Show two inequalities - (context gamma function converges)

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Homework Help Overview

The discussion revolves around inequalities related to the gamma function, specifically focusing on the behavior of the function \(x^{t-1} e^{-x}\) under certain conditions. The original poster expresses uncertainty about proving two specific inequalities and their implications for convergence.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to understand the inequalities \(0 \leq x \leq 1 \implies x^{t-1} e^{-x} \leq x^{t-1}\) and \(x^{t-1} e^{-x} \leq x^{-x/2}\) for \(x \geq 1\). They question why the first inequality cannot be applied for all \(x\) and explore examples to illustrate their confusion.
  • Some participants question the relevance of the inequalities for proving convergence of integrals involving the gamma function, particularly when \(t \geq 0\).
  • There is a request for clarification on the second inequality and its implications, particularly regarding the transformation involving logarithms.

Discussion Status

The discussion is ongoing, with some participants providing insights into the convergence issues related to the inequalities. The original poster acknowledges the feedback received but continues to seek assistance specifically for the second inequality.

Contextual Notes

There is an emphasis on the context of the gamma function and the convergence of integrals, with specific attention to the conditions under which the inequalities are applied. The original poster's concerns about the applicability of the inequalities suggest a need for deeper exploration of the assumptions involved.

binbagsss
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Homework Statement



I'm not after another proof.
I've just got a couple of inequalities I don't know how to show when following a given proof in my book.

These are:

Q1) ## 0\leq x \leq 1 \implies x^{t-1} e^{-x} \leq x^{t-1} ##

So this is obvioulsy true, however I think I'm being dumb because surely this is true for all ##x##

(the integral over the gamma function is split into ##\int ^{\infty} _1 ## and ## \int^1_0 ##) and so this inequality is used in the latter, but I don't see why it can't be used in the former, e.g ##2^{t-1}/e^{2} \leq 2^{t-1} ## is also true isn't it? and as ## x \to \infty ## this is also true, because the LHS ## \to 0 ##..

Q2) that ##x^{t-1}e^{-x}\leq x^{-x/2}##, for ##x \geq 1##
I am unsure how to show this, or understand why it holds,

and then secondly I need to show that ##x^{t-1}e^{-x}\leq x^{-x/2} \iff x^{t-1}\leq e^{x/2}##

I can write ## x^{-x/2} = e^{(-x/2) ln (x) } ##
I am unsure what to do next or whether this helps

Homework Equations



see above

The Attempt at a Solution



see above
 
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binbagsss said:

Homework Statement



I'm not after another proof.
I've just got a couple of inequalities I don't know how to show when following a given proof in my book.

These are:

Q1) ## 0\leq x \leq 1 \implies x^{t-1} e^{-x} \leq x^{t-1} ##

So this is obvioulsy true, however I think I'm being dumb because surely this is true for all ##x##

(the integral over the gamma function is split into ##\int ^{\infty} _1 ## and ## \int^1_0 ##) and so this inequality is used in the latter, but I don't see why it can't be used in the former, e.g ##2^{t-1}/e^{2} \leq 2^{t-1} ## is also true isn't it? and as ## x \to \infty ## this is also true, because the LHS ## \to 0 ##..

\int_1^R x^{t-1}\,dx = \begin{cases} \ln R &amp; t = 0, \\<br /> \frac{R^t - 1}{t} &amp; t \neq 0\end{cases}. For t \geq 0 these do not converge as R \to \infty. You're trying to prove convergence so this bound is of no assistance.
 
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pasmith said:
\int_1^R x^{t-1}\,dx = \begin{cases} \ln R &amp; t = 0, \\<br /> \frac{R^t - 1}{t} &amp; t \neq 0\end{cases}. For t \geq 0 these do not converge as R \to \infty. You're trying to prove convergence so this bound is of no assistance.

thank you, makes sense !
Q2 anyone?
 
binbagsss said:
thank you, makes sense !
Q2 anyone?
bump
 

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