Looking for the integral:
With $b>0, L>0, \alpha_0>b,\sigma>0, x>L$ ,
$$\phi(x;\alpha_0,\sigma_)=\int_b^\infty\alpha L^{\alpha } x^{-\alpha -1} \frac{e^{-\frac{\left(\log (\alpha -b)-\log (\alpha_0-b)+\frac{\sigma ^2}{2}\right)^2}{2 \sigma ^2}}}{\sqrt{2 \pi } \sigma (\alpha -b)}d\alpha$$
$\textbf{Background}$: This is the density of a Pareto distribution $\alpha L^{\alpha } x^{-\alpha -1} $ with its tail exponent $\alpha$ distributed according to a shifted lognormal with mean $\alpha_0$ and lower bound $b$. In other words $\alpha + b$ follows a $\text{LogNormal}\left(\log (\alpha_0-b)-\frac{\sigma ^2}{2},\sigma \right)$.
Best Answer
The integral doesn't appear to admit a closed form solution. But, with b=1 (which is the lower bound for b),we get:
$$\phi(x;\alpha_0,\sigma)=\frac{1}{x^2 }\sum _{i=0}^\infty \frac{1}{i!}L (\alpha_0-1)^i e^{\frac{1}{2} i (i-1) \sigma ^2} \frac{(i+\log (\frac{L}{x})}{\log (\frac{L}{x})^{1-i} }$$
This result is obtained by expanding $\alpha$ around its lower bound $b$ (which we simplified to $b=1$) and integrating each summand.
We know that as $x \to \infty$ the density $\phi$ collapses to power laws with a tail $\alpha=1$, i.e., where K is a norming constant, $\phi(x)= K x^{-2}$.
$\textbf{Expectation: }$ Even in the absence of a fully explicit density, we can extract the explicit expectation as follows. By integrating first with respect to $x$: $$ \int_b^\infty \int_L^\infty x \phi(x;\alpha)\,\mathrm{d}x \mathrm{d}\alpha= \int_L^\infty \frac{\alpha L \exp \left(-\frac{\left(2 \log (\alpha -b)-2 \log (\text{$\alpha $0}-b)+\sigma ^2\right)^2}{8 \sigma ^2}\right)}{\sqrt{2 \pi } (\alpha -1) \sigma (\alpha -b)}\mathrm{d}\alpha$$ With, again, b=1: $$ \mathbb{E}(X)= \frac{L \left(\alpha_0+e^{\sigma ^2}-1\right)}{\alpha_0-1}$$