Show two inequalities - (context gamma function converges)

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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