Expectation about Generalized exponential distribution I would like to ask this question here too.
The probability density function 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}
I need to show that
\begin{eqnarray*}
E\left(\frac{\left(X-\mu\right)e^{-\left(X-\mu\right)\beta}}{1-e^{-\left(X-\mu\right)\beta}}\right) & = & \int_{\mu}^{\infty}\frac{\left(x-\mu\right)e^{-\left(x-\mu\right)\beta}}{1-e^{-\left(x-\mu\right)\beta}}\cdot f\left(x;\alpha,\beta,\mu\right)\enspace dx\\
 & = & {\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]}\tag{1}
\end{eqnarray*}
($\psi$ is digamma function)
When i applied the transformation $t=e^{-\left(x-\mu\right)\beta}$ the integral 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 to the next step?
 A: I verified you substituion and got the same answer so we will pick up where you left off.  We have
$$
I=-\frac{\alpha}{\beta}\int_0^1(\log t) t (1-t)^{\alpha-2}\,\mathrm dt.
$$
Using $\partial_x t^x=t^x\log t$ gives
$$
I(x)=-\frac{\alpha}{\beta}\partial_x\int_0^1t^x (1-t)^{\alpha-2}\,\mathrm dt,
$$
where $I=I(1)$. Notive that $I(x)$ is very closely related to the integral definition of the beta function leaving us with
$$
I(x)=-\frac{\alpha}{\beta}\partial_x\frac{\Gamma(\alpha-1)\Gamma(x+1)}{\Gamma(\alpha +x)}.
$$
Since $\partial_z\log\Gamma(z):=\psi(z)=\Gamma^\prime(z)/\Gamma(z)\implies\partial_z\Gamma(z)=\psi(z)\Gamma(z)$ we are able to evaluate the derivative yielding
$$
I(x)=-\frac{\alpha\Gamma(\alpha-1)}{\beta}\frac{\Gamma(x)}{\Gamma(\alpha+x)}%
\left(1+x\psi(x)-x\psi(a+x)\right).
$$
Substituting $x=1$ and using the properties $z\Gamma(z)=\Gamma(z+1)$, $\psi(1)=-\gamma$ (Euler-Mascheroni constant), and $H_z=\psi(z+1)-\gamma$ (generalized harmonic numbers) gives after some simplification
$$
I=\frac{H_\alpha-1}{(\alpha-1)\beta}.
$$
This is clearly different from $(1)$ given in your post. If we use the abovementioned properties we can simplify $(1)$ to
$$
\frac{\psi(\alpha)+\frac{1}{\alpha}+\gamma-1}{(\alpha-1) \beta}.
$$
Then using the relationship $\psi(z+1)=\psi(z)+\frac{1}{z}$ and the definition of the generalized harmonic numbers shows that the solution for $I$ agrees with the form $(1)$ given in your post.
