How to prove $\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^{+\infty} f\left(x - \frac{1}{x}\right)dx?$ If $f(x)$ is a continuous function on $(-\infty, +\infty)$ and $\int_{-\infty}^{+\infty} f(x) \, dx$ exists. How can I prove that
$$\int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{+\infty} f\left( x - \frac{1}{x} \right) \, dx\text{ ?}$$

Here is another way to prove the identity courtesy of @achillehui. Let us generalise the identity using the following lemma.
Lemma :

\begin{align}
\int_{-\infty}^\infty f\left(x\right)\,dx=\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\,dx\qquad,\qquad\text{for }\, a>0.
\end{align}

Proof :
Let $\displaystyle\;u(x) = x-\frac{a}{x} $. As $x$ varies over $\mathbb{R}$, we have


*

*$u(x)$ increases monotonically from $-\infty$ at $-\infty$ to $+\infty$ at $0^{-}$.

*$u(x)$ increases monotonically from $-\infty$ at $0^{+}$ to $+\infty$ at $+\infty$.


This means as $x$ varies, $u(x)$ covered $(-\infty,\infty)$ twice.
Let $x_1(u) < 0$ and $x_2(u) > 0$ be the two roots of the equation for a given $u$:
$$u = u(x) =  \frac{x^2-a}{x} \quad\iff\quad x^2 - ux - a = 0\quad\implies\quad x(u)_{1,2}=\frac{1}{2}\left(\;u\pm\sqrt{u^2+4a}\;\right)$$
we have
$$x_1(u) + x_2(u) = u
\quad\implies\quad
\frac{dx_1}{du} + \frac{dx_2}{du} = 1.
$$
From this, we find
$$\begin{align}
\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\,dx
&= \left( \int_{-\infty}^{0^{-}} + \int_{0^{+}}^{+\infty}\right) f\left(u(x)\right)\, dx\\
&= \int_{-\infty}^{\infty} f\left(u\right)\,\left(\frac{dx_1}{du} + \frac{dx_2}{du}\right) du\\
&= \int_{-\infty}^{\infty}f\left(u\right)\, du\qquad;\qquad u\mapsto x\\
&= \int_{-\infty}^{\infty}f\left(x\right)\, dx\qquad\qquad\square
\end{align}
$$

Thus, by setting $a=1$, we have

\begin{align}
\int_{-\infty}^\infty f\left(x\right)\,dx=\int_{-\infty}^\infty f\left(x-\frac{1}{x}\right)\,dx
\end{align}

