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On my own time, I've been trying to learn as much as I can about the upper levels of mathematics. I recently came across the Gamma function: $$\Gamma(n) = (n-1)! = \int_{0}^{\infty}(t^{x-1}e^{-x})dt = \int_{0}^{1}(-\ln(t))^{x-1}dt$$ Therefore, obviously $x! = \int_{0}^{1}(-\ln(t))^{x-1}dt$ (this can also be verified by graphing both functions). This reminded me of something I thought about a long time ago: $f(x)=(x!)^{\frac{1}{x}}$. Now that I understand more about mathematics, I (like many others both in general and on stackoverflow) was able to prove that $(x!)^{\frac{1}{x}}$ diverges to $\infty$. However, when calculating $\lim_{x\to0}(x!)^{\frac{1}{x}}$, my intial guess that $\lim_{x\to0}(x!)^{\frac{1}{x}}=\gamma=0.57721...$ was proven wrong. I found that $\lim_{x\to0}(x!)^{\frac{1}{x}}\approx0.5615...$. This leads me to my first question. Does this number have any significance? Might it have any importance other than being the arbitrary number that's the answer to this question? [ANSWERED BY R_Berger]

Moving on to my second question. When I graphed this to verify my solution, I was surprised to see that the graph was practically a straight line (I expected more than a negligible curve of some type). Taking the derivative of this function would obviously be shown as $$\frac{d}{dx}\left(\int_0^1\left(-\ln\left(t\right)\right)^{x}dt\right)^{\frac{1}{x}}.$$ I was unable to find the above derivative either by hand or by using a calculator. So, is there a way to take the above derivative or any other variant of $x!$ (or $\Gamma(x-1)$)? [NO POINTERS GIVEN YET]

Lastly, most importantly, and my primary reason for asking this question... My ultimate goal is finding $\lim_{x\to\infty}\frac{d}{dx}\left(\int_0^1\left(-\ln\left(t\right)\right)^{x}dt\right)^{\frac{1}{x}}.$ If you have any pointers or small hints on anything I could look into, that would very much be appreciated. [NO POINTERS GIVEN YET]

If you take the time to read and answer this, I thank you very much in advance.

supersmarty1234
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  • For the first question: $\Gamma'(1)=-\gamma$, not sure if it helps much. – Raphael J.F. Berger Mar 27 '18 at 21:31
  • Simply doing a derivation yields:$$ [\Gamma(x+1)^{-x}]'= -\Gamma(x+1)^{-x}(\ln \Gamma(x+1) + x\frac{\Gamma'(x+1)}{\Gamma(x+1)}) $$ and for $x=0$ everything is well defined and evaluates to $$ -\Gamma(1)^{0}(\ln \Gamma(1) + 0\frac{\Gamma'(1)}{\Gamma(1)}) = -1(\ln 1 + 0\frac{\gamma}{1})=0.$$ – Raphael J.F. Berger Mar 27 '18 at 21:54
  • CAS (MMA) evaluates $\lim_{x\to0}\Gamma(x+1)^{\frac{1}{x}}=1.$ – Raphael J.F. Berger Mar 27 '18 at 22:06
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    The number you find is $\lim_{x\to 0} (x!)^{1/x} = e^{-\gamma} = 0.56145\cdots$. You can use Stirlings approximation to show that it convergest to a straight line of slope $1/e$ as $x\to\infty$, i.e. $\lim_{x\to\infty} \frac{(x!)^{1/x}}{x} = \frac{1}{e}$. – Winther Mar 27 '18 at 22:20
  • @R_Berger, Thanks for your input. I did find myself that $\gamma(1)=−\gamma$, so its good have someone else affirm this fact. From what I can tell, the derivation in your second comment of $[\Gamma(x+1)^{-x}]'$ is correct, however, the function I was looking to differentiate was $[\Gamma(x+1)^{\frac{1}{x}}]'$. Sorry :(. Also, I don't understand how $\lim_{x\to0}\Gamma(x+1)^{\frac{1}{x}}=1$ could possibly be correct, I believe this can be seen by just graphing the function. – supersmarty1234 Mar 28 '18 at 03:05
  • @Winther, in my research I became aware of Stirling's approximation, but never took the time to further research it. Out of my own personal curiosity, I have to verify the solution myself, and I will absolutely be taking the time now to learn about Stirling's approximation. Prior to this, I did know that $\lim_{x\to\infty}\frac{x}{x!^{\frac{1}{x}}}=e$, so by just that alone, I would think it would stand to reason that $\lim_{x\to\infty}\frac{x!^{\frac{1}{x}}}{x}=\frac{1}{e}$. Thank you so much! – supersmarty1234 Mar 28 '18 at 03:14
  • I should say that Stirlings is a bit of overkill to show this limit as it can be done without it. So if you managed to show it without it that's a better way – Winther Mar 28 '18 at 03:17
  • @Winther Stirling's Approximation is certainly good information for me to know. I am, however, a little confused about how to prove that $\lim_{x\to0}(x!)^{\frac{1}{x}}=e^{-\gamma}$. – supersmarty1234 Mar 28 '18 at 03:43
  • @supersmarty1234: Yes you are right. – Raphael J.F. Berger Mar 28 '18 at 07:01
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    I have confused $x^{-1}$ with $-x$ initially and then just copied your formula. Mathematica confirms @Winther's limit of $e^{-\gamma}$. Moreover it gives any finite series development of $\Gamma(x+1)^{x^{-1}}$ at $x=0$ (with linear term $\frac{\pi^2}{12}e^{-\gamma}$). – Raphael J.F. Berger Mar 28 '18 at 07:10
  • When you want to use derivations of the Gamma function, at some stage you have to work with Polygamma functions, then you can get some expressions for the derivatives. I think thats not simpler than Sterling for this problem. With the Sterling approximation the $x\to\infty$ should be straightforward. – Raphael J.F. Berger Mar 28 '18 at 14:50

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For limits of $f(x)^{1/x}$ sometimes one looks at logarithms, so we could look at $$\ln \Gamma(x+1)^{1/x} = \frac{\ln\Gamma(x+1)}{x}.$$ For $x\to0$ this looks like a case for de l'Hospital, for the denominator we get $1$ and for the numerator we have:

$$ \ln \Gamma(x+1)' = \frac{\Gamma(x+1)'}{\Gamma(x+1)}$$ at $x=0$ this evaluates (as you know) to $$\frac{-\gamma}{1}=-\gamma.$$

By exponentiating we recieve our $$e^{-\gamma}=\lim_{x\to0}\exp(\ln\Gamma(x+1)^{1/x})=\lim_{x\to0}\Gamma(x+1)^{1/x}$$

Edit

The case of $x\to\infty$: Using the Sterling approximation in terms of the $\Gamma$ function $$ \Gamma(n+1)\approx \sqrt{2\pi n}\big(\frac{n}{e}\big)^n$$ the expression of interest giving the slope is $$\lim_{n\to\infty}\frac{\Gamma(n+1)^\frac{1}{n}}{n}=\lim_{n\to\infty} \big(\sqrt{2\pi n}\big(\frac{n}{e}\frac{1}{n}\big)^n\big)^\frac{1}{n}=\lim_{n\to\infty} \big(2\pi n\big)^{\frac{1}{n}-\frac{1}{2}} \lim_{n\to\infty} \frac{\big(\frac{n^n}{e^n}\big)^\frac{1}{n}}{n}=$$ $$\lim_{n\to\infty} (2\pi)^\frac{1}{n} \lim_{n\to\infty}(n)^{\frac{1}{n}} \lim_{n\to\infty} \frac{1}{e}=e^{-1}.$$

Raphael J.F. Berger
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