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When i read an article about Generalized exponential distribution, i came across an expectation and i couldn't figure out how the authors obtain that result. The pdf is \begin{equation} f\left(x;\alpha,\beta,\mu\right)=\alpha\beta\left(1-e^{-\left(x-\mu\right)\beta}\right)^{\alpha-1}e^{-\left(x-\mu\right)\beta},\ x>\mu,\ \alpha>0,\ \beta>0 \end{equation} and log-likelihood for the sample $X_{1},X_{2},\ldots,X_{n}$ is \begin{equation} n\log\alpha+n\log\beta-\sum_{i=1}^{n}\beta\left(x_{i}-\mu\right)+\left(\alpha-1\right)\sum_{i=1}^{n}\log\left(1-e^{-\left(x_{i}-\mu\right)\beta}\right) \end{equation} and after finding loglikelihood, one of the expectation (for Fisher Information) is \begin{eqnarray*} E\left(\frac{\partial^{2}L}{\partial\alpha\partial\beta}\right) & = & nE\left(\frac{\left({\color{red}X}-\mu\right)e^{-\left(X-\mu\right)\beta}}{1-e^{-\left(X-\mu\right)\beta}}\right)\\ & = & {\color{blue}{n\left[\frac{\alpha}{\beta\left(\alpha-1\right)}\left(\psi\left(\alpha\right)-\psi\left(1\right)\right)-\frac{1}{\beta}\left(\psi\left(\alpha+1\right)-\psi\left(1\right)\right)\right]}} \end{eqnarray*} ($\psi$ is digamma function)

Can somebody explain me, how the author obtain the last equality?

I can find the expectation without the term ${\color{red}X}$. But how can i handle this $X$ term?

When i applied the transformation $t=e^{-\left(x-\mu\right)\beta}$ the integral for expectation became $$ E\left(\frac{\left(X-\mu\right)e^{-\left(x-\mu\right)\beta}}{1-e^{-\left(x-\mu\right)\beta}}\right)=-\frac{\alpha}{\beta}\int_{0}^{1}t\left(1-t\right)^{\alpha-2}\log t\ dt. $$ how can i go the next step?

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