While thinking about the relation:
$$\zeta(s)\Gamma(s)=\int_0^\infty \frac{x^{s-1}}{e^x-1}~dx $$
I started with a Mellin transform on bounded support:
$$ f(s)= \int_0^1 x^{s-1}\bigg(-\frac{1}{e^{\frac{1}{\ln(x)}}-1}-1\bigg)~dx $$
Then the integrand can be rewritten using a series:
$$ =\sum_{n=1}^\infty \int_0^1 e^{\frac{n}{\ln(x)}}x^{s-1}~dx $$
Then each term can be written as a Mellin transform, each of which equals a modified Bessel function of the second kind, $K_1$. Summing these up we get:
$$ f(s)=\sum_{n=1}^\infty 2\sqrt{\frac{n}{s}}K_1\bigg(2\sqrt{sn}\bigg)=\int_0^1 x^{s-1}\bigg(-\frac{1}{e^{\frac{1}{\ln(x)}}-1}-1\bigg)~dx$$
Then using the Mellin inversion theorem:
$$f(x)=\sum_{n=1}^\infty e^{\frac{n}{\ln(x)}} = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{f(s)}{x^s}~ds $$
And we see that $f(x)$ and $f(s)$ form a Mellin pair.
Are all the steps correct?
For what values of $s$ does this complex integral converge? Does this integral help to analytically continue $f(x)?$
Where does $f(s)$ converge? For real $s>0$?