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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.

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  • $\begingroup$ $$\int_0^\infty ln(t)t^{z-1}e^{-t}dt.$$ Have a look at the properties of Digamma, anyway. $\endgroup$ – Yves Daoust Jul 12 '15 at 17:08
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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 $$\frac{\Gamma'(z)}{\Gamma(z)} = -C - \frac{1}{z} + \sum_{n=1}^{\infty}\frac{z}{n(z+n)}$$

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$$\psi(n+1)=H_n-\gamma$$

See harmonic numbers and the Euler-Mascheroni constant for more information.

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