How Does Substitution Simplify the Gamma Function Integral?

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Discussion Overview

The discussion revolves around the simplification of the Gamma function integral through substitution, specifically examining the integral representation of \(\Gamma(\frac{1}{2})\) and the implications of canceling terms in the integrand.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant presents the Gamma function and attempts to simplify the integral by substituting \(t = x^2\) and expresses confusion about canceling \(x\) in the integrand.
  • Another participant confirms the correctness of the substitution and suggests that the cancellation of \(x\) is valid.
  • A different participant questions the validity of canceling \(x\) due to the lower limit of the integral being 0.
  • One participant explains that the functions are defined on intervals that do not include the point 0, which justifies the cancellation of \(x\) in the denominator.

Areas of Agreement / Disagreement

Participants express differing views on the validity of canceling \(x\) in the integrand, particularly concerning the implications of the lower limit being 0. The discussion remains unresolved regarding the justification for this cancellation.

Contextual Notes

There is an assumption about the behavior of the integrand at the lower limit, and the discussion does not resolve the implications of the cancellation at that point.

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\Gamma(x)=\int^{\infty}_0t^{x-1}e^{-t}dt

\Gamma(\frac{1}{2})=\int^{\infty}_0\frac{e^{-t}}{\sqrt{t}}dt=

take t=x^2

dt=2xdx

x=\sqrt{t}

=\int^{\infty}_0\frac{e^{-x^2}}{x}2xdx

Why here we can here reducing integrand by x?
 
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What is your question? You've done the substitution correctly, so you see how the x's cancel. What is the problem?
 
Lower limit is 0. Why I may cancel x's?
 
Because the functions are defined on intervals not containing the point 0, that's why you can have the x in the denominator and simplify it through.
 

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