Melin transform of the floor function [x]

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SUMMARY

The discussion focuses on the application of the Mellin transform to the floor function [x] for complex numbers z in the context of analytic continuation. The integral $$ \int_{0}^{\infty}[x]x^{s-1}dx $$ is highlighted, with a direct relationship established to the Riemann zeta function, specifically $$ - \frac{\zeta (-s)}{s} $$. Participants emphasize the importance of including the differential in integrals and suggest evaluating the integral through a series expansion to clarify its connection to the zeta function.

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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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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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