Let $f\in L^1(\Bbb R)$ How can I prove that: $$\int_\Bbb R f(x)dx = \int_\Bbb R f(x-\frac{1}{x})dx$$
In fact the changes of variables $u=\phi(x) = x-\frac{1}{x}$ seems correct but the image of zero : $\phi(0)$ does not exists.
How can I avoid this issues? thanks
This result (and a generalisation) is proved by Glasser, I transcribe the first part here following the notation in the original post.
From a graph of the relation $\phi = x - 1/x$, one sees that the solution for $x$ has two branches $$x_\pm = \frac{1}{2} \left(\phi \pm \sqrt{\phi^2 + 4}\right), \quad -\infty < \phi < \infty.$$ Thus $$I = \int_{-\infty}^\infty f(\phi) \, dx = \int_{-\infty}^{0^-} f(\phi) \, dx_- + \int_{0^+}^{-\infty} f(\phi) \, dx_+=\int_{-\infty}^\infty f(\phi) (x_-'+x_+')\, d\phi.$$ But $$x_-' + x_+' = \frac{1}{2} \left(1 - \frac{\phi}{\sqrt{\phi^2+4}}+1 + \frac{\phi}{\sqrt{\phi^2+4}}\right) = 1.$$