Convergence of $\int_0^\infty R(x) \log x$ where $R(x) = \frac1{x^2}R(\frac1x)$

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In some old notes of mine I found the following interesting result

Proposition: $$ I:= \int_0^\infty R(x) \log x \, \mathrm{d}x = 0, $$

when $R(x)$ is a function satisfying the functional equation

$$R(x) = \frac{1}{x^2} R\Bigr(\frac{1}{x}\Bigl).$$

For instance if $R(x) = 1$ then the result is obviously false. Is it enough to require that $\int_0^\infty R(x) \,\mathrm{d}x$ converges to ensure $I$ converges?

  • What are the weakest restrictions we need to impose upon $R(x)$ to ensure that $I$ convergences?