Compute $\lim_{s\to 0} \left(\int_0^1 (\Gamma (x))^s\space\mathrm{dx}\right)^{1/s}$ 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!
 A: 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 $\sqrt{2\pi}$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\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}

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 !!!.
