The following can be shown without much hassle.

Suppose $R$ is a rational function satisfying the following functional equation. \begin{align} \frac{1}{x^2} R\left( \frac{1}{x} \right) = R(x) \qquad \forall \: x \in \mathbb{R} \backslash \{0\} \tag{1} \end{align}

Then $$ \int_0^\infty R(x) \cdot \log x \,\mathrm{d}x = 0 $$ if it converges.

The proof is done by splitting the integral from $0$ to $1$ and $1$ to $\infty$. And then map $[1,\infty) \to [0,1]$ by $u \mapsto 1/x$. There are many rational functions satisfying this equation. A few examples below

$$ R(x) = \frac{1}{x}\,,\ \frac{x}{(1+x^2)^2} \,,\ \frac{1}{x^2 + x + 1}\,,\ \frac{1}{x^2+1} $$

My question is: Can one find groups or families of functions satisfying $(1)$?

  • 2
    $\begingroup$ Is it a conjecture or easily proven? $\endgroup$ Jul 21, 2013 at 22:28
  • $\begingroup$ Sorry, I fixed the statement! $\endgroup$ Jul 21, 2013 at 22:35
  • $\begingroup$ Well, your example of $\frac{1}{x}$ is already outside that group, isn't it? $\endgroup$
    – Lord Soth
    Jul 21, 2013 at 23:08
  • $\begingroup$ Writing $\forall x \in \mathbb R$ in (1) would include $x=0$, right? $\endgroup$
    – GEdgar
    Jul 21, 2013 at 23:17
  • $\begingroup$ If $R$ is a solution, then so is $aR$ for any real number $a$. $\endgroup$
    – user14972
    Jul 21, 2013 at 23:57

5 Answers 5


Observation: $R(x)$ satisfies $xR(x)=x^{-1}R(x^{-1})$ iff $A(x):=xR(x)$ satisfies $A(x)=A(x^{-1})$.

The map $\overline{p(x)}=x^{-\deg p}p(x^{-1})$ satisfies the following three properties:

$$\overline{\overline{p(x)}}=p(x)\quad{\rm and}\quad \overline{p(x)q(x)}=\overline{p(x)}\,\,\overline{q(x)} \quad{\rm and}\quad \overline{\lambda}=\lambda.$$

Suppose that $\overline{\wp}=\lambda\wp$ for some scalar $\lambda$. Then $\wp=\overline{\overline{\wp}}=\overline{\lambda\wp}=\lambda\overline{\wp}=\lambda^2\wp$, so $\lambda=\pm1$. Suppose we have a rational function $A(x)\in F(x)$ invariant under $x\leftrightarrow x^{-1}$. As $F[x]$ is a UFD, $A$ factors as

$$\begin{array}{ll} A(x) & =x^n\left[\prod_\pi \pi(x)^{e(\pi)}\left(x^{\deg \pi}\pi(x^{-1})\right)^{\ell(\pi)}\right]\prod_\wp \wp(x)^{h(\wp)} \\ & = x^{-n}\left[\prod_\pi \pi(x^{-1})^{e(\pi)}\left(x^{-\deg \pi}\pi(x)\right)^{\ell(\pi)}\right]\prod_\wp \wp(x^{-1})^{h(\wp)} \\ & = \pm\, x^{\large\left(-n-\sum\limits_\pi [e(\pi)+\ell(\pi)]\deg \pi-\sum\limits_\wp h(\wp)\deg\wp\right)}\left[\prod_\pi \pi(x)^{\ell(x)}\left(x^{\deg \pi}\pi(x^{-1})\right)^{e(\pi)}\right]\prod_\wp \wp(x)^{h(\wp)}\end{array}$$

$$\begin{array}{ll} \iff & e(\pi)=\ell(\pi),\quad n=-\sum_\pi e(\pi)\deg\pi-\frac{1}{2}\sum_\wp h(\wp)\deg\wp,\quad \prod_\wp(\underbrace{\wp^{-1}\overline{\wp}}_{\pm1})^{h(\wp)}=1 \\ \iff & A(x)=\left[\prod_\pi \left(\pi(x)\pi(x^{-1})\right)^{e(\pi)}\right]\prod_\wp \left(x^{-(\deg\wp)/2}\wp(x)\right)^{h(\wp)}\end{array} $$

up to rescaling. Here we have grouped terms so that $\pi,\overline{\pi},\wp$ exhaust all irreducibles (up to scaling) each exactly once, and $\wp$ covers all irreducibles with $\overline{\wp}=\pm\wp$. Clearly the $\wp$ and their exponents $h(\wp)$ must be chosen in such a way that $\prod_\wp(\wp^{-1}\overline{\wp})^{h(\wp)}=1$ and $\sum_\wp h(\wp)\deg\wp$ is even.

Now let's narrow our focus down to $F={\bf R}$. The irreducibles are all linear or quadratic. One quickly checks that the only irreducibles with $\overline{\wp}=\wp$ are $x^2+ax+1$ with $a\in(-2,2)$ or $x+1$, and the only irreducible with $\overline{\wp}=-\wp$ is $x-1$, up to scaling. Thus all $A$ are of the form



with $a_i^2-b_i<0$ and $b_i\ne1$ for each $i$, $c_j\ne\pm1$ for each $j$, $|u_k|<2$ for each $k$, and $s+t$ even.

Note that $x(x+1)^{-2}$ is not an example of an $R$ satisfying $xR(x)=x^{-1}R(x^{-1})$, but it is easily checkable that the other three examples derive from the final form above.

(We have worked entirely with elements from an abstract algebraic structure, thereby circumventing the need to pay attention to domains of rational functions.)

  • 3
    $\begingroup$ Amusingly, in an alternate branch of the multiverse I simply wrote "Hint: F[x] is a UFD" for an answer. $\endgroup$
    – anon
    Jul 22, 2013 at 1:27

A few remarks:

As noted by the other posters, if $R(1/x)/x^2 = R(x)$ then $f(x) := xR(x)$ satisfies $f(1/x) = f(x)$. All solutions have the form $R(x) = (1/x)f(x)$ for such an $f$.

If $f(x)=f(1/x)$ for all $x$ then $f(x) = \frac{1}{2}(f(x) + f(1/x)$). Conversely, if $h(x)$ is a rational function then $f(x) := h(x) + h(1/x)$ has the property that $f(x) = f(1/x)$.

Therefore, all the solutions are of the form $$R(x) = \frac{1}{x} (h(x) + h(1/x))$$ for some rational function $h$. More interestingly, you can generate solutions by defining $$R(x) = \frac{1}{x} g(h(x) + h(1/x))$$ for any rational functiond $g$ and $h$. For example, $$R(x) = \frac{1}{x}\frac{1}{1 + x + 1/x}$$ is on your list, with $h(x) = x$ and $g(x) = 1/(1+x)$. This does give an answer to your question, but anon gives a less obvious classification of all possible solutions in the answer above.

Finally, notice that the set of all $f$ satisfying $f(x) = f(1/x)$ forms a subfield of $\mathbb{R}(x)$. In other words, you can combine any two such $f$ by multiplication, divison, addition, subtraction and scalar multiplication, and you will get another such $f$, and another possible choice of $R$. There is probably a tidy handwavy description of what this field is in terms of schemes, but I don't know enough algebraic geometry to say.


I have been thinking about your problem again and came up with an alternative characterisation.

Notice that in each of your examples, the numerator and denominator have coefficients that read the same backwards as they do forwards. For example,

$$\frac{1}{(1+x)^2} = \frac{1}{1+2x+x^2}$$

and the coefficients in the denominator are $(1,2,1)$; a palindrome. Call a polynomial $p(x)$ a palindrome of index $k$ if

$$p(x) = x^k p(1/x)$$

For example, $p(x) = x^4 + 3x^3 + 3x^2 + x$ is a palindrome of index $5$. Then I claim that all the rational functions $r(x)$ which satisfy $r(1/x)=r(x)$ have the form $$r(x) = p(x)/q(x)$$ where $p$ and $q$ are palindromes of the same index. It follows that your functions are precisely those of the form $p(x)/q(x)$ where $p(x)$ is a palindrome of index $k$ and $q(x)$ is a palindrome of index $k+2$, for some $k$. For example, $1/(1+2x+x^2)$ is a palindrome of index $0$ divided by a palindrome of index $2$.

To prove the claim, suppose that $r(x) = p(x)/q(x)$ satisfies $r(1/x)= r(x)$ and work over $\mathbb{C}(x)$ so that $p(x)$ and $q(x)$ may be assumed to have no common linear factors. Choose $k$ to be the smallest possible power so that $x^k p(1/x)$ and $x^k q(1/x)$ are both polynomials. Let $P(x) = x^kp(1/x)$ and $Q(x) = x^kq(1/x)$. Then $$r(x) = p(x)/q(x) = P(x)/Q(x).$$ The polynomials $P$ and $Q$ have no common factor, since if $P(a) = Q(a) = 0$ then $a^kp(1/a) = a^k q(1/a) = 0$. But then either $a=0$, in which case $P$ and $Q$ have a common factor of $x$, which contradicts that $k$ is as small as possible, or else $p(1/a)=q(1/a)=0$, which contradicts that $p$ and $q$ have no common factor.

Then from $p(x)Q(x) = q(x)P(x)$, by factorisation into linear factors it follows that $p(x) = c_1P(x)$ and $q(x) = c_2Q(x)$ for some constants $c_1$ and $c_2$. But then $c_1/c_2=1$ and so $p(x) = cP(x)$, $q(x)=cQ(x)$. But by putting $x=1$, it follows that $(c-1)p(1) = (c-1)q(1) = 0$. Since $p(1)$ and $q(1)$ cannot both be zero, it follows that $c=1$ and therefore $p(x) = P(x) = x^kp(1/x)$ and $q(x) = Q(x) = x^kq(1/x)$ are palindromes of index $k$, as claimed.

Now you can write down all the functions satisfying $(1)$ by just writing down palindromes, for example

$$R(x) = \frac{x^4 + 3x^3 + 3x^2 + x}{2x^7 - x^4 - x^3 +2}$$

satisfies $R(1/x) = x^2R(x)$ and you can get them all this way.


Let us, first of all, look for solutions for $x>0$, and hope that they extend to the wanted solutions. In that case we can substitute $x=e^t$, and the equation becomes $$e^{-2t}R(e^{-t})=R(e^t),$$ or $$e^{-t} R(e^{-t})=e^t R(e^t).$$ We see that $e^t R(e^t)$ is even, and therefore has the form $$e^t R(e^t)=\varphi(|t|), $$ In terms of $x$ we find $$R(x)=\frac{1}{x} \varphi(|\log x|)$$

You should look for functions $\varphi$ such that the resulting $R(x)$ is rational for $x>0$, and extended it for $ x \in \mathbb R \setminus \{ 0 \}$.


If $\frac{1}{x^2} R\left( \frac{1}{x} \right) = R(x)$, $x^2 R(x) = R(1/x)$.

If $R(x) = A(x)/B(x)$, where $A$ and $B$ are relatively prime polynomials of respective degrees $n$ and $m$, $A(1/x)=a(x)/x^n$ and $B(1/x) = b(x)/x^m$, where $a(x)$ and $b(x)$ are the reciprocal polynomials of $A$ and $B$, respectively.


$\begin{align} x^2 A(x)/B(x) &= A(1/x)/B(1/x)\\ &= (a(x)/x^n)/(b(x)/x^m)\\ &= (a(x)/b(x))x^{m-n}\\ \end{align} $

so $x^{2+n} A(x)b(x)=a(x)B(x)x^m $.

If the degrees of $a(x)$ and $b(x)$ are the same as $A$ and $B$ (i.e., the constant terms of $A$ and $B$ are non-zero), the left-hand side has degree $2+2n+m$ and the right-hand side has degree $n+2m$, so $2+2n+m=2m+n$, or $m = n+2$. In this case, $A(x)b(x)=a(x)B(x) $. If, in addition, $A(x) = a(x) = 1$, so $n=0$ and $m=2$, $B(x) = b(x)$, so $B$ is a symmetric polynomial of degree $2$.

Conversely, if $R(x) = 1/B(x)$ where $B$ is a quadratic symmetric polynomial, $R(1/x) = 1/B(1/x) = 1/(B(x)/x^2) =x^2/B(x) = x^2R(x)$.

I will leave it at this for now, and possibly work out later the cases when $a$ and $b$ are not the same degree as $A$ and $B$.


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