I am working with the log of negative binomial distribution $NegBin(r,p)$. I need to differentiate the following with respect to $r$ such that at the end, I am NOT left with $r$ in factorial form.
$$\frac{d \;log( \Gamma(x + r) )}{dr} = \;\; ?$$
I read about digamma function, but I have $x+r$ within the bracket in my case. Can anyone show me how this can be done ?
$\endgroup$1 Answer
$\begingroup$As mentioned in this answer, $$ \frac{\mathrm{d}}{\mathrm{d}x}\log(\Gamma(x))=\frac{\Gamma'(x)}{\Gamma(x)}=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1}\right) $$ where $\gamma$ is the Euler-Mascheroni Constant.
It is also mentioned there, that when $x$ is a positive integer, $$ \sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1}\right)=\sum_{k=1}^{x-1}\frac1k=H_{x-1} $$ where $H_n$ is the $n^\text{th}$ Harmonic Number.
Since you ask about a slightly different form, we simply apply the chain rule: $$ \frac{\mathrm{d}}{\mathrm{d}r}\log(\Gamma(x+r))=-\gamma+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x+r-1}\right) $$
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