Proving that $\int_{-\infty}^{\infty} f(x - \frac{1}{x}) dx = \int_{-\infty}^{\infty} f(x) dx.$ 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?
 A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mc{J} & =
\color{#f00}{\int_{-\infty}^{\infty}\mrm{f}\pars{x - {1 \over x}}\,\dd x}\ =\
\overbrace{\int_{0}^{\infty}\bracks{\mrm{f}\pars{x - {1 \over x}} +
\mrm{f}\pars{-x + {1 \over x}}}\,\dd x}^{\ds{=\ \mc{J}}}
\\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=} \,\,\,
\int_{\infty}^{0}\bracks{\mrm{f}\pars{{1 \over x} - x} +
\mrm{f}\pars{-\,{1 \over x} + x}}\,\pars{-\,{\dd x \over x^{2}}}
\\[5mm] & =
\underbrace{\int_{0}^{\infty}\bracks{\mrm{f}\pars{{1 \over x} - x} +
\mrm{f}\pars{-\,{1 \over x} + x}}\,{1 \over x^{2}}\,\dd x}_{\ds{=\ \mc{J}}}
\end{align}

\begin{align}
\mc{J} & = {\mc{J} + \mc{J} \over 2} =
{1 \over 2}\int_{0}^{\infty}\bracks{\mrm{f}\pars{{1 \over x} - x} +
\mrm{f}\pars{-\,{1 \over x} + x}}\,\pars{1 + {1 \over x^{2}}}\,\dd x
\\[5mm] & \stackrel{t\ \equiv\ 1/x - x}{=}\,\,\,
{1 \over 2}\int_{\infty}^{-\infty}\bracks{\mrm{f}\pars{t} + \mrm{f}\pars{-t}}\,
\pars{-\,\dd t} = \color{#f00}{\int_{0}^{\infty}\mrm{f}\pars{x}\,\dd x}
\end{align}
A: 
METHODOLOGY $1$:

First, we split the integral as 
$$\begin{align}
\int_{-\infty}^\infty f\left(x-\frac1x\right)\,dx&=\int_{0}^\infty f\left(x-\frac1x\right)\,dx +\int_0^\infty f\left(-x+\frac1x\right)\,dx \tag 1
\end{align}$$
In the first integral on the right-hand side of $(1)$, we enforce the substitution $x\to e^x$ and in the second integral, we enforce the substitution $x\to e^{-x}$.  Proceeding, we find that 
$$\begin{align}
\int_{-\infty}^\infty f\left(x-\frac1x\right)\,dx&=\int_{-\infty}^\infty f\left(2\sinh(x)\right)\,e^x\,dx +\int_{-\infty}^\infty f\left(2\sinh(x)\right)\,e^{-x}\,dx \\\\
&=\int_{-\infty}^\infty f(2\sinh(x))\,2\cosh(x)\,dx\tag2
\end{align}$$
Finally, we enforce the substitution $2\sinh(x)\to x$ in the integral on the right-hand side of $(2)$ to obtain
$$\int_{-\infty}^\infty f\left(x-\frac1x\right)\,dx=\int_{-\infty}^\infty f(x)\,dx$$
as was to be shown!


METHODOLOGY $2$:

We begin with the expression given in $(1)$.  In the first integral on the right-hand side of $(1)$, we enforce the substitution $x-\frac1x\to x$, while in the second integral on the right-hand side of $(1)$ we enforce the substitution $-x+\frac1x \to x$.  Proceeding, we find that
$$\begin{align}
\int_{-\infty}^\infty f\left(x-\frac1x\right)\,dx&=\int_{-\infty}^\infty f\left(x\right)\,\left(\frac12+\frac{x}{2\sqrt{x^2+1}}\right)\,dx +\int_{-\infty}^\infty f\left(x\right)\,\left(\frac12-\frac{x}{2\sqrt{x^2+1}}\right)\,dx \\\\
&=\int_{-\infty}^\infty f(x)\,dx
\end{align}$$
as expected!
A: $x-\frac1x=u\implies\left(1+\frac1{x^2}\right)\mathrm{d}x=\mathrm{d}u$
$$
\begin{align}
&\int_{-\infty}^\infty f\left(x-\frac1x\right)\mathrm{d}x\tag{1}\\
&=\int_{-\infty}^0f\left(x-\frac1x\right)\mathrm{d}x+\int_0^\infty f\left(x-\frac1x\right)\mathrm{d}x\tag{2}\\
&=\int_0^\infty f\left(x-\frac1x\right)\frac1{x^2}\,\mathrm{d}x+\int_{-\infty}^0f\left(x-\frac1x\right)\frac1{x^2}\,\mathrm{d}x\tag{3}\\
&=\frac12\int_0^\infty f\left(x-\frac1x\right)\left(1+\frac1{x^2}\right)\mathrm{d}x
+\frac12\int_{-\infty}^0f\left(x-\frac1x\right)\left(1+\frac1{x^2}\right)\mathrm{d}x\tag{4}\\
&=\frac12\int_{-\infty}^\infty f(u)\,\mathrm{d}u+\frac12\int_{-\infty}^\infty f(u)\,\mathrm{d}u\tag{5}\\
&=\int_{-\infty}^\infty f(u)\,\mathrm{d}u\tag{6}
\end{align}
$$
Explanation:
$(2)$: split up domain of integration
$(3)$: substitute $x\mapsto-\frac1x$
$(4)$: average $(2)$ and $(3)$
$(5)$: substitute $x-\frac1x=u$
$(6)$: algebra
