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

Solution 1:

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.

Solution 2:

Here is another way to prove the identity courtesy of @achillehui. Let us generalise the identity using the following lemma.

Lemma :

\begin{align} \int_{-\infty}^\infty f\left(x\right)\,dx=\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\,dx\qquad,\qquad\text{for }\, a>0. \end{align}

Proof :

Let $\displaystyle\;u(x) = x-\frac{a}{x} $. As $x$ varies over $\mathbb{R}$, we have

• $u(x)$ increases monotonically from $-\infty$ at $-\infty$ to $+\infty$ at $0^{-}$.
• $u(x)$ increases monotonically from $-\infty$ at $0^{+}$ to $+\infty$ at $+\infty$.

This means as $x$ varies, $u(x)$ covered $(-\infty,\infty)$ twice.

Let $x_1(u) < 0$ and $x_2(u) > 0$ be the two roots of the equation for a given $u$:

$$u = u(x) = \frac{x^2-a}{x} \quad\iff\quad x^2 - ux - a = 0\quad\implies\quad x(u)_{1,2}=\frac{1}{2}\left(\;u\pm\sqrt{u^2+4a}\;\right)$$ we have $$x_1(u) + x_2(u) = u \quad\implies\quad \frac{dx_1}{du} + \frac{dx_2}{du} = 1. $$ From this, we find

$$\begin{align} \int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\,dx &= \left( \int_{-\infty}^{0^{-}} + \int_{0^{+}}^{+\infty}\right) f\left(u(x)\right)\, dx\\ &= \int_{-\infty}^{\infty} f\left(u\right)\,\left(\frac{dx_1}{du} + \frac{dx_2}{du}\right) du\\ &= \int_{-\infty}^{\infty}f\left(u\right)\, du\qquad;\qquad u\mapsto x\\ &= \int_{-\infty}^{\infty}f\left(x\right)\, dx\qquad\qquad\square \end{align} $$

Thus, by setting $a=1$, we have

\begin{align} \int_{-\infty}^\infty f\left(x\right)\,dx=\int_{-\infty}^\infty f\left(x-\frac{1}{x}\right)\,dx \end{align}

Solution 3:

Here is yet another way forward. First, we begin with the integral $I$ given by

$$\begin{align} I&=\int_{-\infty}^{\infty}f\left(x-\frac1x\right)\,dx \tag 1\\\\ &=\int_{-\infty}^{0}f\left(x-\frac1x\right)\,dx +\int_{0}^{\infty}f\left(x-\frac1x\right)\,dx \tag 2 \end{align}$$

We enforce the substitution $x\to -1/x$ in the integrals on the right-hand side of $(2)$ to obtain

$$\begin{align} I&=\int_{0}^{\infty}f\left(x-\frac1x\right)\,\left(\frac{1}{x^2}\right)\,dx +\int_{-\infty}^{0}f\left(x-\frac1x\right)\,\left(\frac{1}{x^2}\right)\,dx \tag 3 \end{align}$$

Adding $(2)$ and $(3)$ reveals

$$\begin{align} 2I&=\int_{0}^{\infty}f\left(x-\frac1x\right)\,\left(1+\frac{1}{x^2}\right)\,dx +\int_{-\infty}^{0}f\left(x-\frac1x\right)\,\left(1+\frac{1}{x^2}\right)\,dx \tag 4 \end{align}$$

Next, we enforce the substitution $x-1/x \to x$ in the integrals on the right-hand side of $(4)$ and obtain

$$\begin{align} 2I&=\int_{-\infty}^{\infty}f\left(x\right)\,dx +\int_{-\infty}^{\infty}f\left(x\right)\,dx \\\\ &=2\int_{-\infty}^{\infty}f\left(x\right)\,dx \tag 5 \end{align}$$

Finally, dividing both sides of $(5)$ by $2$ and using $(1)$ we arrive at

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}f\left(x-\frac1x\right)\,dx=\int_{-\infty}^{\infty}f\left(x\right)\,dx }$$

as was to be shown!

Solution 4:

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