Signup
Loginevaluate this
23 posts
-
-
Originally posted by valentinoo:
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
-
-
-
I get my answer from here:
http://www.math.umd.edu/~jmr/241/doubleint.html
My Matlab Window:
>> firstint=int('1/(1-x*y)','y',0,1)
firstint =
-1/x*log(1-x)
>> answer=int(firstint,'x',0,1)
answer =
1/6*pi^2
-
-
-
Originally posted by eagle:
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!!!
-
-
-
Originally posted by eagle:
I get my answer from here:
http://www.math.umd.edu/~jmr/241/doubleint.html
My Matlab Window:
hey eagle, thanks for that...do u know of any other way beside this 2?
-
-
-
erm... I guess you ask your lecturers or friends better liao...
Unless some maths major appear to help in this forum... But I guess you can't expect much for that to happen... The numbers are not a lot even in this country, let alone a forum...
Any reason why there's a need for using surface integrals to solve?
-
