Dirichlet series as Mellin transforms

Remarks

For identity 1, $\int_0^\infty x^{s-1} e^{-nx} \, dx = \Gamma(s) n^{-s}$ when $\operatorname{Re} s > 0$, and the sum of the absolute values $\Gamma(\operatorname{Re} s) \sum_n \abs{a(n)} n^{-\operatorname{Re} s}$ is finite when $\operatorname{Re} s > \sigma_a$, so sum and integral may be exchanged. For identity 2, $A(x) = 0$ for $x < 1$, and $\int_n^\infty x^{-s-1} \, dx = n^{-s}/s$; exchanging sum and integral is justified the same way. Example: $a(n) = 1$ gives $\sigma_a = 1$, $f_a(x) = 1/(e^{x} - 1)$ and $A(x) = \floor{x}$, so $$\Gamma(s) \zeta(s) = \int_0^\infty \frac{x^{s-1}}{e^{x} - 1} \, dx, \qquad \frac{\zeta(s)}{s} = \int_1^\infty \floor{x} \, x^{-s-1} \, dx, \qquad \operatorname{Re} s > 1.$$ Non-example: at $s = 1$ with $a(n) = 1$, the condition $\operatorname{Re} s > \sigma_a$ fails, and indeed $\int_0^\infty dx/(e^x - 1)$ diverges at $0$.