Integrate of a function in a box with periodic boundary

Question

Defining a function $g(x,y)$ in a 2 dimensional region $0<x<L$, $0<y<L$ and considering a periodic boundary condition for it so that

$g(x=0,y)=g(x=L,y)$ and $g(x,y=0)=g(x,y=L)$,

is it possible to find a function that satisfies the above boundary condition and gives a non-zero value for the below integral?

$\int_0^L \int_0^L dx dy \partial_x g(x,y) $

Answer 1

$\int_0^L \partial_x g(x,y) dx =g(L,y)-g(0,y)=0.$

Answer 2

I don't think that is possible.

Indeed, for all $y$, the function $h : x \mapsto g(x,y)$ is periodic on $]0,L[$. You want to compute $$ A(y)=\int_0^L h'(x)dx= h(L)-h(0)= g(L,y)-g(0,y) = 0$$

and this is true almost everywhere in $]0,L[$ for y. Therefore

$$\int_0^L A(y) dy =0$$

Answer 3

Perform the x integration first:

$\int\limits_a^b \partial_x f(x) dx =f(b)-f(a)$

Thus

$\int\limits_0^L\int\limits_0^L \partial_x f(x,y) dxdy =\int\limits_0^Lf(L,y)-f(0,y) dy=0$

where the last step is due to the boundary conditions.