# Infinite Gamma Derivative Identity

We have $$\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}\;dx \tag{1}$$ We also have $$\frac{d^n}{dz^n}\, x^{z-1}e^{-x}=\log(x)^n e^{-x}x^{z-1}, \;\; z>1 \tag{2}$$ If we create an operator $$\hat{O}=\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n} = e^{\frac{d}{dz}}$$ and apply it to both sides of $(1)$, we get $$\hat{O}\Gamma(z)=\hat{O}\int_0^\infty x^{z-1}e^{-x}\;dx$$ if we allow this under the integral sign, then $$\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=\int_0^\infty \left(\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}x^{z-1}e^{-x}\right)\;dx$$ $$\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=\int_0^\infty \left(\sum_{n=0}^\infty \frac{\log(x)^n}{n!}\right)x^{z-1}e^{-x}\;dx$$ $$\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=\int_0^\infty x^{z}e^{-x}\;dx$$ $$\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=\Gamma(z+1)$$ this seems to hold numerically for $z>1$, for $z=1$, the value jumps between $1/2$ and $3/2$, which is curious as the average is right, but then $(2)$ doesn't hold anymore. The $n^{th}$ derivative of $\Gamma(z)$ seems to get increasingly complicated with large $n$.There is always a factor of $\Gamma(z)$ as well so it seems more to be a statement that $$\frac{1}{\Gamma(z)}\sum_{n=0}^\infty \frac{1}{n!}\frac{d^n}{dz^n}\Gamma(z)=z \tag{3}$$

Can this be proven in another way?

Is there any information about the derivative of $\Gamma(z)$ that makes this identity more obvious?

I notice that if we replace $\frac{d^n}{dz^n}\Gamma(z)$ with $\Gamma(z) \log(z)^n$, then the equation would still hold. Is that related at all?

I've noticed that if we replace $\psi^{(t)}(z)$ as $z^t$ then $$\frac{1}{\Gamma(z)}\frac{d^n}{dz^n} \Gamma(z) \big|_{\psi^{(t)}(z) \to z^t} = \sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}z^{n-k}$$ with $\left\{\begin{matrix} n \\ k \end{matrix}\right\}$ as Stirling numbers of the second kind.

We end up with a kind of ladder operator $$e^{\frac{d}{dz}}\Gamma(z) = \Gamma(z+1)$$ $$e^{-\frac{d}{dz}}\Gamma(z) = \Gamma(z-1)$$

• Is it not simply the Taylor expansion for $\Gamma(x+1)$? – Paul Enta Jul 1 '17 at 19:34
• @PaulEnta It does look similar but I thought it was missing the $(x-a)^n$ looking term. What point $a$ should the series be taken about? – Benedict W. J. Irwin Jul 2 '17 at 8:12
• @PaulEnta I guess if we set $a=z$ for the expansion of $\Gamma(z+1)$ then all the powers go to $1$ and the identity is reclaimed. Good spot if that's what you meant. However I've never set the expansion point to a function of the main variable before. Is that legal? – Benedict W. J. Irwin Jul 2 '17 at 8:19
• Yes, it is valid to choose arbitrarily the values of $a$ and $b$ in the Taylor expansion of $f(a+b)$ near $a$ or near $b$, provided that the series converges... – Paul Enta Jul 2 '17 at 17:59