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The numbers do not appear to be a clue. Here are the first 1/2, 1/3, 5/16,19/60, 1687/5184, 2881/8640, 625961/1843200, 96314839/279936000, 302440148467/870912000000, 84408107137219/241416806400000, 2798993904047397389/7964120973312000000, 7033234035651624556939/19930212735713280000000
Perhaps you may expand $(\sqrt{z}-2)^{-1}$ in powers of $z$ then multiply by $e^{az} \text{erfc}(az)$ and apply Laplace transform to each term. The resulting terms would be derivatives one of the others. Possibly obtain a not converging series but maybe with some asymptotic properties. Or apply the same idea but expanding in powers of $z^{-1/2}$. The parameter a can be almost eliminated. This will simplify the computations.
In a recent paper by R. P. Brent and J. van de Lune (arXiv:1112.4911) you may find the proof of $$\sum_{n=1}^\infty \mu(n)\frac{x^n}{1+x^n} = x-2x^2$$ Therefore $$\lim_{x\to1^+} \sum_{n=1}^\infty \mu(n)\frac{x^n}{1+x^n} = -1.$$ This appear to indicate a certain bias to the negative of Merten's function.
I have the book, but I followed your link. Then I realized that my formula was not well copied. In fact the formula of Landau has a power of $\rho(x)$ on the integrand. I have corrected it. Now I think your formula can be obtained from that of Landau.
I must say that you integrates $\rho(x)^r$, I have not notice this before. Therefore, now I do not see any way to get your result from that of Landau. But, these are certainly functional equations of the type you wanted.
@Adam Azzam I can cite many many papers on Functional Analysis that uses subspaces in his usual sense. And always says closed subspace when this is the intended meaning. Your remarked phrase is false. You may put it as If B is a Banach space, then B is not the countable union of proper closed subspaces. This is true and you must correct. In the proof you also must put the adjective closed where this is your meaning.
