Let's say for a particular function I have $f$:
$$ f(x) + f(x^2)+ f(x^3)+ \dots +f(x^N) = g(x) $$
We start by considering the integral:
$$ I_f = \int_0^{b} f (e^{-\frac{1}{x}}) dx $$
Using asymptotics and Riemann Limit as a sum:
$$ I_f \sim (f(e^\frac{-1}{\epsilon}) + f(e^\frac{-1}{2\epsilon})+ \dots + f(e^\frac{-1}{N \epsilon}) ) \epsilon$$
with $N \epsilon = b$ where $\epsilon \nearrow 0$ . Now choosing $\epsilon = -\frac{1}{\ln \delta}$:
$$ - \ln (\delta) I_f \sim f(\delta) + f(\delta^{1/2}) + \dots+ f(\delta^{1/N})$$
Now let us replace $\delta$ with $\delta^k$ and sum over $1$ to $N$:
$$ - \frac{N(N+1)}{2}\ln (\delta) I_f \sim g(\delta) + g(\delta^{1/2}) + \dots+ g(\delta^{1/N})$$
Using the same trick as above:
$$ - \frac{N(N+1)}{2}\ln (\delta) I_f \sim - \ln (\delta) I_g$$
Cancelling the redundant terms:
$$ \frac{N(N+1)}{2} I_f \sim I_g$$
Substituting $I_f$ and $I_g$:
$$ \frac{N(N+1)}{2} \int_0^{\frac{-N}{\ln \delta}} f (e^{-\frac{1}{x}}) dx \sim \int_0^{\frac{-N}{\ln \delta}} g (e^{-\frac{1}{x}}) dx $$
Question
I am pessimistic this formula works. Is there any example $f$ and $g$ which satisfy this? How bad is the error?