do the inside first, and thread y like a constant
so, for the inside integral, after integrating, you get
( ln |(1-xy)| )( 1/(-y) ) from 0 to
1
Putting in the values, you get
(-1/y)( ln |1-y| - ln (1)
) = (-1/y)( ln |1-y| )
Our next integral is integrating y from 0 to 1, so 1-y will
always be greater or equals to zero => |1-y| =
(1-y)
this is a special integral, and I admit I forgot how to do it
actually... Found the answer online somewhere, and adapted it
here... we can use power series to solve it
since we integrate (-1/y)( ln (1-y) ) from 0 to
1
http://thesaurus.maths.org/mmkb/entry.html?action=entryById&id=2705
(-1/y)( ln (1-y) ) = x + x^2/2 + x^3/3 + ...)/x = 1 + x/2
+ x^2/3 + ....
Integrating from 0 to 1 we get 1 + 1/2^2 + 1/3^2 + ...
This is the same as summing 1/n^2 from 0 to infinity, and whose
answer is pi^2 / 6
excellent!!!
