If there are no mistakes in my specialization from a formula (if you are interested see the Appendix) involving a definite integral and the calculations that I did can be justified, one has $$\int_0^1 \left(\sum_{n=1}^\infty\frac{\mu(n)}{n}( x^n-x^{n(n+1)-1}) \right) \frac{dx}{\log x}=1,\tag{1}$$ where $\mu(n)$ is the Möbius function. Thus informally seems that there exists a function $\mathcal{I}(x)$ defined over the unit interval satisfying $$\int_0^1\mathcal{I}(x)\frac{dx}{\log (x)} = 1.\tag{2}$$
Question. Provide more examples, examples of simple functions $\mathcal{G}(x)$ defined over the unit interval satisfying $(2)$, that is $$\int_0^1\mathcal{G}(x)\frac{dx}{\log (x)} = 1.$$ Imagine that I've a function such that $\int_0^1\mathcal{H}(x)dx=1$, then I know that it means that such function $\mathcal{H}$ has mean $1$ over our unit interval. Is it feasible a good interpretation of previous condition $(2)$ with mathematical meaning for your examples of functions $\mathcal{G}(x)$? Many thanks.
With Wolfram Alpha online calculator and this code written in Wolfram Language for a cutoff function, one can see a graph for the first factor inside the integrand of $(1)$
plot sum mu(n)/n log(n)(x^n-x^(n(n+1))), from n=1 to 1000, for 0<x<1
Appendix: I did my calculations to get $(1)$ using the first integral of page 537 from this nice article, for which there exist a free access from the AMS:
Jeffrey C. Lagarias, Euler’s constant: Euler’s work and modern developments, Bull. Amer. Math. Soc. 50 (2013).
I've edited the example $(1)$, I'm sorry. Many thanks for the patience all users.
Maybe I am missing something but $\int_0^1 (f(x) \log (x))\frac{dx}{\log (x)} = 1$ for any function $f$ that satisfies $\int_0^1 f(x)\;dx = 1$ and there are very many polynomial examples. So is $g(x) := f(x) \log(x)$ a simple function, and if not, why not?