Melin transform of the floor function [x]

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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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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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