A series expansion for $\left(x^2\,B(x,x)\right)^\frac nx$ or $\frac{1}{(Γ(2x))^\frac nx}$ for $0\lt x\le 1, n\in\Bbb N$

Question

Central Beta function Β(x). Imagine we wanted to find a series expansion for $\left(x^2\,B(x,x)\right)^\frac nx, n\in\Bbb N $ which would be particularly difficult because it uses the Beta function. Let there be the rule that any strategies can be used to find a series expansion for $0\le x\le 1$ as long as the series terms are not too complicated and have some pattern. Here is how I would do it using a Taylor expansion at x=a which all give beautiful results. One simple and amazing expansion would be at x=0, but to include all cases, I will use x=a as all cases for $0\lt a\le 1$ have the radius of convergence which works for $0\lt x\le 1$:

$$\left(x^2\,B(x)\right)^\frac nx=\left(x^2\,B(x,x)\right)^\frac nx =\frac{x!^\frac{2n}{x}}{Γ(2x)^\frac nx}=Γ^{-\frac nx}(2x)\sum_{n=0}^\infty \frac{\left(\frac{d^n}{dx^n}x!^\frac {2n}x\right)_{x=a}(x-a)^n}{n!}$$

Here comes the problem where I am unsure about the radius of convergence for the usual Taylor series of $Γ(2x) ^{-\frac nx}$. The problem is this steep change due to the x in the denominator of $\left(\frac{(2x)!}{2x}\right)^{-\frac nx}$. If it was the factorial instead of the gamma, there would not be such a small derivative at $x≈0$ which would imply the series representation working. For example here is the graph of $\frac{1}{Γ(2x)^\frac 1x}$ for $n=1$:

This all relates to the series of $x^\frac 1x$ of which I still do not know the convergence for. It amazingly has an explicit form for the Taylor Series at $x=1$ at OEIS A008405. How can I do the rest or have a new series?

It does not have to be a Taylor Series.

Please correct me and give me feedback!

Answer 1

$\textstyle\displaystyle{\frac{1}{\Gamma(x)}=x\prod_{k=1}^{\infty}\frac{1+\frac{x}{n}}{\left(1+\frac{1}{n}\right)^x}}$

$\implies\textstyle\displaystyle{x^2\Gamma^2(x)=\prod_{k=1}^{\infty}\frac{\left(1+\frac{1}{n}\right)^{2x}}{\left(1+\frac{x}{n}\right)^2}}$

Using the same formula we can get,

$\textstyle\displaystyle{\frac{1}{\Gamma(2x)}=2x\prod_{k=1}^{\infty}\frac{1+\frac{2x}{n}}{\left(1+\frac{1}{n}\right)^{2x}}}$

Finally their ratio would be

$\textstyle\displaystyle{\frac{x^2\Gamma^2(x)}{\Gamma(2x)}}$

$=x^2B(x,x)$

$\textstyle\displaystyle{=2x\prod_{k=1}^{\infty}\frac{1+\frac{2x}{n}}{\left(1+\frac{x}{n}\right)^2}}$

$\textstyle\displaystyle{\implies\frac{n}{x}\ln(x^2B(x,x))=\frac{n\ln(2x)}{x}+\frac{n}{x}\sum_{k=1}^{\infty}\ln\left(\frac{n(n+2x)}{(n+x)^2}\right)}$

Obviously this not the infinite sum for the function you wanted but rather the log of it. So I would call this a partial answer.

Edit:

Technically you can find the series representation of the function using what I derived previously, but it would produce a messy double summation-

If we take $e$ to be the base on both sides then the $LHS$ would be $(x^2B(x,x))^\frac{n}{x}$ and the $RHS$ would be (using the series representation of $e^x$)-

$\textstyle\displaystyle{(x^2B(x))^\frac{n}{x}=\sum_{j=0}^{\infty}\frac{1}{j!}\left(\frac{n\ln(2x)}{x}+\frac{n}{x}\sum_{k=1}^{\infty}\ln\left(\frac{n(n+2x)}{(n+x)^2}\right)\right)^j}$

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And for the series representation of $\textstyle\displaystyle{\Gamma(2x)^{-\frac{n}{x}}}$, I will do a similar thing, this time I am going to use weierstrass' formula for the gamma function-

So, $\Gamma(2x)^{-\frac{n}{x}}=e^{-\frac{n}{x}\ln(\Gamma(2x))}$

$\textstyle\displaystyle{=\sum_{k=0}^{\infty}\frac{1}{k!}\left(-\frac{n}{x}\ln(\Gamma(2x))\right)^k}$

$\textstyle\displaystyle{=\sum_{k=0}^{\infty}\frac{(-1)^kn^k}{k!x^k}\ln^k\left(\frac{1}{2xe^{2{\gamma}x}}\prod_{j=1}^{\infty}\frac{e^\frac{2x}{j}}{1+\frac{2x}{j}}\right)}$

$\textstyle\displaystyle{=\sum_{k=1}^{\infty}\frac{(-1)^kn^k}{k!x^k}\left(-\ln(2x)-2\gamma x+\sum_{j=1}^{\infty}\left(\frac{2x}{j}-\ln\left(1+\frac{2x}{j}\right)\right)\right)^k}$

If I am able to do better than this, then I would add it to the answer.

Answer 2

• The famous software Mathematica shows that \begin{align} \left[\frac{1}{\Gamma (2 x)}\right]^{n/x} &=\exp \biggl[\frac{n \log (2 x)}{x}+2 \gamma n-\frac{n \pi ^2}{3} x-\frac{4n \psi''(1)}{3} x^2-\frac{2n \pi ^4}{45}x^3\\ &\quad -\frac{4n \psi ^{(4)}(1)}{15} x^4-\frac{32 n \pi ^6}{2835}x^5-\frac{8n \psi ^{(6)}(1)}{315} x^6+O\left(x^7\right)\biggr] \end{align} and \begin{align} \biggl[\frac{1}{\Gamma (2 x)}\biggr]^{n/x} &=1+2 (\gamma -1) n (x-1)+\frac{1}{3} n \bigl[6 \gamma ^2 n+6 n-6 \gamma (2 n+1)-\pi ^2+12\bigr] (x-1)^2\\ &\quad+\frac{1}{3} n \bigl\{4 \gamma ^3 n^2-4 n^2+2 \gamma \bigl[6 n^2-\bigl(\pi ^2-18\bigr) n+3\bigr]\\ &\quad-12 \gamma ^2 (n+1) n+2 \bigl(\pi ^2-12\bigr) n+\pi ^2-12-4 \psi''(2)\bigr\}(x-1)^3\\ &\quad+\frac{1}{90} n \bigl\{60 \gamma ^4 n^3+60 n^3-120 \gamma ^3 (2 n+3) n^2-60 \bigl(\pi ^2-12\bigr) n^2\\ &\quad+60 \gamma ^2 \bigl[6 n^2-\bigl(\pi ^2-24\bigr) n+9\bigr] n-60 \gamma \bigl[4 n^3-2 \bigl(\pi ^2-15\bigr) n^2\\ &\quad+n \bigl(30+4 \psi''(2)-2 \pi ^2\bigr)+3\bigr]+5 n \bigl[\pi ^4-36 \pi ^2+48\bigl(6+\psi''(2)\bigr)\bigr]\\ &\quad-4 \pi ^4-30 \pi ^2+720+120 \psi''(2)\bigr\}(x-1)^4 +O\bigl((x-1)^5\bigr). \end{align} These two finite series expansions imply that it should be very difficult to explicitly and simply expand the function $\bigl[\frac{1}{\Gamma (2 x)}\bigr]^{n/x}$ at $x=0,1$ and any other point $x=a\in(0,1)$.
• The function $\frac1{\Gamma(2z)}$ is an entire complex function on $\mathbb{C}$. Therefore, the function $\bigl[\frac1{\Gamma(2z)}\bigr]^{n/z}$ for $n\in\mathbb{N}$ is analytic on the punctured complex plane $\mathbb{C}\setminus\{0\}$. Accordingly, the function $\bigl[\frac1{\Gamma(2z)}\bigr]^{n/z}$ for $n\in\mathbb{N}$ has Taylor's series expansion at any complex point $z_0\ne0$ with a convergent radius $|z_0|$. Consequently, when the function $\bigl[\frac1{\Gamma(2z)}\bigr]^{n/z}$ for $n\in\mathbb{N}$ is expanded at $z=1$, its convergent radius is $1$, its convergent disc is $|z-1|<1$. This means that, the Taylor series expansion at $x=1$ of the function $\bigl[\frac1{\Gamma(2x)}\bigr]^{n/x}$ for $n\in\mathbb{N}$ is convergent for $|x-1|<1$. By Cauchy's limit theorem, we conclude that the convergent interval is $x\in(0,2]$.