# Did I derive a new form of the gamma function?

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

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.

#### Solutions Collecting From Web of "Did I derive a new form of the gamma function?"

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.

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
$$H_x=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\tag{9}$$