Writing complicated integral in terms of the Gamma function

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Homework Help Overview

The original poster attempts to express the integral ##\int_{0}^{1}x^2(ln\frac{1}{x})^3 dx## in terms of the gamma function, exploring a substitution involving ##x=e^{-u}## to transform the integral.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the substitution method and its implications for the limits of integration. There are questions about the correctness of the substitution and the resulting integral's convergence.

Discussion Status

Some participants have provided feedback on the substitution steps, noting potential errors in the limits and the differential transformation. There is an ongoing exploration of the integral's behavior and its divergence.

Contextual Notes

Participants are examining the implications of the substitution on the limits of integration and the validity of the resulting integral, which has been noted to diverge.

davidbenari
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Homework Statement


Write ##\int_{0}^{1}x^2(ln\frac{1}{x})^3 dx## in terms of the gamma function

2. Relevant equation
##\Gamma(p+1)=p\Gamma(p)##

The Attempt at a Solution


Say ##x=e^{-u}## one would eventually obtain the integral

##\int_{-\infty}^{0} u^3 e^{-u} du##

STEPS:
##x=e^{-u}## ##e^{u}=1/x## ##u=ln(1/x)## ##du=xdx##

##\int_{0}^{1}x^2(ln\frac{1}{x})^3 dx=\int_{0}^{-\infty} e^{-u} x u^3 \frac{du}{x}=\int_{-\infty}^{0} u^3 e^{-u} du##

Which wants to look like a gamma function but isn't because of the limits.
 
Last edited:
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I don't know what went wrong with the substitution (can you show the steps?), but the last integral diverges.
 
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sure I'll edit to add.
 
I've added it now mfb.
 
davidbenari said:
##du=xdx##
I don't think that is right.
And your integral limits look wrong, too.
x=1 corresponds to u=0, but x=0 corresponds to a different value for u.
 
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Yeah I've corrected it now. Thanks!
 

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