Is $ \ln \delta \int_0^1 \frac{f(z)}{z(\ln z)^2} dz \sim \sum_{r=1}^\infty \mu(r) f(\delta^{1/r}) $? Background
I think I can show something interesting:
$$  \ln \delta  \int_0^1 \frac{f(z)}{z(\ln z)^2} dz \sim \sum_{r=1}^\infty \mu(r) f(\delta^{1/r}) $$
where $\delta \to 0$ and $\int_0^1 \frac{f(z)}{z(\ln z)^2} dz$ converges. However, there is something that worries me. Let $\delta \to \delta^{1/k}$:
$$  \frac{1}{k}\ln \delta  \int_0^1 \frac{f(z)}{z(\ln z)^2} dz \sim 
 \sum_{r=1}^\infty \mu(r)f(\delta^{1/(kr)})  $$
Summing both sides $k=1$ to $\infty$:
$$  \sum_{k=1}^\infty \frac{1}{k}\ln \delta  \int_0^1 \frac{f(z)}{z(\ln z)^2} dz  \sim f(\delta) $$
Obviously the L.H.S goes to $\infty$ where as the R.H.S is finite. Can someone spot the error (in the contradiction/proof)?
Proof
Consider:
$$ f(u) +f(u^{1/2}) + f(u^{1/3}) + \dots = g(u) $$
Now consider:
$$ I = \int_0^\infty f(e^{-1/x}) dx $$
Writing using asymptotics and using Riemann limit of a sum with $\epsilon \to 0$:
$$ I \sim (f(e^{-1/\epsilon}) + f(e^{-1/ 2 \epsilon}) + f(e^{-1/3 \epsilon}) + \dots )\epsilon$$
Now using $\epsilon = \frac{-1}{\ln \delta} \to 0$. Hence:
$$ -( \ln \delta )I \sim f(\delta) +f(\delta^{1/2}) + f(\delta^{1/3}) + \dots$$
Substituting $g(\delta)$ and $I$:
$$ -( \ln \delta ) \int_0^\infty f(e^{-1/x}) dx \sim g(\delta)$$
Using mobius inversion:
$$ -( \ln \delta ) \int_0^\infty f(e^{-1/x}) dx \sim \sum_{r=1}^\infty \mu(r)f(\delta^{1/r}) $$
Now, let $z=\exp(-1/x)$ and $dx = \frac{dz}{-z (\ln z)^2}$. Hence,
$$  \ln \delta  \int_0^1 \frac{f(z)}{z(\ln z)^2} dz \sim \sum_{r=1}^\infty \mu(r)f(\delta^{1/r}) $$
where $\delta \to 0$ and $\int_0^1 \frac{f(z)}{z(\ln z)^2} dz$ converges.
 A: You write
$$
-(\log \delta) I\sim \left(f(\delta)+f(\delta^{1/2})+f(\delta^{1/3})+\ldots\right).
$$
Then
$$
-(\log\delta)\int^{\infty}_{0}f(e^{-1/x})dx\sim g(\delta)
$$
which is right. But then you say: Using Moebius inversion
$$
-(\log\delta)\int^{\infty}_{0}f(e^{-1/x})dx\sim \sum^{\infty}_{r=1}\mu(r)f(\delta^{1/r}).
$$
This is not true since it is again
$$
-(\log\delta)\int^{\infty}_{0}f(e^{-1/x})dx\sim \sum^{\infty}_{r=1}f(\delta^{1/r})
$$
and you perform no inversion. The right formula is
$$
\int^{1}_{0}\frac{f(t)}{t\log^2 t}dt=-\frac{1}{\log \delta}\sum^{\infty}_{r=1}f(\delta^{1/r})+o_{\delta}(1)\textrm{, }\delta\rightarrow 0^{+}
$$
for $\frac{f(t)}{t\log^2 t}$ continuous and bounded in $[0,1]$. Say $f(t)=\sin^2(\pi t)$, $f(t)=(t(t-1))^2$. In general holds the next
Theorem (see here). If $s_n$, $n\geq 1$ is uniformly distributed in $[a,b]$, $a<b$ and $f(x)$ is continuous and bounded in $[a,b]$, then
$$
\lim_{n\rightarrow+\infty}\frac{1}{n}\sum^{n}_{j=1}g\left(s_j\right)=\frac{1}{b-a}\int^{b}_{a}g(t)dt.
$$
And, also
$$
\int^{\infty}_{0}f(e^{-1/t})dt=\int^{b}_{0}f(e^{-1/t})dt+\int^{\infty}_{b}f(e^{-1/t})dt\sim\epsilon\sum^{N}_{k=1}f(e^{-1/k\epsilon})+\int^{\infty}_{b}f(e^{-1/t})dt,
$$
where $\epsilon=b/N$
