How to prove $\int_{-\infty}^{+\infty} f(x)dx = \int_{-\infty}^{+\infty} f\left(x - \frac{1}{x}\right)dx?$

Question

If $f(x)$ is a continuous function on $(-\infty, +\infty)$ and $\int_{-\infty}^{+\infty} f(x) \, dx$ exists. How can I prove that $$\int_{-\infty}^{+\infty} f(x) \, dx = \int_{-\infty}^{+\infty} f\left( x - \frac{1}{x} \right) \, dx\text{ ?}$$

Answer 1

For $x\ge0$ write the following substitution which maps the positive real domain to the whole of the real domain: $$ x=1/2\,y+1/2\,\sqrt {{y}^{2}+4}, $$ $${\it dx}= \left( 1/2+1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right) {\it dy},$$ and likewise for $x<0$ write the following substitution which maps the negative real domain to the whole of the real domain:$$ x=1/2\,y-1/2\,\sqrt {{y}^{2}+4}, $$ $${\it dx}= \left( 1/2-1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right) {\it dy},$$ and in both cases you then have: $$x-\frac{1}{x}=y,$$ and you then get: $$\int\limits_{0}^{+\infty} f\left(x-\dfrac{1}{x}\right)dx=\int\limits_{-\infty}^{+\infty} f\left(y\right)\left( 1/2+1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right)dy,$$

$$\int\limits_{-\infty}^{0} f\left(x-\dfrac{1}{x}\right)dx=\int\limits_{-\infty}^{+\infty} f\left(y\right)\left( 1/2-1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right)dy,$$ and therefore: $$\int\limits_{-\infty}^{+\infty} f\left(x-\dfrac{1}{x}\right)dx=\int\limits_{-\infty}^{+\infty} f\left(y\right)\left( 1/2+1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right)dy+\int\limits_{-\infty}^{+\infty} f\left(y\right)\left( 1/2-1/2\,{\frac {y}{\sqrt {{y}^{2}+4}}} \right)dy,$$ $$=\int\limits_{-\infty}^{+\infty} f\left(y\right)dy$$

Update: See Cauchy-Schlomilch transformation and Glasser's Master Theorem.

Answer 2

We can write \begin{align} \int_{-\infty}^{\infty}f\left(x-x^{-1}\right)dx&=\int_{0}^{\infty}f\left(x-x^{-1}\right)dx+\int_{-\infty}^{0}f\left(x-x^{-1}\right)dx\\ &=\int_{-\infty}^{\infty}f(2\sinh\theta)\,e^{\theta}d\theta+\int_{-\infty}^{\infty}f(2\sinh\theta)\,e^{-\theta}d\theta\\ &=\int_{-\infty}^{\infty}f(2\sinh\theta)\,2\cosh\theta\,d\theta\\ &=\int_{-\infty}^{\infty}f(x)\,dx. \end{align} To pass from the first to the second line, we make the change of variables $x=e^{\theta}$ in the first integral and $x=-e^{-\theta}$ in the second one.

Answer 3

Here is a neat way of showing the claim without using any "weird" change of variables. Assume $f$ is nice enough to do all that follows. I think $f\in C_c^\infty(R-\{0\})$ is enough. Let $$F(a) = \int_{-\infty}^\infty f\left(x-\frac{a}{x} \right)dx $$ so that $$F'(a) = \int_{-\infty}^\infty -\frac{1}{x}f'\left(x-\frac{a}{x} \right)dx.$$

[Please note if there is an error in the following, I've gotten it wrong a couple times.] Split up the integral and do a change of variables $x=a/y$ so that

$$F'(a)=\int_{\infty}^0 \left[-\frac{y}{a}f'\left(\frac{a}{y}-y\right) \cdot - \frac{a}{y^2} \right]dy+\int_{0}^{-\infty} \left[-\frac{y}{a}f'\left(\frac{a}{y}-y\right) \cdot - \frac{a}{y^2}\right]dy$$ or $$F'(a)= \int_{-\infty}^\infty -\frac{1}{y}f'\left(\frac{a}{y}-y\right) dy.$$

Here's the tricky part. The original claim is true for odd functions $f$, since both integrals integrate to zero. Thus, only the even part of $f$ matters. We can then assume that $f$ is even, and hence $f'$ is odd. Moving a minus sign out from the argument of $f'$ in the above integral and we get $F'(a)=-F'(a)$. This means that $F'(a)=0$ for all $a$ after appealing to the continuity of $F'(a)$. Hence $F(a)=F(0)$ is constant and $$F(1)=\int_{-\infty}^\infty f\left(x-\frac{1}{x} \right)dx = \int_{-\infty}^\infty f\left(x\right)dx=F(0).$$ Approximate any $f$ with a $C_c^\infty$ function supported away from the origin, and the original claim holds.

Answer 4

Curiously, I happen to come across this a few days after finding this question, where @achille hui proves the theorem that

Given such a meromorphic function $ϕ(z)$ and any Lebesgue integrable function $f(x)$ on $\mathbb{R}$, we have following identity: $$∫^∞_{−∞}f(ϕ(x))dx=∫^∞_{−∞}f(x)dx$$

In that question's scope, it is used to show that $$∫^∞_{−∞}f(x-\cot(x))dx=∫^∞_{−∞}f(x)dx$$

Here, $\phi(x)=x-\frac1x$, so to show that $$∫^∞_{−∞}f\left(x-\frac1x\right)dx=∫^∞_{−∞}f(x)dx$$ we simply have to just check the conditions of the quoted theorem.

We can see that $\phi(x)$ is indeed meromorphic, as its only pole is at the origin and it is holomorphic.

Then, by the quoted theorem, assuming $f(x)$ is Lebesgue integrable over all reals, your case follows as a corollary.