How to derive the expected value of $X^\alpha\log X$ Let $X$ follow a Weibull distribution, with density $$f(x)=\frac{\alpha}{\theta}x^{\alpha-1}e^{-\frac{x^{\alpha}}{\theta}}\quad  x>0 .$$
How can I find the following expectation?
$$E[X^{\alpha}\log X]$$
The answer given in the paper by Debasis Kundu "Estimation of $P(Y\le X)$ for Weibull distribution" (page 9) is given below. Paper link: http://home.iitk.ac.in/~kundu/paper112.pdf
$$\frac{1}{\alpha}\theta[\ln(\theta)+\Gamma(2)]$$
Any suggestions? Many thanks.
 A: Take the moment generating function of the Weibull distribution,
$\mathbb{E}[X^\alpha]$, and differentiate it once with respect to $\alpha$.
A: I get the following result:
$$E(X^{\alpha} \log{X}) = \frac{\theta}{\alpha} (\Gamma'(2) + \log{\theta})$$
get this from
$$\begin{align}E(X^{\alpha} \log{X}) &= \frac{\alpha}{\theta}\underbrace{\int_0^{\infty} dx \, \log{x} \, x^{2 \alpha-1} \, e^{-x^{\alpha}/\theta}}_{y=x^{\alpha}}\\&= \frac{1}{\alpha \theta} \underbrace{\int_0^{\infty} dy \, y\, \log{y} \, e^{-y/\theta}}_{\text{integrate by parts}}\\ &= \frac{1}{\alpha} \int_0^{\infty} dy \, e^{-y/\theta} (1+\log{y})\\&=\frac{\theta}{\alpha} + \frac{1}{\alpha} \underbrace{\int_0^{\infty} dy \log{y} \, e^{-y/\theta}}_{y=u \theta}\\&= \frac{\theta}{\alpha}+\frac{\theta}{\alpha} \log{\theta} + \frac{\theta}{\alpha} \underbrace{\int_0^{\infty} du \, \log{u} \, e^{-u}}_{\text{this is equal to } -\gamma} \\ &= \frac{\theta}{\alpha} (1-\gamma + \log{\theta})\\ &= \frac{\theta}{\alpha} (\Gamma'(2) + \log{\theta}) \end{align}$$
where $\gamma \approx 0.5772$ is the Euler-Mascheoni constant.
