It is well known how altering the integral for the Gamma function:

$$\displaystyle \Gamma(s) = \int_0^\infty t^{s-1} e^{-t}\,dt$$

through substituting $t=nx$,

$$\displaystyle \Gamma(s)\frac{1}{n^s} = \int_0^\infty x^{s-1} e^{-n\,x}\,dx$$

and summing both sides, will "give birth" to the $\zeta$ function for $\Re(s) \gt 1$.

$$\displaystyle \Gamma(s)\zeta(s)=\Gamma(s)\sum_{n=1}^{\infty}\frac{1}{n^s} = \int_0^\infty x^{s-1} \sum_{n=1}^{\infty}e^{-n\,x}\,dx$$

This can be extended further by introducing the Mobius function $\mu(n)$ on both sides:

$$\displaystyle \dfrac{\Gamma(s)}{\zeta(s)}=\Gamma(s)\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s} = \int_0^\infty y^{s-1} \sum_{n=1}^{\infty}\mu(n)\,e^{-n\,y}\,dy$$

Multiplying both functions together annihilates $\zeta(s)$ and gives the following double integral for $\Gamma(s)^2$:

$$\displaystyle \Gamma(s)^2 = \iint_0^\infty \dfrac{(x\,y)^{s-1}}{e^x-1} \sum_{n=1}^{\infty}\frac{\mu(n)}{e^{n\,y}}\,dx\, dy, \qquad \Re(s)>1$$

**Questions:**

Could this double integral be simplified any further?

Is there a way to equate it to $\displaystyle \left(\int_0^\infty t^{s-1} e^{-t}\,dt\right)^2$ and thereby 'solve' the $\displaystyle \sum_{n=1}^{\infty}\frac{\mu(n)}{e^{n\,y}}$ factor?