10

Is there any explicit series, product, integral, continued fraction or other kind of expression for the point at which $\Gamma(x)$ has a minimum in $(0,1)$?

The decimal value can be found here http://oeis.org/A030169, but I haven't found much information there, or in other sources.

This is the point $a \in (0,1)$ such that:

$$\psi(a)=0$$

Using the integral representation of digamma function:

$$\psi(1+s)=-\gamma+\int_0^1 \frac{1-t^s}{1-t}dt$$

Introducing $b=a-1$, we can deduce that:

$$b \sum_{k=0}^\infty (-1)^k \zeta(k+2) b^k=\gamma$$

Inverting this power series, we can get a power series for $b$ in terms of Euler-Mascheroni constant, however, it's a long and tedious process and doesn't give us an explicit formula.

There's a very old related question, however I'm asking for a more particular result.

Yuriy S
  • 32,728

1 Answers1

3

It is known that $$\psi\left(\dfrac32\right) = \gamma - 2\ln2 + 2 = y_0 \approx 0.03648\ 99740,\tag1$$ $$\psi'\left(\dfrac32\right) = \dfrac12\pi^2-4 = d_1\approx 0.93480\ 22005,\tag2$$ $$\psi^{(n)}\left(\dfrac12\right) = (-1)^{(n+1)}n!\left(2^{n+1}-1\right)\zeta(n+1),\quad n= 1,2,\dots\tag3$$ $$\psi^{(n)}(z+1) = \psi^{(n)}(z) + (-1)^n\dfrac{n!}{z^{n+1}},\tag4$$ wherein $(3)-(4)$ leads to $$\psi^{(n)}\left(\dfrac32\right) = (-1)^{(n+1)}n!\left(2^{n+1}-1\right)\zeta(n+1)+(-1)^nn!2^ {n+1}= d_n,\quad n = 2,3,\dots,\tag5$$ $$d_2=16-14\zeta(3)\approx -0.82879\ 66442,\quad d_3 = \pi^4-96\approx 1.40909\ 10340.\tag6$$ Let $v(y)$ is the inverse function to $\psi\left(x\right),$ then $$\left\{\begin{align} &v(y_0) = \dfrac32\\ &v'_y(y)= \dfrac{dv}{dy} = \dfrac1{\psi'(x)},\\ &v'(y_0) = \dfrac1{\psi'(x)}\bigg|_{x=3/2} \ = \dfrac1{d_1}\\ &v''_{y}(y)= \left(v'_y(y)\right)^{'}_y = \left(\dfrac1{\psi'(x)}\right)^{'}_x\dfrac1{\psi'(x)} = -\dfrac{\psi''(x)}{(\psi'(x))^3},\\ &v''(y_0) = -\dfrac{\psi''(x)}{(\psi'(x))^3}\bigg|_{x=3/2} = -\dfrac{d_2}{d_1^3}\\ &v'''_{y}(y)= \left(v''_y(y)\right)^{'}_y = \left(-\dfrac{\psi''(x)}{(\psi'(x))^3}\right)^{'}_x\dfrac1{\psi'(x)} = \dfrac{3\psi''(x) - \psi'''(x)\psi'(x)}{(\psi'(x))^5},\\ &v'''_{y}(y_0)= \dfrac{3\psi''(x) - \psi'''(x)\psi'(x)}{(\psi'(x))^5}\bigg|_{x=3/2} = \dfrac{3d_2-d_1d_3}{d_1^5}\dots \end{align}\right.\tag7$$ Therefore, $$v(y)=\dfrac32 + \dfrac{y-y_0}{d_1} - \dfrac12\dfrac{d_2}{d_1^3}(y-y_0)^2 + \dfrac16\dfrac{3d_2-d_1d_3}{d_1^5}(y-y_0)^3 +\dots,\tag8$$ $$a = v(0) \approx 1.46168.$$ Exact value of $a$ is $1.46163\ 21449\ 68362\dots$

Yuri Negometyanov
  • 28,835