Consider $f : \mathbb{R} \rightarrow \mathbb{R}$ to be a continuous function such that $\int_{-\infty}^{\infty} f(x) dx$ exists. I would like to prove that $$\int_{-\infty}^{\infty} f(x - \frac{1}{x}) dx = \int_{-\infty}^{\infty} f(x) dx.$$ I decided to tackle it in cases. In particular, I started out with the case where $f$ was linear. We see that $$\int_{-\infty}^{\infty} f\left(x-\frac{1}{x}\right) dx = \int_{-\infty}^{\infty} f(x) - f\left(\frac{1}{x}\right) dx = \int_{-\infty}^{\infty} f(x) dx - \int_{-\infty}^{\infty} f\left(\frac{1}{x} \right) dx.$$ This problem then boils down into proving that $$\int_{-\infty}^{\infty} f\left(\frac{1}{x}\right) dx = 0.$$ we set $u = \frac{1}{x} \implies du = \frac{-1}{x^2} dx.$ However, I am having problems simplifying the term inside the integral into a function of $u$. Any suggestions on this case? And any recommendations on the case where $f$ is not linear?
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1The case $f$ is linear could be good to get a sense for the problem, but it's unlikely to help because the case "$f$ is not linear" is probably as hard as the original question. – Caleb Stanford Sep 19 '16 at 22:04
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2Hint. Split the original integral into to parts, one from $-\infty$ to $0$ (which we call $I_1$) and the other from $0$ to $\infty$ (which we call $I_2$). Now apply the substitution $x \mapsto -1/x$ to $I_1$ and simplify the resulting expression for $I_1 + I_2$ using another substitution. Also I am pretty sure this is asked several times before with some nice generalizations, so you may want to check them. – Sangchul Lee Sep 19 '16 at 22:24
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1@SangchulLee While you were writing your comment I was writing my answer. It's the same idea. – Felix Marin Sep 19 '16 at 22:32
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The only linear functions $\mathbb{R}\to\mathbb{R}$ are $f(x)=mx$ (or in elementary algebra we sometimes say $f(x)=mx+b$ is because the graph is a line, but this isn't additive), in which case $\int_{-\infty}^\infty$ doesn't exist. If you replace "linear" with "additive" then we'd probably have to involve measure theory (and $f$ wouldn't be continuous). – anon Sep 20 '16 at 04:01
3 Answers
$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mc{J} & = \color{#f00}{\int_{-\infty}^{\infty}\mrm{f}\pars{x - {1 \over x}}\,\dd x}\ =\ \overbrace{\int_{0}^{\infty}\bracks{\mrm{f}\pars{x - {1 \over x}} + \mrm{f}\pars{-x + {1 \over x}}}\,\dd x}^{\ds{=\ \mc{J}}} \\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=} \,\,\, \int_{\infty}^{0}\bracks{\mrm{f}\pars{{1 \over x} - x} + \mrm{f}\pars{-\,{1 \over x} + x}}\,\pars{-\,{\dd x \over x^{2}}} \\[5mm] & = \underbrace{\int_{0}^{\infty}\bracks{\mrm{f}\pars{{1 \over x} - x} + \mrm{f}\pars{-\,{1 \over x} + x}}\,{1 \over x^{2}}\,\dd x}_{\ds{=\ \mc{J}}} \end{align}
\begin{align} \mc{J} & = {\mc{J} + \mc{J} \over 2} = {1 \over 2}\int_{0}^{\infty}\bracks{\mrm{f}\pars{{1 \over x} - x} + \mrm{f}\pars{-\,{1 \over x} + x}}\,\pars{1 + {1 \over x^{2}}}\,\dd x \\[5mm] & \stackrel{t\ \equiv\ 1/x - x}{=}\,\,\, {1 \over 2}\int_{\infty}^{-\infty}\bracks{\mrm{f}\pars{t} + \mrm{f}\pars{-t}}\, \pars{-\,\dd t} = \color{#f00}{\int_{0}^{\infty}\mrm{f}\pars{x}\,\dd x} \end{align}
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I cannot seem to be able to reproduce the last step... could you expand? – b00n heT Sep 19 '16 at 22:48
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@b00nheT Note that $ \int_{-\infty}^{\infty},\mathrm{f}\left(t\right),\mathrm{d}t = \int_{-\infty}^{\infty},\mathrm{f}\left(-t\right),\mathrm{d}t $ – Felix Marin Sep 19 '16 at 22:54
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Thanks @Dr.MV I guess sometimes I lost some time because I try to write clearly everything. – Felix Marin Sep 20 '16 at 02:11
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Felix, I just posted a solution with another way forward. It's always amazing to see that two different approaches lead to the same end point. -Mark – Mark Viola Sep 20 '16 at 02:25
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METHODOLOGY $1$:
First, we split the integral as
$$\begin{align} \int_{-\infty}^\infty f\left(x-\frac1x\right)\,dx&=\int_{0}^\infty f\left(x-\frac1x\right)\,dx +\int_0^\infty f\left(-x+\frac1x\right)\,dx \tag 1 \end{align}$$
In the first integral on the right-hand side of $(1)$, we enforce the substitution $x\to e^x$ and in the second integral, we enforce the substitution $x\to e^{-x}$. Proceeding, we find that
$$\begin{align} \int_{-\infty}^\infty f\left(x-\frac1x\right)\,dx&=\int_{-\infty}^\infty f\left(2\sinh(x)\right)\,e^x\,dx +\int_{-\infty}^\infty f\left(2\sinh(x)\right)\,e^{-x}\,dx \\\\ &=\int_{-\infty}^\infty f(2\sinh(x))\,2\cosh(x)\,dx\tag2 \end{align}$$
Finally, we enforce the substitution $2\sinh(x)\to x$ in the integral on the right-hand side of $(2)$ to obtain
$$\int_{-\infty}^\infty f\left(x-\frac1x\right)\,dx=\int_{-\infty}^\infty f(x)\,dx$$
as was to be shown!
METHODOLOGY $2$:
We begin with the expression given in $(1)$. In the first integral on the right-hand side of $(1)$, we enforce the substitution $x-\frac1x\to x$, while in the second integral on the right-hand side of $(1)$ we enforce the substitution $-x+\frac1x \to x$. Proceeding, we find that
$$\begin{align} \int_{-\infty}^\infty f\left(x-\frac1x\right)\,dx&=\int_{-\infty}^\infty f\left(x\right)\,\left(\frac12+\frac{x}{2\sqrt{x^2+1}}\right)\,dx +\int_{-\infty}^\infty f\left(x\right)\,\left(\frac12-\frac{x}{2\sqrt{x^2+1}}\right)\,dx \\\\ &=\int_{-\infty}^\infty f(x)\,dx \end{align}$$
as expected!
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+1. We usually use this substitution in gaussian integrals too. Nice job. – Felix Marin Sep 20 '16 at 02:28
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$x-\frac1x=u\implies\left(1+\frac1{x^2}\right)\mathrm{d}x=\mathrm{d}u$
$$
\begin{align}
&\int_{-\infty}^\infty f\left(x-\frac1x\right)\mathrm{d}x\tag{1}\\
&=\int_{-\infty}^0f\left(x-\frac1x\right)\mathrm{d}x+\int_0^\infty f\left(x-\frac1x\right)\mathrm{d}x\tag{2}\\
&=\int_0^\infty f\left(x-\frac1x\right)\frac1{x^2}\,\mathrm{d}x+\int_{-\infty}^0f\left(x-\frac1x\right)\frac1{x^2}\,\mathrm{d}x\tag{3}\\
&=\frac12\int_0^\infty f\left(x-\frac1x\right)\left(1+\frac1{x^2}\right)\mathrm{d}x
+\frac12\int_{-\infty}^0f\left(x-\frac1x\right)\left(1+\frac1{x^2}\right)\mathrm{d}x\tag{4}\\
&=\frac12\int_{-\infty}^\infty f(u)\,\mathrm{d}u+\frac12\int_{-\infty}^\infty f(u)\,\mathrm{d}u\tag{5}\\
&=\int_{-\infty}^\infty f(u)\,\mathrm{d}u\tag{6}
\end{align}
$$
Explanation:
$(2)$: split up domain of integration
$(3)$: substitute $x\mapsto-\frac1x$
$(4)$: average $(2)$ and $(3)$
$(5)$: substitute $x-\frac1x=u$
$(6)$: algebra
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This answer got my vote for being the cleanest, both notationally and computationally. – heropup Sep 20 '16 at 03:58