Invariance of the Lebesgue integral. Problem
Let $f\in L^1(\mathbb{R})$. Show that $\int_{\mathbb{R}}f(x)dx=\int_{\mathbb{R}}f(x-\frac{1}{x})dx$.
Discussion I know the Lebesgue integral is translation invariant (as the Lebesgue measure is), but I have never encountered the above invariance. I thought maybe if I rewrote both integrals as the measure of a set I could show both sets had the same measure, or I could use a change of variables, but nothing has worked yet. The question is a small part of a bigger problem related to fourier transforms.
 A: Let us assume that $f\in C_0^\infty(\mathbb{R})$ so that all of the following integrals exist.
The statement then follows by density.
Note that the function $(0,\infty)\rightarrow\mathbb{R}$, $x\mapsto x-\frac{1}{x}$ is smooth and bijective.
By a change of variables $x=y-\frac{1}{y}$ we see
$$
\begin{align}
\int_{\mathbb{R}} f(x) dx &= \int_0^\infty f\left(x-\frac{1}{x}\right)\left(1+\frac{1}{x^2}\right)dx\\
&=\int_0^\infty f\left(x-\frac{1}{x}\right)dx+\int_0^\infty f\left(x-\frac{1}{x}\right)\frac{1}{x^2}dx
\end{align}$$
But by a substitution $-\frac{1}{x}=y$ in the last integral we see that
$$\int_0^\infty f\left(x-\frac{1}{x}\right)\frac{1}{x^2} dx=\int_{-\infty}^0 f\left(x-\frac{1}{x}\right)dx$$
which gives the desired identity.
A: I'll give another solution (well, mostly the same, but presented in a different light). It is longer, but it does not rely on any trick. Let $T(x) := x-1/x$. Remark that the equation $T(x) = y$ has exactly two solutions, given by :
$$\frac{y \pm \sqrt{4+y^2}}{2}.$$
Or, another fancy way to state this fact is that $T^{-1}$ has two branches. I'll denote them by $T_-^{-1}$ and $T_+^{-1}$, with the corresponding sign in the formula above. Note that $T \circ T_-^{-1} = T \circ T_+^{-1} = id$ on $\mathbb{R}^*$, while $T_-^{-1} \circ T = id$ on $\mathbb{R}_-^*$ and $T_+^{-1} \circ T = id$ on $\mathbb{R}_+^*$.
Then I just use (formally) the change of variables $y := T(x)$.
$$\int_{\mathbb{R}} f(T(x)) dx = \int_{\mathbb{R}_-^*} \frac{f}{|T'|\circ T_-^{-1}}(T(x)) |T'(x)| dx + \int_{\mathbb{R}_+^*} \frac{f}{|T'|\circ T_+^{-1}}(T(x)) |T'|(x) dx,$$
whence:
$$\int_{\mathbb{R}} f(T(x)) dx = \int_{\mathbb{R}} \sum_{\pm} \frac{1}{|T'|\circ T_\pm^{-1}}(y) f(y) dy.$$
This formula generalizes the change of variable to transformations which are not bijections. All we need to prove is that:
$$\sum_{\pm} \frac{1}{|T'|\circ T_\pm^{-1}} \equiv 1.$$
At this point, everything is explicit and you only need to check that this condition holds. It is tedious, but it does not require any insight. For the sake of completeness, I'll give a proof with relatively few computations.
Note that $T' (x) = 1+x^{-2}$, so that:
$$\frac{1}{|T'|\circ T_\pm^{-1}} = \frac{1}{1+ \frac{1}{(T_\pm^{-1})^2}} = \frac{(T_\pm^{-1})^2}{1+ (T_\pm^{-1})^2}.$$
I'll now write $x_\pm$ instead of $T_\pm^{-1} (y)$. Note that $x_\pm$ are the solution to the quadratic equation $x^2 -yx-1 = 0$, so that $x_- x_+ = -1$ and $(x_- x_+)^2 = 1$. Then:
$$\sum_{\pm} \frac{1}{|T'|\circ T_\pm^{-1}} = \sum_{\pm} \frac{x_\pm^2}{1+ x_\pm^2} = \frac{x_-^2+2 (x_- x_+)^2+x_+^2}{1+ x_-^2 + x_+^2 + (x_- x_+)^2} 
= \frac{2+x_-^2+x_+^2}{2+ x_-^2 + x_+^2} = 1.$$
