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Let $f\in L^1(\Bbb R)$ How can I prove that: $$\int_\Bbb R f(x)dx = \int_\Bbb R f(x-\frac{1}{x})dx$$

In fact the changes of variables $u=\phi(x) = x-\frac{1}{x}$ seems correct but the image of zero : $\phi(0)$ does not exists.

How can I avoid this issues? thanks

Guy Fsone
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    This result goes back to Cauchy and a generalization is due to Glasser (Glasser, M. L. "A Remarkable Property of Definite Integrals." Math. Comput. 40, 561-563, 1983. ). See: http://mathworld.wolfram.com/GlassersMasterTheorem.html Note that it has to be interpreted in a principal value to be true in general – Winther Dec 07 '17 at 19:51
  • The trick is the change of variable $x = 1/y$ to obtain $\int_{-\infty}^\infty f(x-\frac{1}{x})dx = \int_{-\infty}^\infty f(\frac1y-y)\frac{dy}{y^2}$ and to note with $F'(x) = f(x)$ that $\int_a^b f(x-\frac{1}{x})(1+\frac{1}{x^2})dx = F(x-\frac{1}{x})|_a^b$. – reuns Dec 07 '17 at 19:57
  • @GuyFsone "$\phi(0)$ does not exists" is a nonsense. $\int_0^\infty g(x)dx = \lim_{n \to 0} \int_{1/n}^n g(x)dx$ and the change of variable you mentioned works fine, – reuns Dec 07 '17 at 20:00

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This result (and a generalisation) is proved by Glasser, I transcribe the first part here following the notation in the original post.

From a graph of the relation $\phi = x - 1/x$, one sees that the solution for $x$ has two branches $$x_\pm = \frac{1}{2} \left(\phi \pm \sqrt{\phi^2 + 4}\right), \quad -\infty < \phi < \infty.$$ Thus $$I = \int_{-\infty}^\infty f(\phi) \, dx = \int_{-\infty}^{0^-} f(\phi) \, dx_- + \int_{0^+}^{-\infty} f(\phi) \, dx_+=\int_{-\infty}^\infty f(\phi) (x_-'+x_+')\, d\phi.$$ But $$x_-' + x_+' = \frac{1}{2} \left(1 - \frac{\phi}{\sqrt{\phi^2+4}}+1 + \frac{\phi}{\sqrt{\phi^2+4}}\right) = 1.$$

B. Mehta
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