I was recently fiddling with the moebius function and realized I could do some manipulations on it.
To Prove
$$ f(x) + f(x^2) + \dots = x $$
$$\implies \left(1 - \frac{13}{12 n^2}\right) \int_{0^+}^\infty (-x)e^{-x\left(1 + \frac{1}{3 n^2}\right)} f'(e^{-x\left(1 + \frac{1}{3 n^2}\right)}) dx = \frac{1}{n(n+1)} \tag{1}$$
Background and Manipulations
Consider the series:
$$ f(x) := \sum_{r=1}^\infty \mu(r) x^r \tag{2}$$
This satisfies:
$$ \sum_{r=1}^\infty f(x^r) = f(x) + f(x^2) + \dots = x \tag{3}$$
We will use $x = 1 - \epsilon = 1- \frac{1}{n}$ where $n$ is a large integer.
$$ \sum_{r=1}^\infty f((1- \frac{1}{n})^r) = 1- \frac{1}{n} \tag{4}$$
We choose $n$ such that $(1- \frac{1}{n})^r = e^{r \ln(1- \frac{1}{n}) } \sim e^{\frac{-r}{n} - \frac{r}{2 n^2} - \frac{r}{3 n^3}} \tag{5}$
Thus:
$$ \lim_{k \to \infty} \sum_{r=1}^{nk} f(e^{\frac{-r}{n} - \frac{r}{2 n^2} - \frac{r}{3 n^3} - \frac{r}{4 n^4}} ) = 1- \frac{1}{n} \tag{6}$$
We now change $n \to n+1$
$$ \lim_{k \to \infty} \sum_{r=1}^{(n+1)k} f(e^{\frac{-r}{n+1} - \frac{r}{2 (n+1)^2} - \frac{r}{3(n+1)^3 } - \frac{r}{4 (n+1)^4} } ) = 1- \frac{1}{n+1} \tag{7}$$
Re expressing in terms of $\frac{r}{n}$
$$ \implies \lim_{k \to \infty} \sum_{r=1}^{(n+1)k} f(e^{\frac{-r}{n} \cdot (1 + \frac{1}{n})^{-1} - \frac{r}{2n^2}(1+\frac{1}{n})^{-2} - \frac{r}{3n^3} \ (1 + \frac{1}{n})^{-3} - \frac{r}{4 n^4} } ) = 1- \frac{1}{n+1} \tag{8} $$
$$ \implies \lim_{k \to \infty} \sum_{r=1}^{(n+1)k} f(e^{\frac{-r}{n} \cdot (1 - \frac{1}{n} + \frac{1}{n^2} - \frac{1}{n^3} ) - \frac{r}{2n^2}(1-\frac{2}{n}+\frac{3}{n^2}) - \frac{r}{3n^3} \ (1 - \frac{3}{n}) - \frac{r}{4 n^4} } ) = 1- \frac{1}{n+1} \tag{9}$$
$$ \implies \lim_{k \to \infty} \sum_{r=1}^{(n+1)k} f(e^{\frac{-r}{n} + \frac{r}{2n^2} - \frac{r}{3n^3} - \frac{5r}{6 n^4} } ) = 1- \frac{1}{n+1} \tag{10}$$
Subtracting the $n$ and simplified $n+1$ equation:
$$ \lim_{k \to \infty} \sum_{r=1}^{nk} (f(e^{\frac{-r}{n} - \frac{r}{2 n^2} - \frac{r}{3 n^3} - \frac{r}{4 n^4} } ) - f(e^{\frac{-r}{n} + \frac{r}{2n^2} - \frac{r}{3n^3} - \frac{5r}{6 n^4} } )) + \sum_{r=kn+1}^{k(n+1)}f(e^{\frac{-r}{n+1} - \frac{r}{2 (n+1)^2} - \frac{r}{3(n+1)^3 } - \frac{r}{4 (n+1)^4} } )) = \frac{1}{n(n+1)} \tag{11}$$
Taking common factor of $\exp({\frac{-r}{n} - \frac{r}{3 n^3}})$ and resuming the second series:
$$ \lim_{k \to \infty} \sum_{r=1}^{nk} (f(e^{\frac{-r}{n} - \frac{r}{3 n^3}}e^{ - \frac{r}{2 n^2} - \frac{r}{4 n^4} } ) - f(e^{\frac{-r}{n} - \frac{r}{3n^3}} e^{ \frac{r}{2n^2} - \frac{5r}{6 n^4} } )) + \sum_{r=1}^{k} f(e^{\frac{-(kn+r)}{n+1} - \frac{(kn+r)}{2 (n+1)^2} - \frac{(kn+r)}{3(n+1)^3 } - \frac{(kn+r)}{4 (n+1)^4} } ) = \frac{1}{n(n+1)} \tag{12}$$
Expanding the first series inside the bracket and taking $kn + r$ in the second series common:
$$ \lim_{k \to \infty} \sum_{r=1}^{nk} (f(e^{\frac{-r}{n} - \frac{r}{3 n^3}}{(1 - \frac{r}{2 n^2} - \frac{r}{4 n^4} + \frac{r^2}{8 n^4} } )) - f(e^{\frac{-r}{n} - \frac{r}{3n^3}} {(1+ \frac{r}{2n^2} - \frac{5r}{6 n^4} + \frac{r^2}{8 n^4} } )) + \sum_{r=1}^{k} f(e^{\frac{-(kn+r)}{n+1} (1 - \frac{1}{2 (n+1)} - \frac{1}{3(n+1)^2} - \frac{1}{4 (n+1)^3} )} ) = \frac{1}{n(n+1)} \tag{13}$$
Taylor expanding and simplifying while multiplying the second series with $n+1$ in the numerator and denominator:
$$ \lim_{k \to \infty} \sum_{r=1}^{nk} (f'(e^{\frac{-r}{n} - \frac{r}{3 n^3}}) {e^{\frac{-r}{n} - \frac{r}{3 n^3}}( - \frac{r}{ n^2} - \frac{13r}{12 n^4} } )) + (n+1) \sum_{r=1}^{k} f(e^{\frac{-kn}{n+1} (1 - \frac{1}{2 (n+1)} - \frac{1}{3(n+1)^2} - \frac{1}{4 (n+1)^3})} \cdot e^{\frac{-r}{n+1} (1 - \frac{1}{2 (n+1)} - \frac{1}{3(n+1)^2} - \frac{1}{4 (n+1)^3})} \frac{1}{n+1} = \frac{1}{n(n+1)} \tag{14}$$
Writing as an integral:
$$\implies \int_{0^+}^\infty (-x)e^{-x(1 + \frac{1}{3 n^2})} f'(e^{-x(1 + \frac{1}{3 n^2})}) dx - \frac{13}{12 n^2} \int_{0^+}^\infty (-x)e^{-x(1 + \frac{1}{3 n^2})} f'(e^{-x(1 + \frac{1}{3 n^2})} dx) + (n+1) \lim_{k \to \infty} \int_{0^+}^k f(e^{- k \lambda -x \lambda} ) dx = \frac{1}{n(n+1)} \tag{15}$$
Where $\lambda$ is a function of $n$. Taking the integral common:
$$\implies (1 - \frac{13}{12 n^2}) \int_{0^+}^\infty (-x)e^{-x(1 + \frac{1}{3 n^2})} f'(e^{-x(1 + \frac{1}{3 n^2})}) dx + (n+1) \lim_{k \to \infty} \int_{0^+}^k f(e^{- k \lambda -x \lambda} ) dx = \frac{1}{n(n+1)} \tag{16}$$
Using a change of limits in the second integral $x + k \to z$:
$$\implies (1 - \frac{13}{12 n^2}) \int_{0^+}^\infty (-x)e^{-x(1 + \frac{1}{3 n^2})} f'(e^{-x(1 + \frac{1}{3 n^2})}) dx + (n+1) \lim_{k \to \infty} \int_{k}^{2k} f(e^{x \lambda} ) dx = \frac{1}{n(n+1)} \tag{17}$$
Which makes the second integral go to $0$:
$$\implies (1 - \frac{13}{12 n^2}) \int_{0^+}^\infty (-x)e^{-x(1 + \frac{1}{3 n^2})} f'(e^{-x(1 + \frac{1}{3 n^2})}) dx = \frac{1}{n(n+1)}\tag{18} $$
Can we say something $f(x)$ using the above equation?
Question
Is it possible to make the above proof rigorous? (I've tried to provide some sort of error estimate) Can the error be improved upon? Is there a useful error bound approximation for the converting an summation into an integral? Can anyone computationally show which step of mine is wrong? (It does not seem intuitive to me)
P.S: This post has been heavily (twice) edited to include an error. Hopefully it is now immune to some of the comments made(?)