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.
Best Answer
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.