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Topic: Odd integration problem (Read 812 times) |
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NightBreeze
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Posts: 26
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Odd integration problem
« on: Jun 21st, 2008, 11:17am » |
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Evaluate
   {x/y}{y/x} dxdy
where {x} denotes the fractional part of x: {x} = x -  x
The answer is rather surprising.
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| « Last Edit: Jun 25th, 2008, 3:47pm by NightBreeze » |
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Miles
Junior Member
Cambridge, England
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Posts: 95
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Re: Odd integation problem
« Reply #1 on: Jun 24th, 2008, 7:00am » |
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| 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)
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So the answer is 1 - pi^2 / 12.
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NightBreeze
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Posts: 26
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Re: Odd integation problem
« Reply #2 on: Jun 24th, 2008, 10:03am » |
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Correct.
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