Evaluate:
1 1
∫ ∫ {x/y}{y/x} dxdy
0 0
where {n}= n  floor(n)
let f(n)=floor(n) for brevity
{x/y}{y/x}=
(x/yf(x/y))*(y/xf(y/x) expanding
1x*f(y/x)/yy*f(x/y)/x+f(x/y)*f(y/x)
now there are three cases to consider
x<y then f(x/y)=0 and it simplifies to
1x*f(y/x)/y
now let f(y/x)=k with k some integer greater than 1
this implies that
k<=y/x<k+1
kx<=y<(k+1)x
now y is further restricted by the range [0,1] thus we need
kx<=1 or x<=1/k
now here there are two ranges to split x into
x in [0,1/(k+1)]
in this range y is in [kx,(k+1)x]
and x in [1/(k+1),1/k]
in this range y is in [kx,1]
thus we have two integrals
int(1xk/y,{x,0,1/(k+1)},{y,kx,(k+1)x})
and
int(1xk/y,{x,1/(k+1),1/k},{y,kx,1})
using mathematica to evaluate these integrals gives us
a nice big equation too complicated for me to type here. I take this expression and sum it for k=1 to infinity and get the value
(1/2)(pi^2/24)
next case is if x=y
in this case f(x/y)=f(y/x)=1 and the expression simplifies to zero
finally, we have x>y
by symmetry this is the same as x<y and thus again we get the value
(1/2)(pi^2/24)
adding these two values get the final value of the integral as
1(pi^2/12)
which is approximately
0.17753296657588678176
which seems to agree with Charlies numerical solution

Posted by Daniel
on 20130303 01:28:12 