I would like some help with calculating the Fisher Information $I_o(\beta)$ and the expected information for a gamma distribution defined by
\begin{align*} f_X(x) = \frac{\beta^\alpha x^{\alpha - 1}e^{-\beta x}}{\Gamma(\alpha)} \; x > 0, \alpha >0, \beta > 0 \end{align*}
Where $\alpha$ is a known value and $\beta$ is the parameter of interest.
Attempt
I have attempted to calculate a likelihood function as follows:
\begin{align} L(\beta | X_i) &= \prod_{i = 1}^{n}f(x_i | \alpha, \beta) \\ &= \prod_{i = 1}^{n}\left( \frac{\beta^{\alpha}}{\Gamma(\alpha)}x_{i}^{\alpha - 1}\mathrm{exp}{\{-\beta x\}}\right) \\ &= \left(\frac{\beta^{\alpha}}{\Gamma(\alpha)}\right)^{n}\prod_{i = 1}^{n} \left(x_{i}^{\alpha-1}\right) \mathrm{exp}\{-\beta\sum_{i = 1}^{n} x_i\} \\ L(\beta | X_i) &= \left(\frac{\beta^{\alpha}}{\Gamma(\alpha)}\right)^{n}\left(\prod_{i = 1}^{n} x_{i}\right)^{\alpha-1} \mathrm{exp}\{-\beta\sum_{i = 1}^{n} x_i\} \end{align}
Thus the log-likelihood would be the following:
\begin{align} \ell(\beta | x_i) &= \ln\left((L(\beta | X_i)\right)) \\ &= n\alpha\ln(\beta) - n\ln(\Gamma(\alpha)) + (na-n)\ln(x_i) - \beta \sum_{i = 1}^{n} x_i \\ \end{align}
I understand that the information is found by taking the 2nd derivative of any of the likelihood functions where $I_o(\beta) = -\frac{\mathrm{d}^{2}{\ell}}{\mathrm{d}{\beta}^{2}} $
The derivatives calculated were as follows: \begin{align} &= n\alpha\ln(\beta) - n\ln(\Gamma(\alpha)) + (na-n)\ln(x_i) - \beta \sum_{i = 1}^{n} x_i \\ &= \frac{n\alpha}{\beta} - \beta \\ &= - \frac{n \alpha}{\beta^2} - 1 \end{align}
Thus the information would be
\begin{align} I_0(\beta) = \frac{n \alpha}{\beta^2} + 1 \end{align}
I am unsure what to do once I get to the expectation.
\begin{align} \mathbb{E}\{I_0(\beta)\} &= \mathbb{E}\left(\frac{n \alpha}{\beta^2} + 1\right) \end{align}
Would I have made a mistake within the derivation process? Any insight would be very much appreciated.