Let for integers $n\geq 1$ the Möbius function $\mu(n)$, you can see the definition of this arithmetic function from this MathWorld, and let $\Gamma(s)$ the Gamma function. I would like to know how to justify, if it is feasible, that the integral sign and the infinite series can be interchanged here
$$\int_0^1\left(\sum_{n=1}^{\infty}\frac{\mu(n)}{n^2}\log\Gamma(x+n)\right)dx.\tag{1}$$
Notice that if it is possible to provide a justification that the integral sign and summation can be interchanged, then symbolically one can to calculate a closed-form of the resulting $RHS$, using a theorem due to Raabe ([1], or well the Wikipedia's article dedicated to Joseph Ludwig Raabe) and standard series involving the Möbius function (this includes the prime number theorem).
Using experiments I try to see if I can swap the order or the summation and the integration. For example I believe (my experiments were sparsed and the quantities are oscillating around the mentioned closed-form) that $$\int_0^1\left(\sum_{n=1}^{N=70}\frac{\mu(n)}{n^2}\log\Gamma(x+n)\right)dx\approx -0.4379544$$ but with upper limit $N=80$ in the sum of previous integrand I know that my approximation gets worse again.
Question. Is it possible to interchange the integral sign and summation here $$\int_0^1\left(\sum_{n=1}^{\infty}\frac{\mu(n)}{n^2}\log\Gamma(x+n)\right)dx\,?$$ Many thanks.
I think that it isn't in the literature, but feel free if it was to refer answering this as a reference request. I know how to calculate the mentioned closed-form, thus I am asking exclusively about how to justify/discard that the integral sign and the series can be interchanged.
References:
[1] T. Amdeberhan, M. Coffey, O. Espinosa, C. Koutschan, D. Manna, and V. Moll, Integrals of powers of loggamma, Proc. Amer. Math. Soc., 139(2):535–545, (2010).