Is Integration by Parts Applicable for X^3/((e^x)-1) from -infinity to infinity?

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Homework Help Overview

The discussion revolves around the integral of the function X^3/((e^x)-1) from -infinity to infinity, with participants questioning the applicability of integration by parts and the convergence of the integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are considering whether integration by parts is suitable for the given integral and discussing the convergence of the integral as x approaches -infinity. There are also considerations about changing the limits of integration and the implications of the function's properties.

Discussion Status

The discussion is active, with participants exploring different interpretations of the integral's limits and convergence. Some guidance has been offered regarding the treatment of the integral, but there is no explicit consensus on the approach to take.

Contextual Notes

There are questions about the convergence of the integral at -infinity and whether the function can be treated as even for simplification purposes. The original poster's assumptions about the limits and the applicability of integration techniques are under scrutiny.

silvermane
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Homework Statement


Integrating X^3/((e^x)-1), where we integrate from -infinity to infinity

The Attempt at a Solution


We thought about it, and we're not entirely sure if integration by parts is applicable here. Is there a table with this function perhaps?
 
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That integral doesn't converge as x->(-infinity), does it? The denominator is near 1 and the numerator is near x^3. Are you sure about the limits?
 


it doesn't matter if it converges at -infinity you can just make the upper and lower bounds 0 to infinity and put a coefficient of 2 in front of the integrand
 


SgtSniper90 said:
it doesn't matter if it converges at -infinity you can just make the upper and lower bounds 0 to infinity and put a coefficient of 2 in front of the integrand

That only works if it's an even function, which this isn't.
 

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