An Analogous Riemann Integral $$1=\sum_{n=2}^\infty (\zeta (n)-1)$$ is a fairly well known result W|A validates this result Is there a closed form to the analogous integral: $$\text{?}=\int_2^\infty \text{d}x \, (\zeta(x)-1)$$ I have managed to prove that the integral converges, but can get nowhere beyond a numeric approximation.

Since $\zeta(x) - 1 = \sum_{n=2}^\infty n^{-x}$, your integral is $$ \sum_{n=2}^\infty \int_2^\infty n^{-x}\; dx = \sum_{n=2}^\infty \dfrac{1}{n^2 \ln n}$$

I don't think this has a closed form.