quote Modify

Evaluate

{x/y}{y/x} dxdy

where {x} denotes the fractional part of x: {x} = x - x

 « Last Edit: Jun 25th, 2008, 3:47pm by NightBreeze » IP Logged
Miles
Junior Member

Cambridge, England

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Posts: 95
 Re: Odd integation problem   « Reply #1 on: Jun 24th, 2008, 7:00am » Quote Modify

 hidden: The trick is to find a convenient way to cut up the unit square (0 < x < 1, 0 < y < 1) and sum the integral on each piece.   By symmetry, we can integrate over the half of the square where x > y and double the result.  This ensures that {y/x} = y/x.  We chop this half into regions bounded by y = x / n & y = x / (n+1) & x = 1, for each integer n >=1.  Note that in this region, n < x/y < n+1 so {x/y} = x/y - n.  Putting it all together the result we want is          the sum for n>=1          of the integral of        2.(y/x).(x/y - n) = 2(1 - ny/x)        with respect to x and y        over the region where 0 < x < 1, x/n < y < x/(n+1).   The integration gives me 1/2n - 1/(n+1) + n / (n+1)^2 which rearranges to      (1/2).[1/n - 1/(n+1) - 1/(n+1)^2]   On summing over n>=1, the first two terms in the sum telescope into 1/2 and using a standard result the last term gives   -(1/2).(pi^2 / 6 - 1)

So the answer is 1 - pi^2 / 12.
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NightBreeze
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Posts: 26
 Re: Odd integation problem   « Reply #2 on: Jun 24th, 2008, 10:03am » Quote Modify

Correct.
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