Euler gamma function differential

Do you have any piece of advice on how to calculate a differential of an Euler Gamma function for $ x \in R- Z $?Thank you for all your answers.

Solution 1:

Since $\Gamma(x)$ is a convex function of $x$, $$ \scriptsize\overbrace{\frac{\log(\Gamma(x+n))-\log(\Gamma(x+n-1))}{(x+n)-(x+n-1)}}^{\log(x+n-1)} \le\overbrace{\frac{\Gamma'(x+n)}{\Gamma(x+n)}}^{\frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x+n))} \le\overbrace{\frac{\log(\Gamma(x+n+1))-\log(\Gamma(x+n))}{(x+n+1)-(x+n)}}^{\log(x+n)}\tag{1} $$ Thus, for a fixed $x$, $(1)$ says $$ \frac{\Gamma'(x+n)}{\Gamma(x+n)}=\log(n)+O\!\left(\frac1n\right)\tag{2} $$ The recursion for $\Gamma(x)$ yields $$ \begin{align} \Gamma(x+n)&=(x+n-1)(x+n-2)\cdots x\,\Gamma(x)\tag{3}\\[6pt] \log(\Gamma(x+n))&=\log(x+n-1)+\log(x+n-2)+\cdots+\log(x)+\log(\Gamma(x))\tag{4}\\ \frac{\Gamma'(x+n)}{\Gamma(x+n)}&=\frac1{x+n-1}+\frac1{x+n-2}+\cdots+\frac1x+\frac{\Gamma'(x)}{\Gamma(x)}\tag{5} \end{align} $$ Rewriting $(5)$ and applying $(2)$, we get $$ \begin{align} \frac{\Gamma'(x)}{\Gamma(x)} &=\frac{\Gamma'(x+n)}{\Gamma(x+n)}-\left(\frac1x+\frac1{x+1}+\cdots+\frac1{x+n-1}\right)\tag{6}\\ &=\log(n)+O\!\left(\frac1n\right)-\left(\frac11+\frac12+\cdots+\frac1n\right)\tag{7}\\ &+\left(\frac11+\frac12+\cdots+\frac1n\right)-\left(\frac1x+\frac1{x+1}+\cdots+\frac1{x+n-1}\right)\tag{8}\\ &=-\gamma+O\!\left(\frac1n\right)+\sum_{k=0}^{n-1}\left(\frac1{k+1}-\frac1{k+x}\right)\tag{9}\\ &=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+x}\right)\tag{10} \end{align} $$