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I'm just unable to comprehend what would $$f(x) = (\log(n))!$$ be.

Most of what I saw online talks about $$f(x) = (\log(n!)).$$

Any idea into the intuition for this would be helpful!

Gary
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10Minhelp
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    It may be worth noting that $\log(n)$ generally isn't a whole number, so you can't really take its factorial... – Raad Shaikh Sep 27 '22 at 01:41
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    You can use the Gamma function, which is the analytic continuation of the factorial (basically just the factorial but also for non-integers). Most graphing calculators actually have this built in, so you can use e.g. Desmos' graphic calculator to see what your expression looks like by just typing it in. I don't think $(\log(n))!$ is particularly meaningful though - but I'm not entirely sure if applications for it have been found. –  Sep 27 '22 at 01:44
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    What is the context for this? $\log(n!)=\log n+\log( n-1)+...+\log 1$ is completely different from $(\log(n))!$ which is the factorial of a non-integer. – Suzu Hirose Sep 27 '22 at 01:49
  • @vrugtehagel Using Demos $\log(n!)$ looks more interesting – Henry Sep 27 '22 at 02:03
  • @SuzuHirose yea i need to understand what (())! would be like – 10Minhelp Sep 27 '22 at 02:35
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    But $\log(n)$ is not an integer even if $n$ is an integer, so why do you want to use the symbol $n$, which usually means a natural number or integer? As the above comment says the Gamma function is the well-known analytic continuation of the factorial. But there is nothing special about $\log(n)$ where $n$ is an integer that would affect it. That's why I asked about context. – Suzu Hirose Sep 27 '22 at 02:42
  • I'm not sure how much there is to say here, but if we interpret $y! = \Gamma(y + 1)$, Stirling's approximation yields the peculiar-looking asymptotic formula $$(\log n)! = \sqrt{2 \pi} \frac{(\log x)^{\log x + \frac{1}{2}}}{x} + O ((\log n)^{-1}) .$$ As others have written, $\log(n!)$ comes up more frequently in applications; see, e.g., https://en.wikipedia.org/wiki/Gamma_function#The_log-gamma_function. – Travis Willse Sep 27 '22 at 02:47

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The closest thing to the factorial for non-integer numbers is the Gamma function $\Gamma(x)$ which, if $x$ is a positive integer, is equal to $(x-1)!$, and if $x$ is positive and not an integer, lies between the values of the factorial. You can then attach a meaning to $\log(n)!$ of $\Gamma(\log(n)+1)$. But the natural logarithm $\log(n)$ is not an integer for integer $n$ for any except the "trivial" case $\log(1)=0$.

Suzu Hirose
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I'm not how much there is to say here. At any rate in order to make sense of the function, we should interpret $y!$ as $\Gamma(y + 1)$, where $\Gamma$ denotes the gamma function, and hence our function $x \mapsto (\log x)!$ as $x \mapsto \Gamma(\log x + 1)$, with which it coincides on the nonnegative integers and which is defined everywhere except at the nonpositive integers. (This extension is the best one possible in a sense made precise by the Bohr-Mollerup Theorem.)

We can immediately read off properties of $x \mapsto (\log x)!$ from those of $\Gamma$: For example, $x \mapsto (\log x)!$ has simple poles precisely at $e^{-1}, e^{-2}, e^{-3}, \ldots$ and hence domain $(0, \infty) \setminus \{e^{-1}, e^{-2}, e^{-3}, \ldots\}$, and its range is $\Bbb R \setminus \{0\}$. The function's (unique) minimum occurs at $\beta = e^{\alpha - 1} = 1.58666\ldots$, where $\alpha = 1.46163\ldots$ is the location of the unique minimum of $\Gamma$ on $(0, \infty)$, and $x \mapsto (\log x)!$ is strictly monotone on $[\beta, \infty)$.

Stirling's approximation gives the (peculiar-looking) asymptotic formula as $n \to \infty$: $$(\log x)! = \sqrt{2 \pi} \frac{(\log x)^{\log x + \frac{1}{2}}}{x} \left(1 + O((\log x)^{-1})\right) .$$ Using an additional term of the expansion of $\Gamma(y + 1)$ at $\infty$ gives a better approximation for large $x$: $$(\log x)! \sim \sqrt{2 \pi} \frac{\log x^{\log x - \frac{1}{2}} (12 \log x + 1)}{12 x} .$$ In fact, already at $x = 2$ the relative error of the latter formula is smaller than $1$ part in $1\,300$.

Plot of (\log x)! and the two given approximations

The red curve is the graph of $(\log x)!$, the blue curve is the graph of the leading term of the above asymptotic formula, and the green curve is the graph of the second, better approximation.

As some of the commenters have mentioned, it's more common to consider $\log (x!) = \log \Gamma(x + 1)$; indeed $\log \Gamma$ is simply called the log-gamma function.

Travis Willse
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