On the convergence of $-\int_0^1\left(\int_1^\infty \left(\frac{1}{\zeta(x/y)}-1\right)dx\right)dy$ For fixed real numbers $1<x<\infty$ and $0<y<1$ one has that 
$$\frac{1}{\zeta\left(\frac{x}{y}\right)}-1=\sum_{k=2}^\infty \frac{\mu(k)}{k^{x/y}},$$ where $\mu(n)$ is the Möbius function and thus $\zeta(t)$ is the Riemann Zeta function for a real argument $t>1$.
I am asked myself if it is possible to get a good approximation, and thus show also the convergence of
$$I:=-\int_0^1\left(\int_1^\infty \left(\frac{1}{\zeta(x/y)}-1\right)dx\right)dy.$$
Notice that since I've created this exercise I believe that I can write by integration that 
$$I=-\sum_{k=2}^\infty\int_0^1\frac{\mu(k)yk^{-1/y}}{\log k}dy=\sum_{k=2}^\infty\frac{\mu(k)\left(k\log^2 k\left(\operatorname{Chi}(\log k)-\operatorname{Shi}(\log k)\right)-1+\log k\right)}{2k\log k}.$$
See if you need the codes from Wolfram Alpha online calculator to justify previous identity
integrate 1/k^(z/y) dz, from z=1 to z=infinite
integrate  (y k^(-1/y))/(log(k)) dy, from y=0 to 1
Also as remark I believe that $$\sum_{k=2}^\infty\mu(k)\log (k)\left(\operatorname{Chi}(\log k)-\operatorname{Shi}(\log k)\right)$$ is convergent 
sum mu(k)log(k)(Chi(log(k))-Shi(log(k))), from k=2 to 100 

Question. Can you prove that $$-\int_0^1\left(\int_1^\infty \left(\frac{1}{\zeta(x/y)}-1\right)dx\right)dy$$ does converge? Can you provide us a good approximation of $I$? Thanks in advance.

 A: I'll address convergence. Letting $x=yt$ gives
$$\int_0^1\int_1^\infty \left(1-\frac{1}{\zeta(x/y)}\right)dx\,dy = \int_0^1y\int_{1/y}^\infty \left(1-\frac{1}{\zeta(t)}\right)dt\,dy.$$
Since $1/y\ge 1,$ the inner integral on the right is bounded above by
$$\tag 1 \int_{1}^\infty \left(1-\frac{1}{\zeta(t)}\right)dt.$$
It's enough to show $(1)$ is finite. In fact, since the integrand in $(1)$ bounded, it's enough to prove the integral over $[2,\infty)$ is finite.
Note that
$$1-\frac{1}{\zeta (t)} = \frac{\zeta (t)-1}{\zeta (t)} \le \frac{\zeta (t)-1}{1} = \sum_{n=2}^{\infty}\frac{1}{n^t}.$$
For $t\ge 2,$ the last expression is no more than
$$\frac{1}{2^t}\sum_{n=2}^{\infty}\frac{2^t}{n^t}=\frac{1}{2^t}\sum_{n=2}^{\infty}\left (\frac{2}{n}\right)^t \le \frac{1}{2^t}\sum_{n=2}^{\infty}\left (\frac{2}{n}\right)^2 = \frac{1}{2^t}\cdot 4 \sum_{n=2}^{\infty}\left (\frac{1}{n}\right)^2 < \frac{1}{2^t}\frac{4\pi^2}{6}.$$
Since $ 2^{-t} $ is integrable over $[2,\infty),$ we have the desired convergence.
