Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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. –  Chris's sis Jan 30 '13 at 8:29
add comment

1 Answer 1

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 $\sqrt{2\pi}$.

share|improve this answer
1  
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. –  Chris's sis Jan 30 '13 at 8:31
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.