I solved an integral using the variable substitution

$$ x = \frac{u-v}{\sqrt{2}} $$ $$ y = \frac{u+v}{\sqrt{2}} $$

This is interesting because the Jacobian of this transformation is $1$. Particularly, this transformation corresponds to a clockwise rotation of $\pi/4$ around the origin.

The problem involved an integrand containing $xy$, which is transformed to $1/2(u^2-v^2)$.

Now, out of curiosity, are there other transformations in higher dimensions sending $x_1x_2 \cdots x_n$ to $a\cdot(u_1^2 \pm u_2^2 \pm \ldots \pm u_n^2)$ for some constant $a$? What is the Jacobian, if such transformations exist? Do these transformations involve complex coefficients, and what are the transformations with real coefficients?

Edit: As J.G. pointed out in the comments, it should probably be $a\cdot(u_1^n \pm u_2^n \pm \ldots \pm u_n^n)$.

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    $\begingroup$ Are you sure you didn't want the $u_j$ squared? $\endgroup$
    – J.G.
    Commented Dec 25, 2023 at 14:07
  • $\begingroup$ @J.G. yes, thank you. $\endgroup$ Commented Dec 25, 2023 at 14:11
  • $\begingroup$ On second thoughts, they should probably be to the $n$th power. $\endgroup$
    – J.G.
    Commented Dec 25, 2023 at 14:32
  • $\begingroup$ @J.G. mhhh, probably yes. Also, it makes sense under a dimensional point of view. $\endgroup$ Commented Dec 25, 2023 at 14:36
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    $\begingroup$ @Francesco The theory of quadratic forms allows you to transform any expression of the form $\sum_{i,j = 1}^n a_{ij}x_ix_j$ into the form $\sum_{i=1}^n a_i u_i^2$ with a suitable transformation, and with the spectral theorem we can always take this substitution to be an orthogonal transformation (which would have Jacobian $1$). I don't believe that this generalizes nicely to higher exponents. $\endgroup$ Commented Dec 25, 2023 at 22:17

1 Answer 1


There is no such substitution. Moreover, there is no differentiable substitution which transforms the product $x_1 x_2 \cdots x_n$ with $n > 2$ into a sum of the form $f_1(u_1)+ \dots + f_n(u_n)$ where the functions $f_i$ are such that the derivatives $f_i'$ have only finite number of zeros.

Let $x=(x_1, \dots, x_n)$, $u = (u_1, \dots, u_n)$, let $P(x) = \prod_{i = 1}^n x_i$ be the product, and let $F(u)=\sum_{i = 1}^n f_i(u)$.

Assume $F(u) = P(X(u))$ for some function $X \colon \mathbb{R}^n \to \mathbb{R}^n$. Note that if $X_1(u)=0$ and $X_2(u)=0$, then $$\nabla F(u) = \sum_{i = 1}^n \nabla X_i(u)\prod_{j \neq i} X_j(u)=0.$$ Thus we have $\nabla F(u)=0$ for all points $u$ corresponding to the points of the linear subspace $x_1 = x_2 = 0$. If $n > 2$, the number of such points is infinite. On the other hand, $\nabla F(u) = 0$ is just $f_i'(u_i)=0$ for all $i$, and because of our condition on $f_i$, the number of such $u$ is finite. We have a contradiction.

The condition on $f_i$ can be relaxed if instead of cardinality we compare dimension (codimension 2 for $X_1(u) = X_2(u) = 0$ and dimension 0 for $f_1'(u_1) = \dots = f_n'(u_n)=0$). My guess would be that the conditions on $f_i$ and $X$ can be relaxed to just continuity, but I don't know topology well enough to prove this.


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