Compute $\lim_{s\to 0} \left(\int_0^1 (\Gamma (x))^s\space\mathrm{dx}\right)^{1/s}$

Question

Compute $$\lim_{s\to 0} \left(\int_0^1 (\Gamma (x))^s\space\mathrm{dx}\right)^{1/s}$$

This is a problem I thought of these days and I think I know a way although not
completely justified. This is what I have

Firstly take log $$\lim_{s\to 0} \frac{\ln\left(\int_0^1 (\Gamma (x))^s\space\mathrm{dx}\right)}{s}\space \text{(Unjustified part where considering the numerator tends to 0) }$$ and then apply l'Hôpital's rule $$\lim_{s\to 0} \frac{\displaystyle \frac{d}{ds}\ln\left(\int_0^1 (\Gamma (x))^s\space\mathrm{dx}\right)}{\displaystyle \frac{d}{ds}s}\space=$$ $$\lim_{s\to 0} \frac{\displaystyle \frac{d}{ds} \left(\int_0^1 (\Gamma (x))^s\space\mathrm{dx}\right)}{\int_0^1 (\Gamma (x))^s\space\mathrm{dx}}\space=$$ and now differentiate under the integral sign $$\lim_{s\to 0} \frac{\displaystyle \int_0^1 \frac{d}{ds}(\Gamma (x))^s\space\mathrm{dx}}{\int_0^1 (\Gamma (x))^s\space\mathrm{dx}}\space=$$ $$\lim_{s\to 0} \frac{\displaystyle \int_0^1 (\Gamma (x))^s \ln (\Gamma(x))\space\mathrm{dx}}{\int_0^1 (\Gamma (x))^s\space\mathrm{dx}}\space=$$ $$\int_0^1 \ln (\Gamma(x))\space\mathrm{dx} \space \text{(Unjustified - I considered $\lim_{s\to 0} \int_0^1 (\Gamma (x))^s=1$ ) }$$ At this point I'm done since we know to compute $\int_0^1 \ln (\Gamma(x))\space\mathrm{dx}$. So, for
the problematic part I managed to split $$\lim_{s\to 0} \int_0^1 (\Gamma (x))^s \mathrm{dx}$$ into $$\lim_{s\to 0} \left(\int_0^{\epsilon} (\Gamma (x))^s \mathrm{dx}+\int_{\epsilon}^{1} (\Gamma (x))^s \mathrm{dx}\right)$$ and then I'm thinking to use the uniform convergence for the first integral
to prove that it tends to $0$. Am I on the right way? What would you suggest
me to do further? Would you approach the problem in a different manner?
Thanks!

Answer 1

Check my answer here to find a proof of the following:

If $\mu$ is a positive measure on a space $X$, $\mu(X) = 1$ and $\|f\|_p$ is finite for some $p$ then: $$ \lim_{p \to 0} \|f\|_{p} = \exp\left(\int_X \log|f| \,d\mu\right) $$

Mathematica suggests that $\|\Gamma\|_{1/2}$ is finite, but I haven't proved this yet. (Edit: See the comment by @DavidMoews below for a proof.)

Check this answer here to find that:

$$ \int_0^1 \log \Gamma(x) \,dx = \dfrac{1}{2}\log(2\pi) $$

And conclude that the limit you're after is $\sqrt{2\pi}$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\lim_{s \to 0}\bracks{\int_{0}^{1} \Gamma^{\large\, s}\pars{x}\dd x}^{1/s}} \\[5mm] = &\ \root{2\pi}\lim_{s \to 0}\braces{\int_{0}^{1} \bracks{{\pars{x + 1}!} \over \root{2\pi}x\pars{x + 1}}^{s}\dd x}^{1/s} \end{align} However, with the Robbins's Inequality, $$ \int_{0}^{1}\varphi_{-}^{s}\pars{x}\dd x < \int_{0}^{1} \bracks{{\pars{x + 1}!} \over \root{2\pi}x\pars{x + 1}}^{s}\dd x < \int_{0}^{1}\varphi_{+}^{s}\pars{x}\dd x $$ \begin{align} & \mbox{where} \\[1mm] & \left\{\begin{array}{lcl} \ds{\varphi_{-}\pars{x}} & \ds{\equiv} & \ds{{\pars{x + 1}^{\pars{x + 1/2}} \over x} \exp\pars{-\bracks{x + 1 - {1 \over 12x + 13}}}} \\[3mm] \ds{\varphi_{+}\pars{s}} & \ds{\equiv} & \ds{{\pars{x + 1}^{\pars{x + 1/2}} \over x} \exp\pars{-\bracks{x + 1 - {1 \over 12x + 12}}}\,\dd x} \end{array}\right. \end{align}

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Numerically, it's suggested that $$ \lim_{s \to 0}\bracks{\int_{0}^{1} \varphi_{\pm}^{\large\, s}\pars{x}\dd x}^{\large 1/s} = {\large 1}\qquad \substack{% \mbox{which will show that} \\[2mm] \mbox{the coveted limit is}\ \ds{\large\root{2\pi}}} $$ We don't have an analytic proof yet !!!.