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.

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    $\begingroup$ I have seen a theorem that characterizes the measure-preserving rational functions on $\Bbb{R}$, and this is a special case. I am not claiming that either this theorem or its proof sheds a light on our intuitive sense, though. $\endgroup$ – Sangchul Lee Aug 15 '15 at 4:30
  • $\begingroup$ Do you remember the name/source of the theorem? It might be easier to develop intuition by considering the situation in full generality. $\endgroup$ – Samir Khan Aug 15 '15 at 4:37
  • $\begingroup$ @sangchul, given the answer in terms of preimages, there is a criterion for rational functions of degree $k$ that looks only at the first two highest-degree coefficients of the numerator and denominator. $\endgroup$ – ASCII Advocate Aug 15 '15 at 5:25
  • $\begingroup$ Originally I have seen that result in this Math.SE community. Another source refers to the Pólya and Szegö's book Problems and Theorems in Analysis I, II, but I have not checked these books. $\endgroup$ – Sangchul Lee Aug 15 '15 at 14:24

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


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.

  • $\begingroup$ Thank you very much! This answer shed a lot of light on the underlying symmetry. $\endgroup$ – Samir Khan Aug 15 '15 at 5:30
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    $\begingroup$ @SamirKhan You're welcome. My true pleasure!! It was a really tough question, so a big +1 for your asking. I noted that both of us are UC alum! It was a fantastic experience, one that I will always cherish. Such a rich tradition of excellence... and all of those noble prizes! $\endgroup$ – Mark Viola Aug 15 '15 at 5:33
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    $\begingroup$ No. There is a rather precise sense in which the even/odd aspect is irrelevant and just obscuring the actual structure of the problem. First of all, the "not depending on antisymmetric part of $f$" applies to any definite integral of any $f$ on any interval (for the mirror reflection symmetry of the interval), so is too general to carry much weight in this much more specific application. More in the next comment. $\endgroup$ – ASCII Advocate Aug 15 '15 at 15:35
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    $\begingroup$ Now, the symmetry group implicitly used in your solution has order 4, not 2. It is the direct product of a useful symmetry group of order 2 (the one that exchanges the two solutions of $x - 1/x = y$) and an irrelevant even/odd symmetry that changes the sign of $x$. That an even/odd symmetry exists at all, and commutes with the useful symmetry, is somewhat accidental and has to do with the fact that $(x - 1/x)$ is an odd function. It is there in the solution only to compensate for the use of $1/x$ to transform the integral, rather than $-1/x$ that switches the two solutions. $\endgroup$ – ASCII Advocate Aug 15 '15 at 15:44
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    $\begingroup$ @ASCIIAdvocate For those who are not versed in group theory, these comments really might not be of much use I'm afraid. Perhaps, in order to help here, these might have to be taken down a notch or two - or three. Just friendly advice and in my humble opinion only. $\endgroup$ – Mark Viola Aug 15 '15 at 15:49

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 $\Delta a$).

  • $\begingroup$ To connect this to the originating problem, the two solutions (and there are always solutions) of $x - (a/x) = y$ have sum $x_1 + x_2 = y$. $\endgroup$ – ASCII Advocate Aug 15 '15 at 5:24

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 $a = 0, b = 3$:


enter image description here


enter image description here

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.


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 $\epsilon f(a)$ .

  • $\begingroup$ I think this is the same as what @ASCII Advocate means by "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." $\endgroup$ – Keith McClary Aug 15 '15 at 20:04

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