TITLE: A simple proof for Glasser: $\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx, a>0$
QUESTION [9 upvotes]: $$\int_{-\infty}^{\infty} f(x-a/x) dx=\int_{-\infty}^{\infty} f(x) dx~~~~(1)$$
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
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^2}=\pi=\int_{-\infty}^{\infty} \frac{dx}{1+(x-a/x)^2} dx , a >0$$

REPLY [3 votes]: 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}