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Question

Consider the function $$f(x,y) = \frac{xy(x^2-y^2)}{(x^2+y^2)^3}$$

Using the substitution $u=x^2+y^2$ and $du=2y dy$, the following integral becomes

$$\int_0^2 \int_0^1 f(x,y) dy dx = \frac{1}{5}$$ Using the substitution $u=x^2+y^2$ and $du=2x dy$, the following integral becomes

$$\int_0^1 \int_0^2 f(x,y) dy dx = -\frac{1}{20}$$ Why the iterated integration are giving different results? In other words, why Fubini theorem fails for this integral?

Is it because both $\int^\infty dy \int^\infty|f(x,y)| dx$, $\int^\infty dx \int^\infty|f(x,y)| dy $ doesn't converge? If so how do I calcualte the integral (it is so hard to calculate it)?

mathnoob123
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  • Maybe you just messed up while integrating? Try putting it into WolframAlpha. – Logan Clark Nov 23 '17 at 15:38
  • No, it is correct. – mathnoob123 Nov 23 '17 at 15:39
  • You posted this question yesterday, no? Remember, for functions of several variables, you need to do a substitution in that many variables, i.e., let $(u, v) = (x^2 + y^2, x^2 - y^2)$ and find the Jacobian. – Sean Roberson Nov 23 '17 at 15:41
  • Acutally this is the question, that I am supposed to answer that why in this condition is the iterrated integration not working. For that matter of posting the question, i dont think I posted any questions yesterday. – mathnoob123 Nov 23 '17 at 15:45
  • Nonetheless this change of variables is incomplete. – Sean Roberson Nov 23 '17 at 15:47
  • Why is that? Can you please explain? – mathnoob123 Nov 23 '17 at 15:48
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    This question is an exact duplicate of your previous (now deleted) question (https://math.stackexchange.com/questions/2533882/non-existence-of-double-integrals). $$\int_{I\times J}f(x,y),d\mu $$ does not exist because $\int_{I}\int_{J}f(x,y),dx,dy$ and $\int_{J}\int_{I}f(x,y),dy,dx$ are different. In particular $f(x,y)\not\in L^1(I\times J)$. So the integral $\int_{I\times J}f(x,y),d\mu$ makes no sense. – Jack D'Aurizio Nov 23 '17 at 18:32
  • @JackD'Aurizio That's the question. Why are those two integrals different. Why does Fubini's theorem fail on them? – mathnoob123 Nov 23 '17 at 18:36
  • @mathnoob123: because the integrand function does not belong to $L^1(I\times J)$, as you prove in your last lines. – Jack D'Aurizio Nov 23 '17 at 18:38
  • @JackD'Aurizio Apologies but I am not aware of those terminologies. The method I used was to convert the integral in polar coordinates which became $\int \frac{sin 4\theta}{4r}$ and stated the integral doesn't converge on the provided domain. Would it be okay for a reasoning? – mathnoob123 Nov 23 '17 at 18:40
  • NO. If a function is not integrable over $I\times J$, the integral over such domain makes no sense, no matter which coordinates you take for describing $I\times J$. You cannot compute something which is not well-defined. – Jack D'Aurizio Nov 23 '17 at 18:42
  • Let us continue this discussion in chat. – mathnoob123 Nov 23 '17 at 18:43

1 Answers1

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Let's go back to calculus I. Suppose you wish to compute

$$ \int_a ^b f(x) \ dx $$

by means of a change of variable $x = g(u)$. Then you'd write $dx = g'(u) \ du$ and (assuming $g$ is one-to-one), the integral becomes

$$ \int_{g^{-1}(a)} ^{g^{-1}(b)} f(g(u)) g'(u) \ du $$

and you proceed as normal.

For functions of several variables, you do something similar, but a full change of variables requires as many variables as there are present in the integral.

My comment said to consider $(u, v) = (x^2 + y^2, x^2 - y^2)$. Why? The objects on the right of the equals sign here appear in the integrand, and just placing $u = x^2 + y^2$ doesn't quite do a change of variables justice, as you need to map $(x, y)$ into $(u, v)$. What you've done was map $(x, y)$ into $(u, y)$, which doesn't really work.

I've mentioned a Jacobian in the comment. This is a scaling factor that is present in a coordinate change. A definition is that if $(x_1, x_2, \ldots, x_n) = (f_1(u), f_2(u), \ldots, f_n(u))$ where $u = (u_1, \ldots, u_n)$ is the coordinate system you're moving into, then the Jacobian $J$ is

$$ J = \left| \frac{\partial (x_1, \ldots, x_n)}{\partial (u_1, \ldots, u_n)} \right| = \det \left( \frac{\partial x_i}{\partial u_j} \right) $$

where the matrix in question is built by taking the derivative of $x_i$ with respect to $u_j$.

Usually the absolute value is taken, so the new volume measure with this change is $dx = |J| \ du$.

Try to follow along here and compute the integral. I have a feeling it should converge once the right substitution is made.

Sean Roberson
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  • The thing is I dont have to find a way to evaluate the value. I have to find the problem with the way we are asked to evaluate. – mathnoob123 Nov 23 '17 at 16:17
  • The substitution depends on two variables so the limits of integration aren't wholly valid. That's why you do this process. – Sean Roberson Nov 23 '17 at 16:19
  • What I am supposed to answer is why the Fubini theorem doesn't apply for this probelm – mathnoob123 Nov 23 '17 at 16:38
  • I have a feeling it should converge once the right substitution is made. : not at all. This is not a matter of applying the right substitution or computing a Jacobian. $f\not\in L^1(I\times J)$, so $$\iint_{I\times J}f(x,y)$$ makes no sense. Both $\int_I \int_J f(x,y),dx,dy$ and $\int_I \int_J f(x,y),dy,dx$ are well defined and convergent, but they simply are different things. – Jack D'Aurizio Nov 23 '17 at 18:40