Finding Cramer Rao Lower bound for a bivariate parameter

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I am having a hard time figuring out the Cramer Rao lower bound for a random sample of size $n$ from a population with $\Gamma(p, \theta)$ with $p, \theta$ unknown. The problem doesn't say what formulation to use for the gamma function, but I have used

$$ \frac{\theta^p}{\Gamma(p)} x^{p-1} e^{-\theta x} I_{[0,\infty)}(x). $$

So I know that I have to find the Fisher Information matrix and compute the inverse, however the entries I get are not "nice". The derivatives of the logarithm of the pdf is (considering only $x_1$)

$$ \frac{ \partial }{ \partial p} \ln f(x_1|p,\theta) = \ln \theta - \frac{\partial}{\partial p} (\ln \Gamma (p)) + \ln x_1, \\ \frac{\partial}{\partial \theta} \ln f(x_1|p,\theta) = \frac{p}{\theta}-x_1. $$

Now, the entries in the Fisher Information matrix would be, with $\beta = [p, \theta]^\mathrm{T}$,

$$ I_{ij} = n \mathrm{E} \left[ \left( \frac{ \partial}{\partial \beta_i} \ln f(X| p, \theta) \right) \left( \frac{ \partial}{\partial \beta_j} \ln f(X| p, \theta) \right) \right]. $$

However, with the derivatives obtained above the entries become horrible... Apart from $I_{22} = np/\theta^2 $, that is. Am I on the right track here?

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The alternative form of Fisher Information is nearly always easier to use. (https://en.wikipedia.org/wiki/Fisher_information#Matrix_form)

$$ I_{ij} = - n \mathrm E \left[ \frac{\partial^2}{\partial \beta_i \beta_j} \ln f( X \mid p,\theta) \right] $$ In this case (ignoring indicators function), $\ln f( X \mid p, \theta) = p \ln \theta - \ln(\Gamma(p)) + (p-1) \ln X - \theta X $ so that $\frac{\partial}{\partial \theta}\ln f( X \mid p, \theta) = p/\theta - X$ and $\frac{\partial^2}{\partial \theta^2} \ln f( X \mid p, \theta)= - p/\theta^2$ and $\frac{\partial^2}{\partial p \partial \theta} \ln f( X \mid p, \theta)= 1/\theta$.

The difficult one is $\frac{ \partial^2}{\partial p^2} \ln f( X \mid p, \theta) = \dots ?$ I don't know how to differentiate the gamma function to a closed form. But the other ones are easy to run an expectation through.

Hopefully this helps.