Introducing, by means of an artificial way, the Riesz function in the integrand of the offset logarithmic integral one has for $X>2$ that $$\operatorname{Li}(X)=\int_2^X \frac{t}{\log ^2 t}\sum_{n=1}^\infty \operatorname{Riesz}\left(\frac{\log t}{n^2}\right)dt=\int_2^X\frac{t\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(k)t^{-1/(nk)^2}}{(nk)^2}}{\log t}dt.$$
Then as corollary one has that
Fact. Let a real number $t>2$, then the numerator of the integrand in second integral is $1.$
Now I don't know if my calculations to introduce the Riesz function in the integral of the logarithmic function aren't necessary to prove previous fact, and neither I know how think get with the tricks that I know about the theory of Möbius inversion. Can you prove it? That is can you answer this
Question. Let $\mu(n)$ the Möbius function, can you prove (directly, or using a statement that you do know: Möbius inversion or other) that for $t>2$ $$\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{\mu(k)}{(nk)^2t^{1/(nk)^2}}=\frac{1}{t}?$$ If it is well know, you can provide me the name of the theorem and hints to get it. Many thanks.
I believe that my calculations to get the first identity are rights, ask if you need it. Thus if you find a mistake, I am saying if the identity in Question is wrong, tell us.
All instances of $n$ appear multiplied by $k$, which implies reindexing by $nk$ might lead us somewhere. Setting $u=nk$ gives us
$$\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{\mu(k)}{(nk)^2 t^{1/(nk)^2}} = \sum_{u=1}^\infty \frac{\sum_{k|u} \mu(k)}{u^2 t^{1/u^2}} $$
The numerator is the Dirichlet convolution of $\mu(n)$ with the constant function $1(n)$ yielding the identity function of the convolution $\epsilon(n) = [n=1]$. Using that yields the conclusion
$$\sum_{u=1}^\infty \frac{\sum_{k|u} \mu(k)}{u^2 t^{1/u^2}} = \sum_{u=1}^\infty \frac{[u=1]}{u^2 t^{1/u^2}} = \frac{1}{t} $$