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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?

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I think you are at the right track and the answer should be $\sqrt{2\pi}$. I think the unjustified part can be handled using one of the convergence theorems. For $s\rightarrow 0$, $(\Gamma(x))^s \rightarrow 1$ for all $0<x\leq 1$. – Anon Jan 28 '13 at 19:59
With help from @DavidMoews, we now have a complete proof that the limit is indeed $\sqrt{2\pi}$. – Ayman Hourieh Jan 28 '13 at 21:39
@Anon: thanks for your feedback. – user 1618033 Jan 30 '13 at 8:29
up vote 6 down vote accepted

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 $\sqrt{2\pi}$.

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The singularity of $\Gamma(x)$ at $0$ is $O(1/x)$, so (on $[0,1]$) $||\Gamma||_p$ is finite whenever $p<1$. – David Moews Jan 28 '13 at 21:32
@DavidMoews Of course! Thanks for your comment. The proof is now complete. – Ayman Hourieh Jan 28 '13 at 21:37
@AymanHourieh: thanks for you nice solution. However, I still ask myself if I can finish the last part by only using some elementary tools of calculus. – user 1618033 Jan 30 '13 at 8:31

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