# 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.

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To avoid the risk of being misinterpreted, you might remove the word "only" in the sentence starting with "I remember..." –  cardinal Aug 16 '11 at 23:49
You can easily generate families of counterexamples by making $f$ separable. –  anon Aug 17 '11 at 0:05
@anon, what does it mean for a function to be separable? –  Gerry Myerson Aug 17 '11 at 0:45
I mean $f(x,y)=g(x)h(y)$ for some $g$ and $h$. Perhaps I'm fudging standard terminology just a wee bit. –  anon Aug 17 '11 at 1:05

Let $f$ be 1 below the diagonal, -1 above.

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ok, can you tell me if my thoughts on this are correct. intuitively the integral should evaluate to 0. but if I use a short and wide rectangle i would capture more of the negative region, vica versa if I were to use a thin and tall rectangle –  brita Aug 17 '11 at 0:52
@brita, yes, that's the idea. Given any real number $\alpha$, you can find a rectangle $R$ with arbitrarily long sides such the integral over $R$ is $\alpha$. –  Gerry Myerson Aug 17 '11 at 1:39

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.

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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 $$\ell(z)=\int_0^{+\infty}\min\{t,z\}g(t)\,\mathrm dt.$$

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