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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?

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    $\begingroup$ In general, one cannot solve for a two-dimensional integrand if one only has a one-parameter family of identities (in this case, your parameter is $s$), as the problem is severely underdetermined. Basically all one can do here is divide out the $x$ integration, and "solve" for $\sum_n \mu(n)/e^{ny}$ as the inverse Mellin transform of $\Gamma(s)/\zeta(s)$ as per your previous equation. $\endgroup$
    – Terry Tao
    Jan 17, 2015 at 0:34
  • $\begingroup$ Many thanks, Terry. I was already a bit afraid ending up "full circle" here. $\endgroup$
    – Agno
    Jan 17, 2015 at 10:31
  • $\begingroup$ at least where everything is absolutely convergent $$\Gamma(s) \Gamma(z) = \iint_0^\infty x^{s-1}y^{z-1} e^{-x-y} dx dy$$ so if you take any inversible Dirichlet series $A(s)B(s) = (\sum_{n=1}^\infty a_n n^{-s})(\sum_{m=1}^\infty b_m m^{-s}) = 1$ $$A(s)\Gamma(s) B(z)\Gamma(z) = \iint_0^\infty x^{s-1}y^{z-1} \sum_{n=1}^\infty \sum_{m=1}^\infty a_n b_m e^{-nx-my} dx dy$$ and $$\Gamma(s)^2 = \iint_0^\infty (xy)^{s-1} \sum_{n=1}^\infty \sum_{m=1}^\infty a_n b_m e^{-nx-my} dx dy$$ so as @Terry said this is under-determined. $\endgroup$
    – reuns
    Feb 21, 2016 at 0:00

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