Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
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

3 Answers 3

up vote 5 down vote accepted

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

share|improve this answer
    
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

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.