A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$ $$\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx~~~~(1)$$
Several difficult integral can be handled using (1) easily. For instance we know that $$\int_{-\infty}^{\infty} \text{sech}^2 x~dx=2~~~~~~~~~~~~(2)$$ but what is interesting is that one may verify numerically that $$\int_{-\infty}^{\infty} \text{sech}^2(x-a/x)~dx=2, a>0.~~~~~~~~~(3)$$ This interesting property (1) follows from Glasser's Master Theorem. See one more interesting integral and its solutions in MSE post:s
https://en.wikipedia.org/wiki/Glasser%27s_master_theorem.
In (1) what is also interesting is that LHS is independent of $a(>0)$. I have tried proving (1) or even (3) by the ordinary rules of integration ( Real Analysis) without a success. Can someone help me?
EDIT: One more friendly (doable by hand) twin is
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^2}=\pi=\int_{-\infty}^{\infty} \frac{dx}{1+(x-a/x)^2} dx , a >0$$
 A: Since the integrand is even, we have
$$\mathcal{I} \stackrel{def}{=}\int_{-\infty}^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})} dx
= 2\int_0^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})}dx$$
In one copy of the integral on RHS, change variables to $y = \frac{a}{x}$, one get
$$\int_0^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})} =
\int_0^\infty \frac{1}{\cosh^2(\frac{a}{y} - y)} \frac{a}{y^2}dy$$
Renaming $y$ back to $x$ and add back to another copy of integral on RHS, we get
$$\mathcal{I} = \int_0^\infty \frac{1}{\cosh^2(x - \frac{a}{x})}\left(1 + \frac{a}{x^2}\right)dx
= \int_0^\infty \frac{1}{\cosh^2(x - \frac{a}{x})}\frac{d}{dx}\left(x - \frac{a}{x}\right) dx$$
Change variable to $z = x - \frac{a}{x}$, this becomes
$$\mathcal{I} = \int_{-\infty}^\infty \frac{dz}{\cosh^2(z)}$$
i.e. transform your integral $(3)$ to integral $(2)$, the one you already know.
Let's look at the more general identity $(1)$.
As you increases $x$ from $-\infty$ to $0$, $x - \frac{a}{x}$ increase from $-\infty$ to $\infty$ once. If one further increases $x$ from $0$ to $\infty$, $x - \frac{a}{x}$ increases from $-\infty$ to $\infty$ the second time.
For any $t \in \mathbb{R}$, let $x_1(t) < 0$, $x_2(t) > 0$ be the two roots of
$$x - \frac{a}{x} = t \equiv x^2 - tx - a = 0$$
This is a quadratic equation in $x$. By Vieta's formula, we have
$$x_1(t) + x_2(t) = t \implies x_1'(t) + x_2'(t) = 1$$
If you change variable to $t = x - \frac{a}{x}$ for both $(-\infty,0)$ and $(0,\infty)$, we obtain:
$$\begin{align} 
 \int_{-\infty}^\infty f(x - \frac{a}{x}) dx
&= \left(\int_{-\infty}^0 + \int_0^\infty\right) f(x - \frac{a}{x})dx\\
&= \int_{-\infty}^{\infty} f(t) x'_1(t) dt +
 \int_{-\infty}^{\infty} f(t) x'_2(t) dt\\
&= \int_{-\infty}^\infty f(t) (x'_1(t) + x'_2(t))dt\\
&= \int_{-\infty}^\infty f(t)dt
\end{align}$$
This is the identity $(1)$ we seek.
A: $u=x-\frac ax\Rightarrow du=(1+\frac a{x^2})dx$ and so:
$$\int_{-\infty}^\infty f(x-a/x)dx=\int_{-\infty}^\infty\frac{f(u)}{1+\frac a{x^2}}dx$$
now we need to get this expression on the bottom in terms of $u$.
$$\frac{f(u)}{1+a/x^2}=\frac{x^2f(u)}{x^2+a}=f(u)-\frac{a^2}{x^2+a}f(u)$$
So now the challenge we face is showing that:
$$\int_{-\infty}^\infty\frac{a^2}{x^2+a}f(u)du=0$$
we can start by trying to undo the substitution:
$$\int_{-\infty}^\infty\frac{a}{x^2+a}f(x-a/x)(1+a/x^2)dx=a\int_{-\infty}^\infty\frac{1+a/x^2}{x^2+a}f(x-a/x)dx$$
$$=a\int_{-\infty}^\infty\frac1{x^2}\frac{x^2+a}{x^2+a}f(x-a/x)dx=a\int_{-\infty}^\infty\frac{f(x-a/x)}{x^2}dx$$
if we try letting $x\to-x$ and combine we get:
$$I_1=\frac a2\int_{-\infty}^\infty\frac{f(-x+a/x)+f(x-a/x)}{x^2}dx=\frac a2\int_{-\infty}^\infty\frac{f(-(x-a/x))+f(x-a/x)}{x^2}dx$$
now we can try and make our substitution again:
$$u=x-a/x$$
$$I_1=\frac a2\int_{-\infty}^\infty\frac{f(u)+f(-u)}{x^2}\frac{du}{1+a/x^2}dx=\frac a2\int_{-\infty}^\infty\frac{f(u)+f(-u)}{x^2+a}du$$
Now notice how this function is symmetric i.e. $I_1=-I_1\therefore I_1=0$
A: Glasser's formula (1) amounts to the statement that the image measure $\mu$ of the Lebesgue measure $\lambda$ on $\Bbb R$, under the (two-to-one) transformation $\varphi: x\mapsto x-a/x$ is also Lebesgue measure. To verify this it suffices to show that $\mu\{(A,B]\}=B-A$ for each $A<B$. Notice that if we define $\alpha_\pm:={A\pm\sqrt{A^2+4}\over 2}$ and $\beta_\pm:={B\pm\sqrt{B^2+4}\over 2}$ (so that $\varphi(\alpha_\pm) = A$ and $\varphi(\beta_\pm)=B)$ then
$$
\eqalign{
\mu\{(A,B]\}
&=\lambda\{\varphi^{-1}\{(A,B]\}\}=\lambda\{(\alpha_-,\beta_-]\}+\lambda\{(\alpha_+,\beta_+]\}\cr
&=(\beta_--\alpha_-)+(\beta_+-\alpha_+)\cr
&=B-A,\cr
}
$$
because $\beta_++\beta_-=B$ and $\alpha_++\alpha_-=A$.
A: Note
\begin{align}
\int_{-\infty}^{\infty} f\left(x-\frac ax \right) dx
&= \int_{-\infty}^{0} 
\overset{x=-\sqrt a e^{-t}} {f\left(x-\frac ax \right) dx }+ \int_{0}^{\infty} \overset{x=\sqrt a e^t}{f\left(x-\frac ax \right) dx}\\
&= \int_{-\infty}^{\infty} \underset{ x=\sqrt a (e^t-e^{-t})}{f[\sqrt a(e^t-e^{-t})]\sqrt a(e^t +e^{-t})} dt =\int_{-\infty}^{\infty} f(x) dx
\end{align}
