Originally Posted by HallsofIvy You say
but the integral you show is from 1/a to 1/a. Where did that come from? 
The cube is centered at the origin with vertices at coordinates (1/a,1/a,1/a), (1/a,1/a,1/a), (1/a,1/a,1/a), (1/a,1/a,1/a), (1/a,1/a,1/a), (1/a,1/a,1/a), (1/a,1/a,1/a), (1/a,1/a,1/a).
I tried to cheat by calculating one face of the cube only thinking to multiply by 6 afterwards. (cube has 6 faces). I also tried to cheat by dividing a face into four quadrants, ie (x positive, y positive), (x, negative, y positive), (x positive, y negative) and (x negative, y negative). I chose to evaluate with 0 < x < 1/a and 0 < y < 1/a for both positive thinking to multiply the result by 4 as there are 4 quadrants. In other words using symmetry arguments where ever possible.
No, it isn't. The integral of any odd function is even, F(a)= F(a), so that F(a) F(a)= F(a) F(a)= 0. Then integral of any even function is odd, F(a)= F(a), so that F(a) F(a)= F(a)+ F(a)= 2F(a).

Well, if we use an even function like y = x^2 as an example and evaluate it around 1 < x < 1 then I(Y(x)) = x^3/3 where I(Y(1))  I(y(1)) = 1/3   1/3 = 2/3. Ohhh, oops. That's what happens when I try to do this too late at night to battle insomnia. ;)
I tried again but leaving out the constant factor
for brevity.
Then integrating with respect to y
substituting x=1/a for the result and doubling it this time because when we substitute for x = 1/a we are same but for a sign but then we also subtract the two.
substituting in x = 1/a yields
and x = 1/a yields
yielding
which is where I think
SymPy Live led me up the garden path by telling me it was zero.
Doing the same for y = 1/a
and y = 1/a
this time yielding
Ok, that's a lot better than what I had but we still get the total flux is dependent on the area of the sides of the cube and not just the charge inside according to what Gauss's law says so what am I missing here?