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?

  • 3
    $\begingroup$ 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. $\endgroup$ Commented Jan 12, 2012 at 0:49
  • $\begingroup$ I assume that you've seen this paper by Laforgia (also referred to in DLMF). Is there a reason you don't mention it? $\endgroup$
    – t.b.
    Commented Jan 12, 2012 at 4:47
  • $\begingroup$ Laforgia can be downloaded from: ams.org/journals/mcom/1984-42-166/S0025-5718-1984-0736455-1/… $\endgroup$
    – Will Jagy
    Commented Jan 12, 2012 at 5:09
  • $\begingroup$ @J.M. The Laforgia article seems to say that Gautschi requires your $x$ to be a positive integer. I do not like the notation, Laforgia uses the same letter $k$ to be integral sometimes and sometimes real... $\endgroup$
    – Will Jagy
    Commented Jan 12, 2012 at 5:17
  • 2
    $\begingroup$ @J.M.: since the inequality is two-sided, one could call this, with a bit of mispronunciation, the Goat-Cheese Sandwich Theorem. $\endgroup$
    – robjohn
    Commented Mar 18, 2018 at 17:19

3 Answers 3


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 $$ x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+1)^{1-s}\tag{5} $$

  • $\begingroup$ Note that $(5)$ gives $\Gamma(x+s)=\Gamma(x)x^se^{O(1/x)}$. $\endgroup$
    – robjohn
    Commented Dec 29, 2021 at 15:02

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


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


and then by letting


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 $\psi(n)<\log n$.


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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