According to http://functions.wolfram.com/, for every $x\geq 2$ it is $$ \left( \frac{x}{e}\right)^{x-1} \leq \Gamma(x) \leq \left( \frac{x}{2}\right)^{x-1}, $$ where $\Gamma$ is the Gamma function. How can this be proven?

Update 1: According to Qi and Chen (see eq. (10)), the following holds:

$$ \ln \Gamma(x) = (x-1)(\ln x - 1) + \phi(x), $$ where $$ \phi(x) = \int_0^\infty \left(\frac{1}{t}-\frac{1}{e^{t}-1}\right) e^{-t} \frac{1-e^{-(x-1)t}}{t}\mathrm{d}t, $$ thus, (for the left-hand side of the given inequality) it suffices to prove that $\phi(x)\geq 0$ for all $x\geq 2$ because it is equivalently written as $(x-1)(\ln x - 1)\leq \ln \Gamma(x)$.

Update 2: We take the derivative of $\phi$ which is: $$ \phi'(x)=\int_{0}^{\infty}\left(\frac{1}{t}-\frac{1}{e^{t}-1}\right) e^{-t(1+x)}\mathrm{d}t $$ and notice that the integrated function is positive for all $t>0$ and $x\geq 2$, therefore $\phi'(x)>0$ and $\phi$ is increasing with $\phi(2)=0.3069$ (with MATLAB, but I guess there must be some way to show analytically that $\phi(2)>0$). Then, we have proven that $\phi(x)\geq 0$ for all $x\geq 2$. Any better ideas?

Still we need to prove the right-hand side of the given inequality.

  • $\begingroup$ Quite possibly from en.wikipedia.org/wiki/… $\endgroup$ – Travis Willse Nov 26 '14 at 0:19
  • $\begingroup$ @Travis The left-hand side of the Stirling inequality is written as $$\sqrt{2\pi x} \left(\frac{x}{e}\right)^x \leq \Gamma(x+1)$$ for $x\in\mathbb{N}$ which proves something similar to the assertion over the integers with $x\geq 2$. $\endgroup$ – Pantelis Sopasakis Nov 26 '14 at 16:28
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    $\begingroup$ You can directly see that the integrand of $\phi(x)$ is non-negative (for $x\geqslant 1$). $\alpha = x-1 \geqslant 0$, hence $1-e^{-\alpha t} \geqslant 0$ for $t \geqslant 0$. Also, $e^t \geqslant 1+t$ for $t\in\mathbb{R}$, hence for $t > 0$ we have $\frac{1}{t} - \frac{1}{e^t-1} = \frac{e^t-1-t}{t(e^t-1)} \geqslant 0$ (actually, $> 0$). $\endgroup$ – Daniel Fischer Nov 26 '14 at 19:18
  • $\begingroup$ @DanielFischer You're right, thanks a lot! $\endgroup$ – Pantelis Sopasakis Nov 26 '14 at 21:44

Here's a sketch of an alternative proof for the lower bound. First, note that $$ x\cdot\frac{x^{x-1}}{e^{x-1}} = \frac{(x+1)^x}{e^x} \cdot \frac{e}{\big(1+\frac1x\big)^x} \ge \frac{(x+1)^x}{e^x} $$ Since $\Gamma(x+1)=x\Gamma(x)$, this means that the lower bound for $x$ implies the lower bound for $x+1$; so it suffices to show the inequality for $x\in[2,3]$. Now, $(\ln\Gamma)'(2)=1-\gamma$ (see Polygamma function), and $\ln\Gamma(x)$ is convex, so $$ (1-\gamma)(x-2) \le \ln\Gamma(x) $$ Thus it suffices to prove that $$ \ln \Bigl(\Bigl(\frac xe\Bigr)^{x-1}\Bigr)\le (1-\gamma)(x-2) \qquad\text{for $x\in[2,3]$.} $$ And since the LHS here is also convex, it suffices to check this at $x=2$ and $x=3$.


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