<s>[INST] Functional equation: $R(1/x)/x^2 = R(x) $ 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 $(1)$?
 [/INST] 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.
Then
$\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 $B$.
</s>