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Prove that $$Q(x,\ln 2) := \frac{\int_{\ln 2}^{\infty} t^{x-1} e^{-t} dt}{\int_{0}^{\infty} t^{x-1} e^{-t} dt} \geqslant 1 - 2^{-x}$$ for all $x\geqslant 1$.

($Q$ is the regularized gamma function.)

Graph of $Q(x, \ln 2)$ and $(1-2^{-x})$ against $x$

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What is the motivation? You know the result is true? Or is it just your guess? –  Aryabhata Mar 30 '12 at 6:19
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@Aryabhata This is a true result because I obtained as a solution to a probability problem the expression $(1-2^x) + 2^x Q(x,\ln 2)$. I would like to see that this expression is within $[0,1]$ for all $x\geqslant 1$, using algebraic methods only. The upper bound $1$ is easy to show though. –  hwhm Mar 30 '12 at 7:31
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It might be possible to show that $$\left(\frac{a}{a-1}\right)^x \int_{0}^{\log a} t^{x-1} e^{-t} \,dt \geq 2^x \int_{0}^{\log 2} t^{x-1} e^{-t} \,dt$$ for $a \geq 2$, then let $a \to \infty$. –  Antonio Vargas Mar 30 '12 at 8:07
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2 Answers 2

up vote 6 down vote accepted

We have

$$ \frac{\int_{\ln 2}^{\infty} t^{x-1} e^{-t} \,dt}{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt} = \frac{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt - \int_{0}^{\log 2} t^{x-1} e^{-t} \,dt}{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt} = 1 - \frac{\int_{0}^{\log 2} t^{x-1} e^{-t} dt}{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt}, $$

so we need to show that

$$ \frac{\int_{0}^{\log 2} t^{x-1} e^{-t} \,dt}{\int_{0}^{\infty} t^{x-1} e^{-t} \,dt} \leq 2^{-x}, $$

or, equivalently,

$$ 2^x \int_{0}^{\log 2} t^{x-1} e^{-t} \,dt \leq \int_{0}^{\infty} t^{x-1} e^{-t} \,dt. $$

To do this we will show that

$$ 2^x \int_{0}^{\log 2} t^{x-1} e^{-t} \,dt \leq \left(\frac{e^a}{e^a-1}\right)^x \int_{0}^{a} t^{x-1} e^{-t} \,dt \tag{1} $$

for all $a \geq \log 2$, then let $a \to \infty$. In fact, we will show that the quantity on the right-hand side of the above inequality is nondecreasing in $a$ when $a > 0$ for fixed $x \geq 1$ (and strictly increasing in $a$ when $a > 0$ for fixed $x > 1$).

To start, define

$$ f_x(a) = \left(\frac{e^a}{e^a-1}\right)^x \int_{0}^{a} t^{x-1} e^{-t} \,dt. $$

Then

$$ \begin{align} f_x'(a) &= a^{x-1} e^{-a} \left(\frac{e^a}{e^a-1}\right)^x - x \left(\frac{e^a}{e^a-1}\right)^{x-1} \frac{e^a}{(e^a-1)^2} \int_{0}^{a} t^{x-1} e^{-t} \,dt \\ &= e^{ax} \left(e^a-1\right)^{-x-1} \left[a^{x-1} \left(1-e^{-a}\right) - x \int_{0}^{a} t^{x-1} e^{-t} \,dt\right]. \end{align} $$

Since we're only concerned with the sign of the above expression, define

$$ \begin{align} g_x(a) &= e^{-ax}(e^a - 1)^{x+1} f_x'(a) \\ &= a^{x-1} \left(1-e^{-a}\right) - x \int_{0}^{a} t^{x-1} e^{-t} \,dt. \end{align} $$

If $g_x(a) \geq 0$ for all $a > 0$ then $f_x'(a) \geq 0$ for all $a > 0$, and hence $f_x(a) \geq f_x(\log 2)$ for all $a \geq \log 2$, which is $(1)$.

Well, it will certainly be true that $g_x(a) \geq 0$ for all $a > 0$ if

$$ g_x(0) \geq 0 \hspace{1cm} \text{and} \hspace{1cm} g_x'(a) \geq 0 \,\,\text{ for all }\,\, a \geq 0. \tag{2} $$

Indeed, $g_x(0) = 0$, and for $x \geq 1$ we have

$$ g_x'(a) = a^{x-2} e^{-a} (x-1) (e^a - a - 1) \geq 0 $$

since the function $h(a) = e^a - a - 1$ is nondecreasing when $a \geq 0$ and $h(0) = 0$.

By the remarks immediately before $(2)$ this is sufficient to prove $(1)$, from which the result follows.

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I don't think your expression for $g_x^{\prime} (a)$ is correct. In fact, $g_x (a)$ can be negative for $x\approx 1$ (take $x=1$ for example)... –  hwhm Mar 31 '12 at 8:23
    
Thanks @hwhm, there was a typo in $f_x'(a)$ it turns out. Everything should be correct now. –  Antonio Vargas Mar 31 '12 at 16:25
    
Thank you so much, I should have spotted that typo! –  hwhm Apr 1 '12 at 2:26
    
@hwhm You're very welcome. Thank you for the interesting problem! –  Antonio Vargas Apr 1 '12 at 2:57
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Here is a proof for $x\ge2$: $$ \begin{align} \int_0^{\log(2)}t^{x-1}e^{-t}\mathrm{d}t &\le\int_0^{\log(2)}t^{x-1}\mathrm{d}t\\ &=\frac1x\log(2)^x\tag{1} \end{align} $$ Thus, we get that $$ \frac{\int_0^{\log(2)}t^{x-1}e^{-t}\mathrm{d}t}{\int_0^\infty t^{x-1}e^{-t}\mathrm{d}t} \le\frac{\log(2)^x}{\Gamma(x+1)}\tag{2} $$ For $x\ge2$, $$ \frac{(2\log(2))^x}{\Gamma(x+1)}\le1\tag{3} $$ Once we show $(3)$, the result follows because $$ \begin{align} \frac{\int_{\log(2)}^\infty t^{x-1}e^{-t}\mathrm{d}t}{\int_0^\infty t^{x-1}e^{-t}\mathrm{d}t} &=1-\frac{\int_0^{\log(2)}t^{x-1}e^{-t}\mathrm{d}t}{\int_0^\infty t^{x-1}e^{-t}\mathrm{d}t}\\ &\ge1-\frac{\log(2)^x}{\Gamma(x+1)}\\ &\ge1-2^{-x}\tag{4} \end{align} $$


Inequality $(3)$ is equivalent to $$ \log(\Gamma(x+1))\ge x(\log(2\log(2)))\tag{5} $$ Note that $(5)$ holds at $x=2$ since $\log(2)>2\log(2\log(2))$ follows from $\log(2)<\sqrt{1/2}$.

Since $\Gamma$ is log-convex and for $x\ge2$, $\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x+1))\ge\frac32-\gamma>\log(2\log(2))$.

Thus, $(5)$, and therefore $(3)$, hold for $x\ge2$.

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