A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$

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.

$u=x-\frac ax\Rightarrow du=(1+\frac a{x^2})dx$ 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. $I_1=-I_1\therefore I_1=0$

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} f[\sqrt a(e^t-e^{-t})]\sqrt a(e^t +e^{-t}) dt \end{align} Then, substitute $x=\sqrt a (e^t-e^{-t})$ to obtain $$\int_{-\infty}^{\infty} f\left(x-\frac ax \right)dx =\int_{-\infty}^{\infty} f(x) dx$$