61
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ In what sense does the integral exist? Riemann? Lebesgue? $\endgroup$ – Patrick Da Silva Aug 1 '13 at 13:20
  • $\begingroup$ Have you heard of the Glasser's Master theorem? I think that might help. $\endgroup$ – Rohan Shinde Dec 7 '18 at 4:17
78
+100
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Sir, how do you simplify the penultimate step? $\endgroup$ – Kugelblitz Aug 20 '15 at 12:20
  • $\begingroup$ @Kugelblitz Which one? $e^{\theta} + e^{-\theta} =2 \cosh\theta $ ? $\endgroup$ – Start wearing purple Aug 20 '15 at 12:35
  • $\begingroup$ No; I enquire clarification on the transition from the penultimate to the ultimate expression. $\endgroup$ – Kugelblitz Aug 20 '15 at 13:28
  • $\begingroup$ @Kugelblitz Ah, here it is: $d(2\sinh\theta)=2\cosh\theta\,d\theta$. $\endgroup$ – Start wearing purple Aug 20 '15 at 16:18
  • $\begingroup$ Very well done! I just posted another way forward that uses a different substitution. I'd be very curious to hear your thoughts, if you have time. ;-)) Mark $\endgroup$ – Mark Viola Oct 2 '15 at 17:53
20
$\begingroup$

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}

$\endgroup$
  • 1
    $\begingroup$ @GrahamHesketh Wait, I've made a mistake. I'll fix it later, right now I'm in the way to the course place. See ya... $\endgroup$ – Anastasiya-Romanova 秀 Oct 8 '14 at 10:39
  • $\begingroup$ Mr. @GrahamHesketh Okay, done! (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 Oct 8 '14 at 14:42
  • $\begingroup$ Nice.............. :) $\endgroup$ – Graham Hesketh Oct 10 '14 at 15:07
  • $\begingroup$ @GrahamHesketh Thank you... (˘‿˘ʃƪ) $\endgroup$ – Anastasiya-Romanova 秀 Oct 10 '14 at 17:06
  • $\begingroup$ This is a very straightforward approach and the "natural one." So, I just posted another way forward that uses a different substitution. I'd be very curious to hear your thoughts if you have time. ;-)) Mark $\endgroup$ – Mark Viola Oct 2 '15 at 17:55
17
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ My thoughts, I think your approach is quite similar to mine. You have not avoided the "weird" substitution as such, in fact you use the same substitution that I use $x-1/x=y$ (two solutions for x). Is that correct? $\endgroup$ – Graham Hesketh Oct 3 '15 at 9:06
  • $\begingroup$ Well, not quite correct. The last substitution is indeed $x-1/x\to x$, but this comes only after coupling the original integral with itsself, transformed-under-inversion. This provides a way forward that involves a natural "u" substitution. $\endgroup$ – Mark Viola Oct 3 '15 at 14:23
12
$\begingroup$

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

$\endgroup$
  • $\begingroup$ So you assumed Lebesgue integrability I guess? $\endgroup$ – Patrick Da Silva Aug 1 '13 at 14:07
  • $\begingroup$ Why do you say that? $\endgroup$ – Graham Hesketh Aug 1 '13 at 14:17
  • $\begingroup$ @PatrickDaSilva I think this works for improper Riemann integrals too. Just replace $0$ and $\infty$ by variables, do the change, and take the limit as those variables tend to $0$ and $\infty$. $\endgroup$ – Potato Aug 1 '13 at 19:15
  • $\begingroup$ @Graham : I'm saying that because you broke an improper integral into two parts, well I'm only sure everything is fine if you assume the function is Lebesgue integrable, but I'm not sure what happens when you do Riemann integrability. Something like 'the integral of a sum is the sum of the integrals, if those two integrals exist' and the issues with that are similar to 'the limit of a sum is the sum of the limits, if those limits exist'. $\endgroup$ – Patrick Da Silva Aug 2 '13 at 19:20
  • $\begingroup$ This is a very straightforward approach and the "natural one." So, I just posted another way forward that uses a different substitution. I'd be very curious to hear your thoughts if you have time. ;-)) Mark $\endgroup$ – Mark Viola Oct 2 '15 at 17:55
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I like your method but I'd like to object to the epithet "weird". I could obviously use exclusively $x\mapsto - x^{-1}$ but that would make my derivation less symmetric. $\endgroup$ – Start wearing purple Aug 1 '13 at 17:43
  • $\begingroup$ Fair enough. I guess it isn't "weird" in the sense that there are no square roots. Your $x=e^\theta$ change of variables is nice enough. $\endgroup$ – abnry Aug 1 '13 at 17:44
  • $\begingroup$ This is a neat approach. I just posted another way forward that avoids the "weird" substitution and circumvents differentiating under the integral. I'd be very curious to hear your thoughts if you have time. ;-)) Mark $\endgroup$ – Mark Viola Oct 2 '15 at 17:52
6
$\begingroup$

Reference:

Glaisher 1876 On a formula of Cauchy's for the evaluation of a class of definite integrals, Proc. Camb. Phil. Soc. III, p. 5-12

This has the original reference to Cauchy, and an added solution by Cayley using a very original change of variable, "weirder" than those used in previous answers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.