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? $\endgroup$ Oct 8, 2016 at 12:39
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    $\begingroup$ Have you tried plotting it against the usual gamma function? $\endgroup$ Oct 8, 2016 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. $\endgroup$ Oct 8, 2016 at 13:01
  • $\begingroup$ Since you seem to be interested in this sort of thing, you might want to see this, too. $\endgroup$ Oct 9, 2016 at 22:52
  • $\begingroup$ @TheGreatDuck I have no idea how you would implement that $\endgroup$ Feb 11, 2017 at 22:56

3 Answers 3


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. $\endgroup$ Oct 8, 2016 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, 2016 at 13:14
  • $\begingroup$ XD No, wait, but was it new before you?! I'm trying to read up on this. $\endgroup$ Oct 8, 2016 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, 2016 at 13:26
  • $\begingroup$ Sure thing, take your time. $\ddot\smile$ $\endgroup$ Oct 8, 2016 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.. $\endgroup$ Oct 8, 2016 at 14:06
  • $\begingroup$ The extension of the Harmonic numbers you presents is nice, but I honestly chose the integral form for faster convergence. $\endgroup$ Oct 8, 2016 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, 2016 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. $\endgroup$ Oct 8, 2016 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, 2016 at 22:57

This is the Weierstrass definition of the Gamma function in disguise. Note that $$\Gamma(z)=\frac{e^{-\gamma z}}z\prod_{n=1}^\infty\frac{e^{z/n}}{1+\frac zn}$$We can re-write this as $$\frac{e^{-\gamma z}}z\prod_{n=1}^\infty\frac{e^{z/n}}{1+\frac zn}=e^{\log\left(\frac{e^{-\gamma z}}z\prod_{n=1}^\infty\frac{e^{z/n}}{1+\frac zn}\right)}=\frac1ze^{-\gamma z+\sum_{k=1}^\infty(z/k-\log(1+z/k))}$$$-\gamma z=-\int_0^z\gamma dx$ while the infinite sum is $\int_0^zH_xdx$ by using $H_x=\sum_{k=1}^\infty\frac x{n(n+x)}$.

PS: The only way I was able to answer this was because I had written the Gamma function in a similar way to solve $\int_0^1\Gamma(1+z)dz$ and then used the Taylor series expansion of the exponential function, but I realized that method was not going to work.

Edit: I just realized that one of the answerers also thought this was a new formula and I derived it too as I said above. Nice coincidence.


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