Pretty much a circle 3

This solution is not a complete proof, but shows roughly how the answer is obtained.

Let $a$ be the value of $x$ that minimises $\Gamma(x)$ , that is $\Gamma(a)=m$ . As $r$ decreases towards $2m$ (the minimum value of $\Gamma(x)+\Gamma(y)$ ), the region $\Gamma(x)+\Gamma(y)<r$ becomes a smaller and smaller loop around $(x,y)=(a,a)$ , which as it turns out approaches the shape of a circle.

Consider the Taylor series expansion of $\Gamma(x)$ around the point $x=a$ : $\Gamma(x)=\Gamma(a)+\Gamma'(a)(x-a)+\frac12\Gamma''(a)(x-a)^2+\dots$ Now, $\Gamma(a)=m$ , and since $x=a$ is a minimum, $\Gamma'(a)=0$ . We will ignore higher-order terms, and only use a second degree expansion. Intuitively we can do this because as $x$ approaches $a$ the terms $(x-a)^3,(x-a)^4,\dots$ approach zero much faster than $(x-a)^2$ . Actually proving that the limit of the area remains the same if we remove these terms is more complex, and not shown here. This leaves $\Gamma(x)=m+\frac12\Gamma''(a)(x-a)^2$ , so the equation $\Gamma(x)+\Gamma(y)<r$ becomes $m+\frac12\Gamma''(a)(x-a)^2+m+\frac12\Gamma''(a)(y-a)^2<r$ $\implies(x-a)^2+(y-a)^2<\frac{2(r-2m)}{\Gamma''(a)}$ This is a circle of radius $\sqrt{\frac{2(r-2m)}{\Gamma''(a)}}$ centred at $(a,a)$ , which has area $\frac{2(r-2m)}{\Gamma''(a)}\pi$ . Hence, the limit becomes $\lim_{r\to2m^+}\frac1{r-2m}\frac{2(r-2m)}{\Gamma''(a)}\pi=\boxed{\frac{2\pi}{\Gamma''(a)}}$ $\Gamma''(a)$ can be written in terms of the polygamma functions : $\psi^{(0)}(x)=\frac d{dx}\ln\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}$ $\psi^{(1)}(x)=\frac d{dx}\psi^{(0)}(x)=\frac{\Gamma''(x)\Gamma(x)-(\Gamma'(x))^2}{(\Gamma(x))^2}=\frac{\Gamma''(x)}{\Gamma(x)}-\left((\psi^{(0)}(x)\right)^2$ $\implies \Gamma''(a)=\Gamma(a)\left(\left(\psi^{(0)}(a)\right)^2+\psi^{(1)}(a)\right)$ The value of $a$ can then be found numerically on a program such as Mathematica by looking for the positive root of $\psi^{(0)}(x)$ :

a = x /. FindRoot[PolyGamma[0, x], {x, 2}]

p = Gamma[a] (PolyGamma[0, a]^2 + PolyGamma[1, a])

2 Pi/p

This gives an answer of $7.33183$ .