There are several formats of loggamma distributions, such as in this PhD thesis https://macsphere.mcmaster.ca/bitstream/11375/6816/1/fulltext.pdf or here for online discussions
Here, we derive mean and variance of another format of loggamma distribution with the pdf:
\[f(x)=\begin{cases} \frac{1}{\Gamma(\alpha)\beta^\alpha}x^{-(1+\beta)/\beta}(\log x)^{\alpha-1}, & \text{x>1}\\ 0 & \text{elsewhere}. \end{cases}\]
The mean of the pdf can be derived from the following steps: \[\begin{align} E(X)&= \int_1^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}x^{-(1+\beta)/\beta}(\log x)^{\alpha-1}x dx\\ & = \int_1^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}(\log x)^{\alpha-1}x^{-\frac{1}{\beta}}dx\\ \end{align}\]
Next, let \(u=\log x\) \[\begin{align} E(X)&= \int_0^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}u^{\alpha-1}(e^u)^{-\frac{1}{\beta}}e^udu\\ &=\int_0^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}u^{\alpha-1}e^{-\frac{u}{\beta}}e^udu \\ &=\int_0^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}u^{\alpha-1}e^{-\frac{u}{\beta}+u}du\\ &=\int_0^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}u^{\alpha-1}e^{-\frac{1-\beta}{\beta}u}du\\ &=\int_0^\infty \frac{1\times (\frac{\beta}{1-\beta})^\alpha}{\Gamma(\alpha)\beta^\alpha \times (\frac{\beta}{1-\beta})^\alpha }u^{\alpha-1}exp({-u/\frac{\beta}{1-\beta}})du\\ &=(\frac{\beta}{1-\beta})^\alpha/\beta^\alpha\\ &=(1-\beta)^{-\alpha} \end{align}\]
We can calculate the variance of the loggamma distribution using the same method as above for \(E(X^2)\)
\[Var(X)=E\left [(X-\mu)^2\right]=E(X^2)-\left [E(X)\right ]^2\] \[\begin{align} E(X^2)&= \int_1^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}x^{-(1+\beta)/\beta}(\log x)^{\alpha-1}x^2 dx\\ & = \int_1^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}(\log x)^{\alpha-1}x^{-\frac{\beta+1}{\beta}+2}dx\\ &= \int_1^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}(\log x)^{\alpha-1}x^{1-\frac{1}{\beta}}dx\\ &=\int_0^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}u^{\alpha-1}e^{u(2-\frac{1}{\beta})}du \\ &=\int_0^\infty \frac{1}{\Gamma(\alpha)\beta^\alpha}u^{\alpha-1}e^{-\frac{1-2\beta}{\beta}u}du\\ &=(\frac{\beta}{1-2\beta})^\alpha/\beta^\alpha\\ &=(1-2\beta)^{-\alpha} \end{align}\]
Therefore, \[Var(X)=(1-2\beta)^{-\alpha}-\left[(1-\beta)^{-\alpha})\right]^2=(1-2\beta)^{-\alpha}-(1-\beta)^{-2\alpha}\]