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In the context of a measure theory and integration couse the following exercise came up:

Let $$Q := (0,1) \times (0,1) \subset \mathbb{R}^2$$ and consider the map $$ f : Q \to \mathbb{R},\, (x,y) \mapsto \frac{1}{1 - xy}.$$

Assume that $f$ is Lebesgue-integrable and compute the value of the integral $$\int_{Q} f(x,y)\, d\lambda^2(x,y)$$ using the substitution (i.e. linear transformation)

$$\begin{pmatrix} x \\ y \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix}.$$

My first question is: how do you come up with that quite specific substitution?

The second question is about the solution to the problem that I actually received but did not fully comprehend since many steps have been omitted. According to the solution, the first thing one should have done there is to come up with the set $$Q' := \{ (u,v) \in \mathbb{R}^2 \vert 0 < u < \sqrt{2}, -\min(u, \sqrt{2} - u) < v < \min(u, \sqrt{2} - u)\}$$ quite magically $\textit{by definition}$ and then perform the relatively easy change of variables under the linear transformation $\phi : Q' \to Q$ given above. Hence, my second question: How to come up with the set $Q'$ analytically? In the solution it was introduced by definition 'falling from the sky'. Can one derive it in any way? I would have never arrived there frankly just by guessing. I heard that there might be a geometrical reason which is based on the rotation of the cube $Q$ but I couldn't read much sense into that.

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    I find that if possible, sometimes drawing out the new transformed region helps to see what the new bounds are in the uv plane. That is take all four sides of the unit square and actually perform the transformation graphically. Then you can make a more succinct definition of it. – 2017-02-15

2 Answers 2

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Looking at $Q^{'}$ is just changing the perspective from the original cube to the square $Q$ as a set over the diagonal of the cube (of length $\sqrt{2}$) bounded by the graphs of two functions (the ones with the $\min$ in the defintion).

The linear transformation is just a stretched rotation which moves the $x$ axis into the diagonal of the original square. (It has the additional nice property of transforming $xy$ into $(u^2-v^2)/2$)

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    Thank you Thomas. But is there a way to derive those $\min$ terms and the whole set $Q'$ in a formal fashion without making a sketch first and then more or less guess them? – 2017-02-15
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A general rotation matrix in $\mathbb{R}^2$ is given by $\begin{pmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{pmatrix}$. For $\theta = \frac{\pi}{4}$ we get exactly your matrix above. So the transformation is a rotation of $45°$ of some cube $Q'$ to the cube $Q := (0,1) \times (0,1) \subset \mathbb{R}^2$. Now we have to write down the set $Q'$. When you draw a picture like enter image description here then you can define $Q'$ and it is exactly $Q' = \{ (u,v) \in \mathbb{R}^2 \vert 0 < u < \sqrt{2}, -\min(u, \sqrt{2} - u) < v < \min(u, \sqrt{2} - u)\}$ like stated.

The good thing about the above transformation is, that the matrix is orthogonal with determinant 1. So when you transform the integral there is no strange determinant term so can ignore it. on the other hand $xy=(\frac{1}{\sqrt{2}}u-\frac{1}{\sqrt{2}}v)(\frac{1}{\sqrt{2}}u+\frac{1}{\sqrt{2}}v)=\frac{1}{2}(u^2+v^2)$ so you have no product of different variables which makes the integrand more handy.

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    Thanks for the visualization. But how do I derive the bounds of the new set $Q'$. It may be a silly thing to ask, but I don't know how you arrive at $\sqrt{2}$ and $\frac{\sqrt{2}}{2}$? And how do the $\min$-terms get into the equation? – 2017-02-15
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    @Taufi that's elementary geometry (Pythagoras theorem) and basic linear algebra (the equation of a linear function related to it's graph). If you don't 'see' (which requires some experience) you have to write down the eqution for the bounding sides of the square and calculate their intersection. In general there will be always the need to try out things and see where you get, there is no final recipe for each and every problem. – 2017-02-15
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    Like Thomas said, to get the $\sqrt{2}$ and the $\frac{\sqrt{2}}{2}$ use Pythagoras (this is a unit square). When you look on the buttom square of the above visualization you can see that $-u < v < u$ for $0 < u < \frac{\sqrt{2}}{2}$. When $u > \frac{\sqrt{2}}{2}$ then we have to choose $-(\sqrt{2}-u) < v < \sqrt{2} - u$ for $v$. So all in all we have the set with the "$\min$" – 2017-02-15