Melin transform of the floor function [x]

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The discussion centers on defining the floor function [z] for complex numbers z using analytic continuation and Mellin transform properties. The integral $$\int_{0}^{\infty}[x]x^{s-1}dx$$ is highlighted, with a connection made to the Riemann zeta function, specifically $$ - \frac{\zeta (-s)}{s}$$. A suggestion is made to evaluate the integral by expressing it as a series, specifically $$\sum_{n=0}^\infty \int_n^{n+1} [x] x^{s-1} dx$$, to clarify its relationship to the zeta function. The importance of including the differential in the integral notation is also emphasized for clarity. This discussion provides insights into the mathematical treatment of the floor function in complex analysis.
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Homework Statement
using analytic continuation can i compute the mellin transform of the floor function as a riemann zeta function
Relevant Equations
$$ \int_{0}^{\infty}[x]x^{s-1}= - \frac{\zeta (-s)}{s} $$
usig analytic continuation and mellin transform properties
 
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How do you define [z] for z \in \mathbb{C}?
 
Rfael said:
Relevant Equations: $$ \int_{0}^{\infty}[x]x^{s-1}= - \frac{\zeta (-s)}{s} $$
Just a small quibble: although the meaning of your integral is clear from the context of your equation, please don't forget to always explicitly include the differential of the variable you're integrating over:$$\int_{0}^{\infty}[x]x^{s-1}dx$$
 
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Your question could have been phrased more clearly.

Try evaluating your integral by first expressing it as ##\sum_{n=0}^\infty \int_n^{n+1} [x] x^{s-1} dx## and then writing out the resulting series. That should make the connection to ##\zeta (-s)## clear—if that's what you're aiming for.
 
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