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

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    $\begingroup$ In what sense does the integral exist? Riemann? Lebesgue? $\endgroup$ Commented Aug 1, 2013 at 13:20
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    $\begingroup$ Have you heard of the Glasser's Master theorem? I think that might help. $\endgroup$ Commented Dec 7, 2018 at 4:17
  • $\begingroup$ This was Problem B-4 on the 1968 Putnam Exam. $\endgroup$ Commented Apr 27, 2023 at 4:57

8 Answers 8

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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.

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  • $\begingroup$ Sir, how do you simplify the penultimate step? $\endgroup$ Commented Aug 20, 2015 at 12:20
  • $\begingroup$ @Kugelblitz Which one? $e^{\theta} + e^{-\theta} =2 \cosh\theta $ ? $\endgroup$ Commented Aug 20, 2015 at 12:35
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    $\begingroup$ No; I enquire clarification on the transition from the penultimate to the ultimate expression. $\endgroup$ Commented Aug 20, 2015 at 13:28
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    $\begingroup$ @Kugelblitz Ah, here it is: $d(2\sinh\theta)=2\cosh\theta\,d\theta$. $\endgroup$ Commented Aug 20, 2015 at 16:18
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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}

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    $\begingroup$ @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... $\endgroup$ Commented Oct 8, 2014 at 10:39
  • $\begingroup$ Mr. @GrahamHesketh Okay, done! (ô‿ô) $\endgroup$ Commented Oct 8, 2014 at 14:42
  • $\begingroup$ Nice.............. :) $\endgroup$ Commented Oct 10, 2014 at 15:07
  • $\begingroup$ @GrahamHesketh Thank you... (˘‿˘ʃƪ) $\endgroup$ Commented Oct 10, 2014 at 17:06
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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!

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  • $\begingroup$ 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? $\endgroup$ Commented Oct 3, 2015 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
    Commented Oct 3, 2015 at 14:23
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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.

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    $\begingroup$ So you assumed Lebesgue integrability I guess? $\endgroup$ Commented Aug 1, 2013 at 14:07
  • $\begingroup$ Why do you say that? $\endgroup$ Commented Aug 1, 2013 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
    Commented Aug 1, 2013 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'. $\endgroup$ Commented Aug 2, 2013 at 19:20
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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Aug 1, 2013 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
    Commented Aug 1, 2013 at 17:44
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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.

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In general, for any $a>0,$

$$ \int_{-\infty}^{\infty} f\left(x-\frac{a}{x}\right) d x=\int_{-\infty}^{\infty} f(x) d x $$ can be proved by substitution.

Noting that $$ f(x)=\underbrace{\frac{f(x)+f(-x)}{2}}_{=h(x) \text { is even }}+\underbrace{\frac{f(x)-f(-x)}{2}}_{=g(x) \text { is odd }} \text {. } $$ and $$ \begin{aligned}\int_{-\infty}^{+\infty} f\left(x-\frac{a}{x}\right)&=\int_{-\infty}^{+\infty} h(x-\frac{a}{x} ) d x+ \underbrace{ \int_{-\infty}^{\infty} g(x-\frac{a}{x} ) d x}_{=0 } \\&= \int_{-\infty}^{+\infty} h(x-\frac{a}{x}) d x\\&= 2\int_{0}^{+\infty} h(x-\frac{a}{x}) d x \end{aligned}\tag*{} $$ It is suffice to prove $$\int_{0}^{+\infty} h(x-\frac{a}{x}) d x= \int_{0}^{+\infty} h(x) d x $$

Letting $x\mapsto \frac{a}{x}$, then $$ \int_0^{+\infty} h\left(x-\frac{a}{x}\right) d x=\int_0^{+\infty} h\left(\frac{a}{x}-x\right) \frac{-a d x}{x^2}=-\int_0^{+\infty} \frac{a}{x^2} h\left(x-\frac{a}{x}\right) d x $$ Taking the average of the two versions yields $$ \begin{aligned} \int_0^{+\infty} h\left(x-\frac{a}{x}\right) d x & =\frac{1}{2} \int_0^{\infty}\left(1-\frac{a}{x^2}\right) h\left(x-\frac{a}{x}\right) d x \\ & =\frac{1}{2} \int_0^{\infty} h\left(x-\frac{a}{x}\right) d\left(x-\frac{a}{x}\right) \\ & =\frac{1}{2} \int_{-\infty}^{\infty} h(x) d x \end{aligned} $$

Now we can conclude that$$\int_{-\infty}^{\infty} f\left(x-\frac{a}{x}\right) d x=\int_{-\infty}^{\infty} f(x) d x$$

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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.

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