I derived a formula for $[x!]^\prime$. Is it correct?

Question

The starting point was that $ \Gamma'(x+1)=\Gamma(x+1)\psi(x+1)$ where $\psi(x+1)=-\gamma+H_{x}$ . Hence $$ [x!]' = x!\biggl[-\gamma+\sum_{k=1}^{x}\frac{1}{k}\biggl]$$ For example $ [4!]' = 24[-\gamma+1+1/2+1/3+1/4]$ which gives $36.1462$ that, put in the tangent equation, gives us exactly the tangent for $x=4$! Let me know if I made any mistakes in the derivation/generalization!

Answer 1

It is true that $$ \partial_x\Gamma(x)\bigg|_{x=5}=\Gamma(x)\psi(x)\bigg|_{x=5}=4!\left(\frac{25}{12}-\gamma\right). $$ The problem with saying that this is the derivative of the factorial function is that to calculate a derivative you must extend the factorial $x!:\Bbb N_0\to\Bbb N_0$ to a continuos function. The gamma function is one way to do this. But now consider another continuous interpolation of the factorial, namely, the Hadamard gamma function: $$ H(x)=\frac{\psi ^{(0)}\left(1-\frac{x}{2}\right)-\psi ^{(0)}\left(\frac{1}{2}-\frac{x}{2}\right)}{2 \Gamma (1-x)}. $$ Just like the gamma function we have for $n\in\Bbb N_0$ that $H(n)=(n-1)$!. Calculating the derivative and evaluating at $x=5$ gives $$ \partial_xH(x)\bigg|_{x=5}=4!\left(\frac{3}{2}-\gamma +\log 2\right), $$ which differs from the result we got using the gamma function. For this reason, it does not make sense to call one result or the other a derivative of the factorial.