Improper Double Integral Counterexample Let $f: \mathbf{R}^2\to \mathbf{R}$. I want to integrate $f$ over the entire first quadrant, call $D$. Then by definition we have
$$\int \int_D f(x,y) dA =\lim_{R\to[0, \infty]\times[0, \infty]}\int \int_R f(x,y) dA$$
where $R$ is a rectangle. 
I remember vaguely that the above is true if $f$ is positive. In other words, if $f$ is positive, then the shape of the rectangle does not matter.
So this brings me to my question: give a function $f$ such that the shape of the rectangles DO MATTER when evaluating the improper double integral.
 A: Let $f$ be 1 below the diagonal, -1 above. 
A: Observe the following, if $g$ is a function on $\mathbf{R}^2$ with $$g(x,0) = g(0,y) = 0$$ then you have that
$$ \partial_y g(x_0,y) = \partial_yg(0,y) + \int_0^{x_0} \partial^2_{xy}g(s,y) ds $$
So 
$$ g(x,y) + g(0,0) - g(0,y) - g(x,0) = \int_0^x\int_0^y \partial^2_{xy} g(s,t) dtds $$
In other words, it suffices to find a twice continuously differentiable function $g$, vanishing on the coordinate axes, such that $\lim_{r\to\infty} g(r\cos\theta,r\sin\theta)$ is dependent on the angle $\theta$ chosen. 
Let $\phi(r)$ be an arbitrary smooth function such that $\phi(r) = 0$ if $r < 1$ and $\phi(r) = 1$ if $r > 2$. Define 
$$ g(x,y) = \frac{\phi(xy)}{x^2 + y^2} $$
Then for $f(x,y) = \partial^2_{xy} g(x,y)$, you have that for the integrals
$$I(s; a) = \iint_{[0,s]\times [0,as]} f(x,y) dA = \frac{\phi(as^2)}{s^2(1 + a^2)} $$
you have that for any fixed $a > 0$, the limit
$$ \lim_{s\to\infty} I(s;a) = \frac{a}{1+a^2} $$
is dependant on the aspect ratio of the rectangle chosen. 
A: Let $g$ denote an integrable odd function. Define $f$ by $f(x,y)=g(y-x)$. Then, for every nonnegative $z$, the integral of $f$ on the rectangle $(0,x)\times(0,x+z)$ converges to a finite limit $\ell(z)$ when $x\to+\infty$. In general, $\ell(z)$ does depend on $z$, since
$$
\ell(z)=\int_0^{+\infty}\min\{t,z\}g(t)\,\mathrm dt.
$$
