TITLE: Proving that $\int_{-\infty}^{\infty} f(x - \frac{1}{x}) dx = \int_{-\infty}^{\infty} f(x) dx.$
QUESTION [7 upvotes]: Consider $f : \mathbb{R} \rightarrow \mathbb{R}$ to be a continuous
function such that $\int_{-\infty}^{\infty} f(x) dx$ exists. I would like to
prove that
$$\int_{-\infty}^{\infty} f(x - \frac{1}{x}) dx =
  \int_{-\infty}^{\infty} f(x) dx.$$
I decided to tackle it in cases.
In particular, I started out with the case where $f$ was linear. We see that
$$\int_{-\infty}^{\infty} f\left(x-\frac{1}{x}\right) dx
= \int_{-\infty}^{\infty} f(x) - f\left(\frac{1}{x}\right) dx
= \int_{-\infty}^{\infty} f(x) dx - \int_{-\infty}^{\infty} f\left(\frac{1}{x}
\right) dx.$$
This problem then boils down into proving that
$$\int_{-\infty}^{\infty} f\left(\frac{1}{x}\right) dx = 0.$$
we set $u = \frac{1}{x} \implies du = \frac{-1}{x^2} dx.$ However, I am
having problems simplifying the term inside the integral into a function of
$u$. Any suggestions on this case? And any recommendations on the case where
$f$ is not linear?

REPLY [4 votes]: $x-\frac1x=u\implies\left(1+\frac1{x^2}\right)\mathrm{d}x=\mathrm{d}u$
$$
\begin{align}
&\int_{-\infty}^\infty f\left(x-\frac1x\right)\mathrm{d}x\tag{1}\\
&=\int_{-\infty}^0f\left(x-\frac1x\right)\mathrm{d}x+\int_0^\infty f\left(x-\frac1x\right)\mathrm{d}x\tag{2}\\
&=\int_0^\infty f\left(x-\frac1x\right)\frac1{x^2}\,\mathrm{d}x+\int_{-\infty}^0f\left(x-\frac1x\right)\frac1{x^2}\,\mathrm{d}x\tag{3}\\
&=\frac12\int_0^\infty f\left(x-\frac1x\right)\left(1+\frac1{x^2}\right)\mathrm{d}x
+\frac12\int_{-\infty}^0f\left(x-\frac1x\right)\left(1+\frac1{x^2}\right)\mathrm{d}x\tag{4}\\
&=\frac12\int_{-\infty}^\infty f(u)\,\mathrm{d}u+\frac12\int_{-\infty}^\infty f(u)\,\mathrm{d}u\tag{5}\\
&=\int_{-\infty}^\infty f(u)\,\mathrm{d}u\tag{6}
\end{align}
$$
Explanation:
$(2)$: split up domain of integration
$(3)$: substitute $x\mapsto-\frac1x$
$(4)$: average $(2)$ and $(3)$
$(5)$: substitute $x-\frac1x=u$
$(6)$: algebra