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A few answers here on math.SE have used as an intermediate step the following inequality that is due to Walter Gautschi:

$$x^{1-s} < \frac{\Gamma(x+1)}{\Gamma(x+s)} < (x+1)^{1-s},\qquad x > 0,\; 0 < s < 1$$

Unfortunately, the paper that the DLMF is pointing to is not easily accessible. How might this inequality be proven?

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Note: I'm actually planning to answer this question a bit later; I have managed to acquire a copy of Gautschi's paper, and I will type up a summary as an answer. But I want to see how others might go about proving it without seeing Gautschi's route. I'll probably leave this standing for two days before posting a summary of Gautschi's paper. –  J. M. Jan 12 '12 at 0:49
    
Thanks for asking this, J.M. I've wondered about this myself. –  Mike Spivey Jan 12 '12 at 3:30
    
Assuming $x$ is positive real, the strict inequalities are false for $s=1.$ –  Will Jagy Jan 12 '12 at 4:09
    
Whoops, I suppose I should add those restrictions, @Will... –  J. M. Jan 12 '12 at 4:19
    
I assume that you've seen this paper by Laforgia (also referred to in DLMF). Is there a reason you don't mention it? –  t.b. Jan 12 '12 at 4:47
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1 Answer 1

The strict log-convexity of $\Gamma$ (see the end of this answer) implies that for $0< s <1$, $$ \Gamma(x+s)<\Gamma(x)^{1-s}\Gamma(x+1)^s=x^{s-1}\Gamma(x+1)\tag{1} $$ which yields $$ x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}\tag{2} $$ Again by the strict log-convexity of $\Gamma$, $$ \Gamma(x+1)<\Gamma(x+s)^s\Gamma(x+s+1)^{1-s}=(x+s)^{1-s}\Gamma(x+s)\tag{3} $$ which yields $$ \frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+s)^{1-s}<(x+1)^{1-s}\tag{4} $$ Combining $(2)$ and $(4)$ yields $$ x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+1)^{1-s}\tag{5} $$

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It is actually a great prove !!! –  89085731 Feb 22 '12 at 10:29
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