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    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

    Many thanks in advance.
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    (Original post by xfootiecrazeesarax)
    Q1)  0\leq x \leq 1 \implies x^{t-1} e^{-x} \leq x^{t-1}
    This is true, as you can (carefully) cancel off the x^{t-1} from both sides and you are left with something nice and easy.

    Q2) that x^{t-1}e^{-x}\leq x^{-x/2} for  x \geq 1
    This is not true in general. Choose x = 2 and t = 4, for example.
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    (Original post by Gregorius)
    This is true, as you can (carefully) cancel off the x^{t-1} from both sides and you are left with something nice and easy.



    This is not true in general. Choose x = 2 and t = 4, for example.
    ahh okay, thanks. I was trying to follow the proof here:

    http://math.stackexchange.com/questi...onverges-for-z

    the first answer...so this is incorrect? or ok for large  x
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    (Original post by xfootiecrazeesarax)
    ahh okay, thanks. I was trying to follow the proof here:

    http://math.stackexchange.com/questi...onverges-for-z

    the first answer...so this is incorrect? or ok for large  x
    It's incorrect for general x, but true for large x (i.e. as x->infinity). (If you read the entire thread, you'll see various comments to this effect).
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    (Original post by DFranklin)
    It's incorrect for general x, but true for large x (i.e. as x->infinity). (If you read the entire thread, you'll see various comments to this effect).
    Okay thanks.

    And yet the integral inequality still holds because the behaviour at x=\infty dominates over the small x behaviour?
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    anyone on how to prove q2?
 
 
 
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