Intuition behind an integral identity A proof for the identity
$$\int_{-\infty}^{\infty} f(x)\, dx=\int_{-\infty}^{\infty} f\left(x-\frac{1}{x}\right)\, dx,$$
has been asked before (for example, here), and one answer to that question actually generalized this identity to 
$$\int_{-\infty}^{\infty} f(x)\, dx=\int_{-\infty}^{\infty} f\left(x-\frac{a}{x}\right)\, dx,$$
for any $a>0$. I've looked through those answers, and I understand the proofs well enough, but none of them provides any intuition on why this identity is true. It seems very strange to me that you can transform the function like this while preserving the integral, and even stranger that the choice of $a$ doesn't matter either. Just to be clear, I'm not asking to see a rigorous proof - there are plenty of those in the other question's answers. But if anyone could give me an intuitive sense of why this holds, or a heuristic argument, I'd really appreciate it. 
 A: Suppose we have a nice map $g:\mathbb{R} \to \mathbb{R}$ that is surjective and $k$-to-$1$, both properties meant for generic $x$ and not necessarily all $x$. In the problem at hand, $k=2$.
I claim that if

the sum of the $k$ pre-images of $x$ is equal to $x+C$ for a constant $C$, 

then the function $G(x)$ validates the formula

$\int f(x) = \int f(G(x))$ with the integral taken over all $\mathbb{R}$.

This is because the sum of the (oriented) lengths of the intervals between the preimages of nearby values $a$ and $b$ is exactly $b - a$, which is also equal to the length of the interval $[a,b]$.  That precisely equalizes the weights on corresponding values of $f$ in the two integrals.   A picture may help in understanding why the first statement implies the second (taking $b$ to be $a + \Delta a$ for a small value of $\Delta a$).
A: Here is the way I like to think of this. Whenever I try to gain intuition about integration, I always boil it down to step functions since the results extend nicely from there and step functions are oh-so-easy to work with. This is a good technique that can be used to visualize many of the standard identities from calculus and gain some intuition about them.
Let's then consider characteristic function $\chi_{[a,b]}$ and see what happens under this transformation. Well 
$$ \chi_{[a,b]}\left(x-\frac{1}{x}\right) = \begin{cases} 1, & x-\frac{1}{x}\in [a,b] \\ 0, & x-\frac{1}{x}\not\in[a,b]\end{cases}$$
For area purposes, it's best to have a function simply of $x$, not $x-\frac{1}{x}$ which means that we need to consider what the inverse image of $[a,b]$ is.
For now, let's suppose that $0 < a < b$. Then the inverse image of $[a,b]$ under $f(x) = x-\frac{1}{x}$ is nothing more than
$$\left[\frac{b-\sqrt{b^2+4}}{2},\frac{a-\sqrt{a^2+4}}{2}\right]\bigcup \left[\frac{a+\sqrt{a^2+4}}{2},\frac{b+\sqrt{b^2+4}}{2}\right].$$
This can be verified quite simply by noting that for the range of $0 < a < b$, $x-\frac{1}{x}$ is monotone on its preimage. With that in mind, it becomes clear that
$$ \chi_{[a,b]}\left(x-\frac{1}{x}\right)(x) = \chi_{[\frac{b-\sqrt{b^2+4}}{2},\frac{a-\sqrt{a^2+4}}{2}]\bigcup [\frac{a+\sqrt{a^2+4}}{2},\frac{b+\sqrt{b^2+4}}{2}]}(x).$$
The area under this characteristic function is then just
$$ \left(\frac{a-\sqrt{a^2+4}}{2} - \frac{b-\sqrt{b^2+4}}{2}\right)+\left(\frac{b+\sqrt{b^2+4}}{2}-\frac{a+\sqrt{a^2+4}}{2}\right) = b-a$$
which is nothing more than the area under $\chi_{[a,b]}$. A very similar analysis can be done for the case of $a < b < 0$ and then it can be extended quite easily to all $[a,b]$. Here are some plots with $a = 0, b = 3$:
$\chi_{[0,3]}$:

$\chi_{[\frac{3-\sqrt{13}}{2},-1]\cup[1,\frac{3+\sqrt{13}}{2}]}$:

As you can see, we have a nice splitting of the interval into two pieces and if you roughly estimate the area, it does indeed come out to $3$ as expected.
Moreover, it's not hard to see that the choice of $\alpha$ (playing the role of your $a$) is not of consequence here since by employing the quadratic equation, the $\alpha$ would only appear inside the square roots. As such, it would cancel when you evaluate the area.
A: I don't know if this provides the type of intuition sought.  And one might interpret this as just one more approach to proving the identity.  But I thought that this might shed a bit of intuition as to what is going on here.  
To that end, we provide a way forward that exploits symmetry and inversion.  First, let us write the function $f$ in terms of its even and odd parts as 
$$f(x)= f^{e}(x)+f^{o}(x)$$
where $f^{e}(x)=\frac12(f(x)+f(-x))$ and $f^{0}(x)=\frac12(f(x)-f(-x))$ are the even and odd parts of $f$, respectively.
Next, we write the integral of interest $I$ as
$$\begin{align}
I&=\int_{-\infty}^{\infty}f\left(x-\frac1x\right)dx\\\\
&=2\int_0^{\infty}f^{e}\left(x-\frac1x\right)dx \tag1
\end{align}$$
So, we can write $I$ in terms of the even part of $f$.
Now, here is why exploiting the even symmetry of $f^{e}$ is important.  We will now make a transformation under inversion.  Letting $x\to \frac1x$, $(1)$ becomes
$$\begin{align}
\int_{-\infty}^{\infty}f\left(x-\frac1x\right)dx&=2\int_0^{\infty}f^{e}\left(\frac1x-x\right)\frac{1}{x^2}dx \\\\
&=2\int_0^{\infty}f^{e}\left(x-\frac1x\right)\frac{1}{x^2}dx \tag2
\end{align}$$
where in arriving at $(2)$, we used $f^{e}(x)=f^{e}(-x)$.  So, upon adding the right-hand sides of $(1)$ and $(2)$ and dividing by $2$ we obtain 
$$\int_{-\infty}^{\infty}f\left(x-\frac1x\right)dx=\int_{0}^{\infty}f^{e}\left(x-\frac1x\right)\left(1+\frac{1}{x^2}\right)\,dx$$
At this point, all we have is that the integral of interest is decomposed into the sum of an integral of its even part and an integral of an "inverted" even part.  But, now we notice that the "scale factor" is the derivative of the argument of $f^{e}$ and thus we find that 
$$\begin{align}
\int_{-\infty}^{\infty}f\left(x-\frac1x\right)dx&=\int_{-\infty}^{\infty}f^{e}(x)\,dx\\\\
&=\int_{-\infty}^{\infty}f(x)\,dx
\end{align}$$
as expected.
A: Let v,w be the inverse functions to $(x-1/x)$, $f$ an even function and $a>0$. The proposition follows if the integrals of corresponding intervals are equal :
$$ \int_{a}^{a+\epsilon} f(y)\, dy=?\int_{v(a)}^{v(a+\epsilon) } f\left(x-\frac{1}{x}\right)\, dx+\int_{w(a)}^{w(a+\epsilon) } f\left(x-\frac{1}{x}\right)\, dx $$
For small $\epsilon$ the LHS $\approx \epsilon f(a)$.
and the RHS
$$\approx   \epsilon (v'(a)f\left( v(a)-\frac{1}{v(a)}\right) +w'(a)f\left( w(a)-\frac{1}{w(a)}\right) ) = \epsilon(v'(a)+w'(a))f(a)$$
By calculation, $(v+w)' =1$, so the RHS also equals $\epsilon f(a)$ .
