TITLE: Mellin Transform of the floor function
QUESTION [8 upvotes]: I have from integral transform tables that:
$$\operatorname{Mellin}(\lfloor x\rfloor) = -\dfrac{\zeta(-z)}{z},\quad \operatorname{Re}(z)<-1$$
How can this be proved?

REPLY [2 votes]: $$ \int_{0}^{\infty} \lfloor x\rfloor x^{z-1} \ dx = \sum_{k=0}^{\infty} \int_{k}^{k+1} k x^{z-1} \ dx = \sum_{k=1}^{\infty} k  \int_{k}^{k+1} x^{z-1} \ dx $$
$$ = \sum_{k=1}^{\infty} \frac{k}{z} \Big((k+1)^{z} - k^{z} \Big) = \frac{1}{z}\Big( (2^{z} - 1 ) + 2 (3^{z}-2^{z} ) + 3 (4^{z}-3^{z} ) + \ldots\Big)$$
$$ = \frac{1}{z} \Big( -1 -2^{z} - 3^{z} - \ldots) = - \frac{1}{z} \Big(1+2^{z}+3^{z} + \ldots \Big) = - \frac{\zeta(-z)}{z}$$