Evaluate:

     1 1
    ∫ ∫ {x/y}{y/x} dxdy
    0 0
 
where {n}= n - floor(n)

If x=y, the expression evaluates to zero.
then, since x and y are interchangeable, it suffices to solve this only for the case that x>y.  The case x<y will have the same sum so we can just double the answer we get for the whole thing.

Since x>y, y/x<1 so {y/x}=y/x
So we must carefully consider only x/y.

My plan will be to take this in pieces where floor(x/y)=n
There will be an infinity of these pieces which we can then sum up.

The first piece, n=1
{x/y} = x/y -1
{x/y}{y/x} = 1 - y/x
For a fixed x which I will call a
we need to integrate from a/2 to a the expression
(1-y/a) dy
[the limits of integration assure floor(x/y)=1]
which evaluates to a/8
So a = x/8
Now integrate this from 0 to 1
(x/8)dx = 1/16

Now generalize to piece n
{x/y} = x/y - n
{x/y}{y/x} = n - ny/x
For a fixed x which I will call a/n
we need to integrate from a/n to a/(n+1) the expression
(1-yn/a) dy
[the limits of integration assure floor(x/y)=n]
which evaluates to a/[2n(n+1)²]
So a = x/[2n(n+1)²]
Now integrate this from 0 to 1
x/[2n(n+1)²] dx = 1/[4n(n+1)²]

Now to sum this from 1 to infinity
Σ 1/[4n(n+1)²]
This is beyond me so we turn to WolframAlpha which assures me the sum is
1/2 - π²/24

And finally double this to get the whole thing:

1 - π²/12

Posted by Jer on 2013-03-03 23:10:34