Show that $\Gamma\left(3^{-3}+1\right)<\Gamma \left(\operatorname{W}(3)^{-\operatorname{W}(3)}+1\right)$

Question

Hi it's one of my old problem that I cannot solve let me propose it :

Prove that $$\Gamma\left(3^{-3}+1\right)<\Gamma \left(\operatorname{W}(3)^{-\operatorname{W}(3)}+1\right)\quad \tag{1}$$ where $W(\cdot)$ is the Lambert W function and $\Gamma(\cdot)$ is the gamma function.

The central function is :

$$f(x)=\Gamma\left(x^{-x}+1\right)$$

The derivative is :

$$f'(x)= -x^{-x}(\ln(x)+1)\Gamma\left(x^{-x}+1\right)\psi(x^{-x}+1)$$

Unfortunately it doesn't helps since $f(x)$ is increasing around $x=3$ while $f\left(\operatorname{W}(x)\right)$ is decreasing around $x=3$ .

Maybe I think we can do the same as the Claude Leibovici's answer here Show that $\Gamma(\Omega)\leq \Gamma\Big(\operatorname{W}\Big(x^{x}\Big)\Big)<2$ on $(0,1]$ .

My question :

How to show $(1)$ ?

Thanks in advance !

Answer 1

Mking the problem more general, considering $$\Delta(x)=\Gamma \left(\operatorname{W}(x)^{-\operatorname{W}(x)}+1\right)-\Gamma\left(x^{-x}+1\right)$$ around $x=e$, use Taylor expansions to order $O\left((x-e)^{n+1}\right)$

For example $$x^{-x}+1=\left(1+e^{-e}\right)-2 e^{-e} (x-e)+\frac{1}{2} e^{-1-e} (4 e-1) (x-e)^2+O\left((x-e)^3\right)$$ gives $$\Gamma\left(x^{-x}+1\right)=\Gamma \left(1+e^{-e}\right)-2 (x-e) \left(e^{-e} \Gamma \left(1+e^{-e}\right) \psi ^{(0)}\left(1+e^{-e}\right)\right)+\frac{1}{2} e^{-1-2 e} (x-e)^2 \Gamma \left(1+e^{-e}\right) \left(-e^e \psi ^{(0)}\left(1+e^{-e}\right)+4 e^{1+e} \psi ^{(0)}\left(1+e^{-e}\right)+4 e \psi ^{(0)}\left(1+e^{-e}\right)^2+4 e \psi ^{(1)}\left(1+e^{-e}\right)\right)+O\left((x-e)^3\right)$$

Similarly, $$W(x)=1+\frac{x-e}{2 e}-\frac{3 (x-e)^2}{16 e^2}+O\left((x-e)^3\right)$$ $$\log(W(x))=\frac{x-e}{2 e}-\frac{5 (x-e)^2}{16 e^2}+O\left((x-e)^3\right)$$ $$-W(x)\log(W(x))=-\frac{x-e}{2 e}+\frac{(x-e)^2}{16 e^2}+O\left((x-e)^3\right)$$ $$\operatorname{W}(x)^{-\operatorname{W}(x)}=e^{-W(x)\log(W(x))}=1-\frac{x-e}{2 e}+\frac{3 (x-e)^2}{16 e^2}+O\left((x-e)^3\right)$$

The formulae are already messy but workable. Computing the numerical value of $\Delta_n(3)$ as a function of the order $n$ of Taylor expansions, we have

$$\left( \begin{array}{cc} n & 10^6 \times \Delta_n(3) \\ 1 & -4892.218349 \\ 2 & +265.5544022 \\ 3 & +68.15191015 \\ 4 & -13.15976659 \\ 5 & +6.138334905 \\ 6 & +4.440578904 \\ 7 & +4.374006576 \\ 8 & +4.408250996 \\ 9 & +4.405368867 \\ 10 & +4.405111061 \\ 11 & +4.405194838 \\ 12 & +4.405189100 \\ \cdots & \cdots \\ \infty & +4.405188465 \end{array} \right)$$

In fact $\Delta(x)=0$ for $x=3.000042164$.

But, there is another to this equation close to $x=\frac 1e$. A first order Taylor expansion gives it equal to $0.44121$ while the exact solution is $0.42655$.

Answer 2

Remarks: Numerically, $\Gamma(W(3)^{-W(3)} + 1) - \Gamma(3^{-3} + 1) \approx 0.0000044051$. Thus, we need good estimates.

---

First, we estimate $W(3)$.

$W(3)$ is the solution of $y\mathrm{e}^y = 3$. Clearly $1< W(3) < 2$.

Fact 1: $y\mathrm{e}^y \ge \frac{2\mathrm{e}y^2 + 4\mathrm{e}y}{y^2 - 6y + 11}$ for all $y \ge 1$.

Let $g(y) = \frac{2\mathrm{e}y^2 + 4\mathrm{e}y}{y^2 - 6y + 11}$. Then, we have $g(W(3)) \le 3$ which results in $$W(3) \le \frac{-2\mathrm{e} - 9 + \sqrt{4\mathrm{e}^2 + 102\mathrm{e} - 18}}{2\mathrm{e} - 3} < \frac76 . $$

---

Second, we estimate $W(3)^{-W(3)}$.

Fact 2: $x^{-x} \ge 1 - (x - 1) + \frac37(x - 1)^3$ for all $x \in [1, 7/6]$.
(The RHS is based on the Taylor approximation of the LHS.)

Using Fact 2, we have $$W(3)^{-W(3)} > \frac{3621}{3811}.$$

---

Third, since $x\mapsto \Gamma(x + 1)$ is strictly increasing on $x>1/2$, it suffices to prove that $\Gamma(28/27) < \Gamma(7432/3811)$ or $$\frac{1}{27}\Gamma(1/27) < \frac{3621}{3811}\Gamma(3621/3811).$$

Using $\Gamma(x)\Gamma(1 - x) = \frac{\pi}{\sin (\pi x)}$, we have $\Gamma(3621/3811)\Gamma(190/3811) = \frac{\pi}{\sin \frac{190\pi}{3811}}$.

It suffices to prove that $$\frac{1}{27}\Gamma(1/27)\Gamma(190/3811) < \frac{3621}{3811}\, \frac{\pi}{\sin \frac{190\pi}{3811}}$$ or $$- \ln \Gamma(1/27) - \ln \Gamma(190/3811) + \ln \left(\frac{3621}{3811}\, \frac{27 \pi}{\sin \frac{190\pi}{3811}}\right) > 0. \tag{1}$$

Using $\ln(1 + u) - u \ge - \frac12 u^2 + \frac13 u^3 - \frac14 u^4$ for all $u \ge 0$ and $$-\ln \Gamma(x) = \ln x + \gamma x + \sum_{n=1}^\infty [\ln(1 + x/n) - x/n],$$ we have \begin{align*} -\ln \Gamma(1/27) &\ge \ln \frac{1}{27} + \frac{1}{27}\gamma + \sum_{n=1}^\infty \left(- \frac12\,\frac{1}{27^2n^2} + \frac13\,\frac{1}{27^3n^3} - \frac14\,\frac{1}{27^4 n^4}\right)\\ &= \ln \frac{1}{27} + \frac{1}{27}\gamma - \frac{1}{8748}\pi^2 + \frac{1}{59049}\zeta(3) - \frac{1}{191318760}\pi^4, \tag{2} \end{align*} and \begin{align*} &-\ln \Gamma(\tfrac{190}{3811})\\ \ge\, & \ln \frac{190}{3811} + \frac{190}{3811}\gamma + \sum_{n=1}^\infty \left(- \frac12\,\frac{190^2}{3811^2n^2} + \frac13\,\frac{190^3}{3811^3 n^3} - \frac14\,\frac{190^4}{3811^4 n^4}\right)\\ =\,& \ln \frac{190}{3811} + \frac{190}{3811}\gamma - \frac{9025}{43571163}\pi^2 + \frac{6859000}{166049702193}\zeta(3)\\[8pt] &\quad - \frac{32580250}{1898446245172569}\pi^4 \tag{3} \end{align*} where $\gamma$ is the Euler-Mascheroni constant, and $\zeta(3) = \sum_{n=1}^\infty \frac{1}{n^3}$.

Finally, using (2) and (3), the inequality (1) is verified: we get $$- \ln \Gamma(1/27) - \ln \Gamma(190/3811) + \ln \left(\frac{3621}{3811}\, \frac{27 \pi}{\sin \frac{190\pi}{3811}}\right) > 0.0000017.$$

We are done.