Consider the following equation with respect to $\alpha$ defined through a probability density function $f(x) = \exp[x-\exp(x)]$.
$$ \int_{-\infty}^\infty (x-2) e^{\alpha(x-2)}e^{x-e^x}dx = 0. $$
How to solve this equation, please? Moreover, can we find a closed form for the integral in the first place? If so, then it will be easier to solve for $\alpha$ I suppose. Thank you!
Hint. By the change of variable $u=e^x$ you get $$ \begin{align} \int_{-\infty}^{+\infty} (x-2) e^{\alpha(x-2)}e^{x-e^x}dx &= e^{-2\alpha}\int_{0}^{+\infty} (\ln u-2) u^{\alpha}e^{-u}du\\\\ &=e^{-2\alpha}\left(\int_{0}^{+\infty} \ln u \:u^{\alpha}e^{-u}du-2\int_{0}^{+\infty} u^{\alpha}e^{-u}du\right)\\\\ &=e^{-2\alpha}(\Gamma'(\alpha+1)-2\:\Gamma(\alpha+1)). \end{align} $$