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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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    $\begingroup$ 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. $\endgroup$ – J. M. is a poor mathematician Jan 12 '12 at 0:49
  • $\begingroup$ Thanks for asking this, J.M. I've wondered about this myself. $\endgroup$ – Mike Spivey Jan 12 '12 at 3:30
  • $\begingroup$ Assuming $x$ is positive real, the strict inequalities are false for $s=1.$ $\endgroup$ – Will Jagy Jan 12 '12 at 4:09
  • $\begingroup$ Whoops, I suppose I should add those restrictions, @Will... $\endgroup$ – J. M. is a poor mathematician Jan 12 '12 at 4:19
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    $\begingroup$ @J.M.: since the inequality is two-sided, one could call this, with a bit of mispronunciation, the Goat-Cheese Sandwich Theorem. $\endgroup$ – robjohn Mar 18 '18 at 17:19
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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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  • $\begingroup$ It is actually a great prove !!! $\endgroup$ – 89085731 Feb 22 '12 at 10:29
  • $\begingroup$ I already gave a (+1) long time ago, and I miss the opportunity of upvoting this answer more than once, since I keep taking it as a reference for many questions about $\Gamma$ function inequalities. $\endgroup$ – Jack D'Aurizio Oct 21 '16 at 7:48
  • $\begingroup$ Very elegant. A big (+1) $\endgroup$ – Mark Viola Feb 2 '17 at 16:19
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I'll probably leave this standing for two days before posting a summary of Gautschi's paper.

Here is the long-overdue follow-through. I have slightly changed a few notations, but this is otherwise Gautschi's original argument.


What Gautschi actually proves in his paper is the more general inequality

$$\exp((s-1)\psi(n+1))\le\frac{\Gamma(n+s)}{\Gamma(n+1)}\le n^{s-1},\; 0\le s\le1,n\in\mathbb Z^{+}\tag{1}\label{1}$$

where $\psi(n)$ is the digamma function.

Gautschi considers the function

$$f(s)=\frac1{1-s}\log\left(\frac{\Gamma(n+s)}{\Gamma(n+1)}\right)$$

over $0\le s <1$, from which we have $f(0)=\log(1/n)$ and

$$\lim_{s\to 1}f(s)=-\psi(n+1)$$

via l'Hôpital. Then we have

$$(1-s)f'(s)=f(s)+\psi(n+s)$$

and then by letting

$$\varphi(s)=(1-s)(f(s)+\psi(n+s))$$

we have $\varphi(0)=\psi(n)-\log n<0$, $\varphi(1)=0$, and $\varphi'(s)=(1-s)\,\psi ^{(1)}(n+s)$ (where $\psi ^{(1)}(n)$ is the trigamma function).

Now, since $\psi ^{(1)}(n+s)=\psi ^{(1)}(s)-\sum\limits_{k=0}^{n-1}\frac1{(s+k)^2}$ is always positive, we have that $\varphi(s)<0$, from which we deduce that $f(s)$ is monotonically decreasing over $0<s<1$ (i.e., $f'(s)<0$). Therefore

$$-\psi(n+1)\le f(s)\le\log\frac1{n}$$

which is equivalent to $\eqref{1}$. The inequality in the OP can then be deduced from the inequality $\psi(n)<\log n$.

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  • $\begingroup$ A bit under five actually, but it certainly took long! :D $\endgroup$ – J. M. is a poor mathematician Jan 12 '17 at 15:41
  • $\begingroup$ Doh! still getting used to 2017. I was subtracting from 2016. $\endgroup$ – robjohn Jan 12 '17 at 16:17
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Given $a,b\geq 0$, let us consider the function $f(x)=x^{a}(1-x)^{b}$ on the interval $[0,1]$.
Its maximum value is given by $\frac{a^a b^b}{(a+b)^{a+b}}$, since $f'$ only vanishes at $x=\frac{a}{a+b}$.
For any $p>0$ we have $$ \| f\|_p^p = \int_{0}^{1}x^{pa}(1-x)^{pb}\,dx=\frac{\Gamma(ap+1)\,\Gamma(bp+1)}{\Gamma((a+b)p+2)} $$ and the LHS is log-convex with respect to $p$. By considering that $$ \lim_{p\to +\infty}\|f \|_p = \frac{a^a b^b}{(a+b)^{a+b}} $$ Gautschi's inequality turns out to be a simple consequence of interpolation and a suitable choice of the parameters $a,b,p$.

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