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        riddles >> putnam exam (pure math) >> Odd integration problem
        
(Message started by: NightBreeze on Jun 21^st, 2008, 11:17am)

Title: Odd integration problem
Post by NightBreeze on Jun 21^st, 2008, 11:17am

Evaluate

http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub0.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup1.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/int.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sub0.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sup1.gif {x/y}{y/x} dxdy

where {x} denotes the fractional part of x: {x} = x - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/lfloor.gifxhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/rfloor.gif

The answer is rather surprising.

Title: Re: Odd integation problem
Post by Miles on Jun 24^th, 2008, 7:00am

[hideb]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)
[/hideb]

So the answer is [hide]1 - pi^2 / 12.[/hide]

Title: Re: Odd integation problem
Post by NightBreeze on Jun 24^th, 2008, 10:03am

Correct.