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.
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