Integral identity using the transformation formula

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be integrable. I want to show that $$ \int_\mathbb{R}f(x)\,\mathrm{d}\lambda(x) = \int_\mathbb{R} f\left(x-\frac{1}{x}\right)\,\mathrm{d}\lambda(x).$$ I tried to use the transformation formula, but I did not get the identity. I would appreciate any hints.

Answer 1

$\def\d{\mathrm{d}}$Define$$ g_1(x) = x - \frac{1}{x}\ (x < 0), \quad g_2(x) = x - \frac{1}{x}\ (x > 0),$$ then$$ h_1(y) = g_1^{-1}(y) = \frac{1}{2} (y - \sqrt{\smash[b]{y^2 + 4}}),\quad h_2(y) = g_2^{-1}(y) = \frac{1}{2} (y + \sqrt{\smash[b]{y^2 + 4}}). \quad \forall y \in \mathbb{R} $$ Therefore for integrable $f$,\begin{align*} \int_{\mathbb{R}} f\left( x - \frac{1}{x} \right) \,\d x &= \int_{(-\infty, 0)} f(g_1(x)) \,\d x + \int_{(0, +\infty)} f(g_2(x)) \,\d x\\ &= \int_{g_1((-\infty, 0))} f(u) |h_1'(u)| \,\d u + \int_{g_2((0, +\infty))} f(u) |h_2'(u)| \,\d u\\ &= \int_{\mathbb{R}} f(u) · \frac{1}{2} \left( 1 - \frac{u}{\sqrt{u^2 + 4}} \right) \,\d u + \int_{\mathbb{R}} f(u) · \frac{1}{2} \left( 1 + \frac{u}{\sqrt{u^2 + 4}} \right) \,\d u\\ &= \int_{\mathbb{R}} f(u) \,\d u. \end{align*}