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The Mellin transform for:

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

equals:

$$\frac{1}{2\,\pi\,i}\int_{c-\infty\,i}^{c-\infty\,i} s^{-x}\,\Gamma(x)\,\zeta(x) \,dx =\frac{1}{e^s-1}$$

but is only valid for $c>1$, however if we take the Dirichlet $\eta$-function:

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

and transform it into:

$$\frac{1}{2\,\pi\,i}\int_{c-\infty\,i}^{c-\infty\,i} s^{-x}\,\Gamma(x)\,\eta(x) \,dx =\frac{1}{e^s+1}$$

then it is valid for all $c >0$ and now allows for integrating over all complex values with $c=\frac12$.

Assuming the RH, this would mean that the integrand becomes zero for all $x=\rho=\frac12 \pm \gamma\,i$ or in other words: any value of $x$ that could encode information about the distribution of the primes is "knocked out" of the integral to be able to correctly produce $\frac{1}{e^s+1}$.

Wild question: suppose we could prove that there is absolutely no information about the distribution of primes in $\frac{1}{e^s+1}$, would that imply that all non-trivial zeros need to be on the critical line for the Mellin transform to be correct?

Just to share an additional learning I got based on the answer below:

For:

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

the Mellin transform is ($\mu(n)$= the Mobius function):

$$\frac{1}{2\,\pi\,i}\int_{c-\infty\,i}^{c-\infty\,i} s^{-x}\,\dfrac{\Gamma(x)}{\zeta(x)}\,dx =\sum_{n=1}^{\infty}\mu(n)\,e^{-n\,s}$$

and this immediately shows that the integrand would diverge when a non-trivial zero would lie off the critical line. Hence full convergence in $\frac12 < c < 1$ implies the RH.

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Not at all, because the localization of the zeros of $\zeta(s)$ have virtually no impact on $\zeta(s)$ and its related inverse Mellin/Laplace transforms and integral representation.

On the other hand, the localization of the zeros of $\zeta(s)$ have a great impact on $\frac{1}{\zeta(s)}, \frac{\zeta'(s)}{\zeta(s)}, \ln \zeta(s)$ and their related inverse Mellin transforms and integral representation.

That's why finding the abscissa of convergence of $(1-2^{1-s}) \zeta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ is trivial, while finding the abscissa of convergence of $\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s}$ is the Riemann hypothesis.

The conclusion is that you should look at the Hurwitz zeta function $\zeta(s,a) = \sum_{n=0}^\infty (n+a)^{-s}$ (i.e. $\Gamma(s) \zeta(s,a)= \int_0^\infty \frac{x^{s-1} }{e^x-1}e^{(1-a)x}dx$ !!) having many zeros off the critical line (for $a \not \in \{ 1, 1/2\}$), because it has no Euler product (though it has a functional equation), so that the Dirichlet series for $\frac{1}{\zeta(s,a)}$ is much more complicated (and unrelated with the prime numbers).

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  • $\begingroup$ Great answer. Many thanks for these deep insights. Much appreciated. $\endgroup$ – Agno Sep 11 '16 at 17:36

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