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Let $$\int_{-\infty}^{\infty}f(x) dx=1$$ Then, find $$\int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right) dx$$

I subtituted $x=t-\frac{1}{t}$ in the original equation and found $$\int_{-\infty}^{0}f(t-\frac{1}{t}) dt= \frac 12$$

But, I couldn't proceed after this.

Quanto
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Aspirant
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2 Answers2

5

Split $( -\infty, \infty)\to ( -\infty, 0) + ( 0, \infty) $ \begin{align} &\int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right) \overset{x\to -\frac1x}{ dx}\\ =& \int^{\infty}_{0}f\left(x-\frac{1}{x}\right) \frac {dx}{x^2}+ \int^{0}_{-\infty}f\left(x-\frac{1}{x}\right) \frac {dx}{x^2}\\ =&\ \frac12\int_{-\infty}^{0}f\left(x-\frac{1}{x}\right)\left( 1+\frac1{x^2}\right) \overset{x\to x-\frac1x}{dx} + \frac12\int_{0}^{\infty}f\left(x-\frac{1}{x}\right)\left( 1+\frac1{x^2}\right) \overset{x\to x-\frac1x}{dx}\\ =& \int_{-\infty}^{\infty}f(x)dx=1 \end{align}

Quanto
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  • See https://link.springer.com/article/10.1007/s11785-010-0095-0 and https://math.stackexchange.com/questions/1690150/has-this-chaotic-map-been-studied/2578363#2578363 ; the correct answer is $1$ not $1/2$. The map $x\mapsto x-1/x$ is a 2-branched covering. – kimchi lover Jul 28 '20 at 16:55
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Let $a, b \in \mathbb{R}$, $g(x) = x - \frac{1}{x}$. Then

$$\chi_{[a,b]} \circ g = \chi_{\left[\frac{a + \sqrt{a^2 + 4}}{2}, \frac{b + \sqrt{b^2 + 4}}{2}\right]}(x) + \chi_{\left[\frac{a - \sqrt{a^2 +4}}{2}, \frac{b - \sqrt{b^2 + 4}}{2}\right]}(x)$$

But $$\mu\left(\left[\frac{a + \sqrt{a^2 + 4}}{2}, \frac{b + \sqrt{b^2 + 4}}{2}\right] \cup \left[\frac{a - \sqrt{a^2 + 4}}{2}, \frac{b - \sqrt{b^2 + 4}}{2}\right]\right) = b - a$$ so $g$ is measure preserving and it follows that the result is 1.

cha21
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