If $f \in L^1(\mathbb{R})$, prove that $$\int_{-\infty}^{\infty}f(x)\,\mathrm{d}x=\int_{-\infty}^{\infty}f(x-\frac{1}{x})\,\mathrm{d}x.$$
I tried to do substitution like $t=x-1/x$, but could not get the result. Then I tried to use characteristic function to solve this, but stucked. I am wondering if this can be solved by convolution approximation?
Let $x-\frac{1}{x}=t$.
For $x>0$ we have $x=\frac{1}{2}(t+\sqrt{4+t^2})$ and $$dx=\frac{1}{2}\left(1+\frac{t}{\sqrt{4+t^2}}\right)dt,$$ $$\begin{aligned} \int_0^\infty f\left(x-\frac{1}{x}\right)dx=\frac{1}{2}\int_{-\infty}^\infty f(t)\left(1+\frac{t}{\sqrt{4+t^2}}\right)dt. \end{aligned}$$ On the other hand for $x<0$, $x-\frac{1}{x}$ implies $x=\frac{1}{2}(t-\sqrt{4+t^2})$ and $$dx=\frac{1}{2}\left(1-\frac{t}{\sqrt{4+t^2}}\right)dt,$$ $$\begin{aligned} \int_{-\infty}^0 f\left(x-\frac{1}{x}\right)dx=\frac{1}{2}\int_{-\infty}^\infty f(t)\left(1-\frac{t}{\sqrt{4+t^2}}\right)dt. \end{aligned}$$ Thus $$\int_{-\infty}^\infty f\left(x-\frac{1}{x}\right)dx=\int_{-\infty}^\infty f(t)dt.$$