I wish to extend the factorial to non-integer arguments in a unique way, given the following conditions:

  1. $n!=n(n-1)!$

  2. $1!=1$

To anyone interested in viewing the final form before reading the whole post:





$$\frac d{dx}f(x)=\frac d{dx}f(x-1)+\ln(x)$$




for $x\in\mathbb N$:


Euler has a nice extension of the harmonic numbers to non-integer arguments,


from the FTOC we have



And with $f'(0)=-\gamma$, the Euler mascheroni constant, we should get the gamma function. Or we may just let it sit as an unknown parameter.

My questions are if this captures all possible extensions of the factorial with the given conditions, since, if it did, it'd be a pretty good general extension to the factorial?

Given a few more assumptions, it is easy enough to set bounds to what $f'(0)$ might be as well.

Notably, this representation fails when considering $\Re(x)\le-1$, but coupled with the first condition, it is extendable to all $x$, except of course the negative integers.

robjohn♦ notes an extension to the harmonic numbers that converges for $x\in\mathbb C$, except the negative integers:


Any suggestions on things I could've improved and flaws in this would be nice.


Using the second condition and $x=1$, we may have


$$\implies f'(0)=-\int_0^1\int_0^1\frac{1-t^\phi}{1-t}dt\ d\phi$$


where $\gamma$ is the Euler-mascheroni constant.

Using this we get a new form of the gamma function(?):



I'm not sure how to deal with trivial manipulations of this expression, as surely someone is gonna say "hey, just multiply everything by $(1+\sin(2\pi x))$ and it will still satisfy the conditions, right?"

But regardless, I think this is a pretty cool new gamma function?

Also, references to this if it's not new.

If someone could make a graph of this to look at, you would be great.

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    $\begingroup$ Hm... why the vote to close as off-topic? $\endgroup$ – Simply Beautiful Art Oct 8 '16 at 12:39
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    $\begingroup$ Have you tried plotting it against the usual gamma function? $\endgroup$ – Daniel Robert-Nicoud Oct 8 '16 at 12:58
  • $\begingroup$ @DanielRobert-Nicoud I don't know how to... my usual graphing calculators won't except this. For WolframAlpha, the computation time is exceeded. If you could provide a plot, that'd be great. $\endgroup$ – Simply Beautiful Art Oct 8 '16 at 13:01
  • $\begingroup$ Since you seem to be interested in this sort of thing, you might want to see this, too. $\endgroup$ – J. M. isn't a mathematician Oct 9 '16 at 22:52
  • $\begingroup$ @TheGreatDuck I have no idea how you would implement that $\endgroup$ – Simply Beautiful Art Feb 11 '17 at 22:56

Unfortunately, this is not new, though I would like to offer another derivation; \begin{align*} \int_{0}^{t}H_ydy &= \int_{0}^{t}\int_{0}^{1}\frac{1-x^y}{1-x}dxdy\\ &= \int_{0}^{1}\int_{0}^{t}\frac{1-x^y}{1-x}dydx\\ &= \int_{0}^{1}\frac{t}{1-x}+\frac{1-x^t}{(1-x)\ln(x)}dx\\ (1)&= \int_{0}^{1}\frac{t}{1-x}+\sum_{j=0}^{t-1}\frac{x^j}{\ln(x)}dx\\ &=\lim_{x\rightarrow 1^{-}}\left( -t\ln(1-x)+\sum_{j=0}^{t-1}\text{li}(x^{j+1})\right)\\ (2)&=\lim_{x\rightarrow 1^{-}}\sum_{j=0}^{t-1}\left(\text{li}(x^{j+1})-\ln(1-x)\right)\\ &= \gamma t + \sum_{j=0}^{t-1}\ln(j+1)\\ &= \gamma t + \ln(t!)\\ &= \gamma t +\ln\Gamma(t+1) \end{align*}

Now writing $H_y$ as $\sum_{n=1}^{\infty}\frac{y}{n(n+y)}$ gives \begin{align*} \int_{0}^{t}H_ydy &= \gamma t + \ln\Gamma(t+1)\\ &= \int_{0}^{t}\sum_{n=1}^{\infty}\frac{y}{n(n+y)}dy\\ &= \sum_{n=1}^{\infty}\frac{t}{n}-\ln \left(1+\frac{t}{n}\right) \end{align*} solving for $\Gamma(t+1)$ and using $\Gamma(t+1) = t\Gamma(t)$ gives $$\Gamma(t) = \frac{e^{-\gamma t}}{t}\prod_{n=1}^{\infty}\left(1+\frac{t}{n}\right)^{-1}e^{\frac{t}{n}}$$

EDIT First, I should say that I am just starting to study analysis - I just try stuff and see if it works, so I apologize for the lack of rigour.

I'm not too sure which steps I should fill in, so I will try to fill in any apparent holes.

assuming we only know that $\gamma$ shows up as $$\lim_{N \rightarrow \infty}(H_N - \ln(N)) = \gamma,$$ one can show that $\int_{0}^{1}H_ydy=\gamma$ using $H_y=\sum_{k=2}^{\infty}(-1)^k\zeta(k)y^{k-1}$ for $|y|<1$ (I have a crude derivation here).

Knowing that, we get \begin{align*} \gamma = \int_{0}^{1}H_ydy &= \int_{0}^{1}\int_{0}^{1}\frac{1-x^y}{1-x}dxdy\\ &= \int_{0}^{1}\int_{0}^{1} \frac{1}{1-x}-\frac{x^y}{1-x}dydx\\ &= \int_{0}^{1}\frac{1}{1-x}+\frac{1}{\ln(x)}dx\\ (*)&= \lim_{x\rightarrow 1^{-}}(\text{li}(x)-\ln(1-x)),\\ \end{align*} where $\text{li}(x) = \int_{0}^{x}\frac{1}{\ln(t)}dt$ is the logarithmic integral.

Now we need to evaluate $\int_{0}^{t}\frac{x^k}{\ln(x)}dx$ for (1); Make the substitution $\;x=e^u$ to get $$\int_{0}^{t}\frac{x^k}{\ln(x)}dx = \int_{-\infty}^{\ln(t)}\frac{e^{(k+1)u}}{u}du,$$ now let $v=(k+1)u$ and see that \begin{align*} \int_{0}^{t}\frac{x^k}{\ln(x)}dx &= \int_{-\infty}^{(k+1)\ln(t)}\frac{e^x}{x}dx\\ &= \int_{0}^{t^{k+1}}\frac{1}{\ln(x)}dx\\ &=\text{li}(t^{k+1}) \end{align*} after the substitution $x=\ln(u)$.

Next we need to evaluate $\lim_{x\rightarrow 1^{-}}(\text{li}(x^k)-\ln(1-x))$. so by using $(*)$ - \begin{align*} \lim_{x\rightarrow 1^{-}}(\text{li}(x^k)-\ln(1-x)) &= \lim_{x\rightarrow 1^{-}}(\text{li}(x) - \ln(1-\sqrt[k]{x})\\ &= \lim_{x\rightarrow 1^{-}}\left(\text{li}(x) - \ln \left(\frac{1-x}{1+\sqrt[k]{x} + \cdots +\sqrt[k]{x^{k-1}}}\right)\right)\\ &= \gamma + \ln(k) \end{align*}

I just have to say that I really like how you derived it.

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    $\begingroup$ Ah, nice, so we reach the Weierstrass definition of the gamma function? Nice catch. Perhaps a reference would be nice. $\endgroup$ – Simply Beautiful Art Oct 8 '16 at 13:13
  • $\begingroup$ This is something I came up with a few days ago, actually. I feel for OP because I thought it was new too $\endgroup$ – Plopperzz Oct 8 '16 at 13:14
  • $\begingroup$ XD No, wait, but was it new before you?! I'm trying to read up on this. $\endgroup$ – Simply Beautiful Art Oct 8 '16 at 13:15
  • $\begingroup$ Oh sorry. Its just where ive been typing up my notes from when i started studying the harmonic numbers. Im off to work so i will find a way to post it later if you'd like. $\endgroup$ – Plopperzz Oct 8 '16 at 13:26
  • $\begingroup$ Sure thing, take your time. $\ddot\smile$ $\endgroup$ – Simply Beautiful Art Oct 8 '16 at 13:27

Integral of the Logarithmic Derivative of Gamma

The logarithmic derivative of the Gamma function is the digamma function: $$ \frac{\Gamma'(x)}{\Gamma(x)}=\psi(x)=-\gamma+H_{x-1}\tag{1} $$ Therefore, $$ \begin{align} \log(\Gamma(x)) &=\int_1^x\left(-\gamma+H_{\phi-1}\right)\mathrm{d}\phi\\ &=\int_1^x\left(-\gamma+\int_0^1\frac{1-t^{\phi-1}}{1-t}\,\mathrm{d}t\right)\mathrm{d}\phi\\ &=\int_0^{x-1}\left(-\gamma+\int_0^1\frac{1-t^\phi}{1-t}\,\mathrm{d}t\right)\mathrm{d}\phi\tag{2} \end{align} $$

Verification of the Gamma Function

The Bohr-Mollerup Theorem says that the Gamma function is uniquely determined as the log-convex function so that $\Gamma(1)=1$ and $\Gamma(x+1)=x\,\Gamma(x)$.

We can verify these assuming only $H_x-H_{x-1}=\frac1x$ and $H'_x\ge0$.

$\boldsymbol{\Gamma(1)=1}$: $$ \begin{align} \log(\Gamma(1)) &=\int_1^1\left(-\gamma+H_{\phi-1}\right)\mathrm{d}\phi\\ &=0\tag{3} \end{align} $$ $\boldsymbol{\Gamma(x+1)=x\,\Gamma(x)}$: Since $\lim\limits_{n\to\infty}\left(H_n-\log(n)\right)=\gamma$ and $H_n-\frac1n\le\int_{n-1}^nH_\phi\,\mathrm{d}\phi\le H_n$, $$ \begin{align} \log(\Gamma(x+1))-\log(\Gamma(x)) &=\int_x^{x+1}\left(-\gamma+H_{\phi-1}\right)\mathrm{d}\phi\\ &=-\gamma+\lim_{n\to\infty}\left(\int_x^nH_{\phi-1}\,\mathrm{d}\phi-\int_{x+1}^nH_{\phi-1}\,\mathrm{d}\phi\right)\\ &=-\gamma+\lim_{n\to\infty}\left(\int_x^nH_{\phi-1}\,\mathrm{d}\phi-\int_x^{n-1}H_\phi\,\mathrm{d}\phi\right)\\ &=-\gamma+\lim_{n\to\infty}\left(-\int_x^n\frac1\phi\,\mathrm{d}\phi+\int_{n-1}^nH_\phi\,\mathrm{d}\phi\right)\\[6pt] &=-\gamma+\log(x)+\lim_{n\to\infty}\left(-\log(n)+H_n\right)\\[8pt] &=\log(x)\tag{4} \end{align} $$ $\boldsymbol{\Gamma}$ is log-convex: $$ \begin{align} \frac{\mathrm{d}^2}{\mathrm{d}x^2}\log(\Gamma(x)) &=H'_{x-1}\\ &\ge0\tag{5} \end{align} $$

Verifying the Necessary Properties of the Extension $$ \begin{align} H_x-H_{x-1} &=\int_0^1\frac{1-t^x}{1-t}\,\mathrm{d}t-\int_0^1\frac{1-t^{x-1}}{1-t}\,\mathrm{d}t\\ &=\int_0^1t^{x-1}\,\mathrm{d}t\\ &=\frac1x\tag{6} \end{align} $$ $$ \begin{align} H'_x &=\int_0^1\frac{-\log(t)t^x}{1-t}\,\mathrm{d}t\\ &\ge0\tag{7} \end{align} $$

Limitations of the Extension

One limitation of the extension $$ H_x=\int_0^1\frac{1-t^x}{1-t}\,\mathrm{d}t\tag{8} $$ is that it doesn't converge for $\mathrm{Re}(x)\le-1$.

An extension of the Harmonic Numbers that works for all $x\in\mathbb{C}$ is $$ H_x=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag{9} $$

  • $\begingroup$ Ok, but how do I relate the logarithmic derivative of the gamma function to the more general factorial? The use of the gamma function to set a relationship to the harmonic numbers isn't what I was really looking for.. $\endgroup$ – Simply Beautiful Art Oct 8 '16 at 14:06
  • $\begingroup$ The extension of the Harmonic numbers you presents is nice, but I honestly chose the integral form for faster convergence. $\endgroup$ – Simply Beautiful Art Oct 8 '16 at 14:11
  • $\begingroup$ Judging from your question, you know that for $n\in\mathbb{Z}$, $n!=\Gamma(n+1)$, so the natural extension is $x!=\Gamma(x+1)$; that is how you relate the logarithmic derivative of the Gamma function to the more general factorial. Nowhere do I use the Gamma function to set up a relation to the Harmonic Numbers, but as you do, I use the extended Harmonic Numbers and their relationship to the digamma function to derive a formula for the logarithm of the Gamma function. This relationship is well known and so I thought it was relevant to your question. $\endgroup$ – robjohn Oct 8 '16 at 14:40
  • $\begingroup$ I mean, sure, yes, but you assume I speak of $n!=\Gamma(n+1)$. It could've been very well that my extension were not equal to the gamma function at all, making this more of a side note to me. $\endgroup$ – Simply Beautiful Art Oct 8 '16 at 14:42
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    $\begingroup$ @SimpleArt: that is the property we are trying to verify about $\Gamma$ defined in this way. The whole point of that section is verifying that $\Gamma$ is the $\Gamma$ we all know and love. $\endgroup$ – robjohn Oct 8 '16 at 22:57

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