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:

> 
>  $$x!=\exp\left[\int\_0^x\left(-\gamma+\int\_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi\right]$$
> 
> 
> 

---

> 
> $$f(x):=\ln(x!)$$
> 
> 
> 

$$f(x)=\ln(x!)=\ln(x)+\ln((x-1)!)=\ln(x)+f(x-1)$$

$$f(x)=f(x-1)+\ln(x)$$

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

$$f'(x)=f'(x-1)+\frac1x\tag1$$

$$f'(x)=f'(x-2)+\frac1{x-1}+\frac1x$$

$$=f'(0)+1+\frac12+\frac13+\dots+\frac1x$$

for $x\in\mathbb N$:

> 
> $$f'(x)=f'(0)+\sum\_{n=1}^x\frac1n\tag2$$
> 
> 
> 

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

$$f'(x)=f'(0)+\int\_0^1\frac{1-t^x}{1-t}dt\tag{2.1}$$

from the FTOC we have

$$\ln(x!)=\int\_0^x\left(f'(0)+\int\_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi$$

> 
> $$x!=\exp\left[\int\_0^x\left(f'(0)+\int\_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi\right]\tag3$$
> 
> 
> 

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:
> 
> 
> $$\int\_0^1\frac{1-t^\phi}{1-t}dt=\sum\_{n=1}^\infty\left(\frac1n-\frac1{n+\phi}\right)$$
> 
> 
> 

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

---

Edit:

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

$$1=\exp\left[\int\_0^1\left(f'(0)+\int\_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi\right]$$

$$\implies f'(0)=-\int\_0^1\int\_0^1\frac{1-t^\phi}{1-t}dt\ d\phi$$

$$f'(0)=-\gamma$$

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

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

> 
> $$\boxed{x!=\exp\left[\int\_0^x\left(-\gamma+\int\_0^1\frac{1-t^\phi}{1-t}dt\right)d\phi\right]}\tag4$$
> 
> 
> 

$$=\exp\left[\int\_0^x\left(-\gamma+\sum\_{n=1}^\infty\left(\frac1n-\frac1{n+\phi}\right)\right)d\phi\right]$$

---

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.