When I searched for the derivative of the Gamma function I got something of the form:
$$\Gamma'(x)=\Gamma(x) \psi(x)$$
But from the definition of the Digamma function to me it's like writing:
$$\Gamma'(x)=\Gamma'(x)$$
And this doesn't seem very useful to me (if I'm wrong feel free to explain me why) so I'm wondering: is there any other form for the derivative(s) of the Gamma function ? This function is defined by an integral so I think that there could be but I'm not sure on how to deal with this.
Best Answer
Another alternative form for the derivative of the gamma function would be $$\frac{\mathrm{d}^n}{\mathrm{d}x^n} \Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t} \ln^n t \, \mathrm{d}t$$ for $\Re(z) > 0$.
Yet another would be $$\Gamma'(m+1) = m!\left(-\gamma + \sum_{k=1}^{m} \frac{1}{k}\right)$$ as long as $m$ is a positive integer. Where $\gamma$ is the Euler-Masheroni constant.
We can write the gamma function as an infinite product, namely $$\Gamma(z) = e^{-Cz} \frac{1}{z}\prod_{n=1}^{\infty} \frac{e^{z/n}}{1 + z/n}$$ Since $\Gamma(z) > 0$ then taking the logarithm of it yields $$\ln \Gamma(z) = -Cz - \ln z + \sum_{n=1}^{\infty}\left(\frac{z}{n} - \ln \left(1 - \frac{z}{n}\right)\right) $$ Differentiating implicitly yields $$\frac{\Gamma'(z)}{\Gamma(z)} = -C - \frac{1}{z} + \sum_{n=1}^{\infty}\frac{z}{n(z+n)}$$