<s>[INST] Integrating the tail exponent of a Pareto Distribution (difficult integrals). Looking for two difficult integrals, whichever is solved first:
With $\alpha>0,L>0, \mu>0,\sigma>0, a>0, b>0, x>L$ ,
$$\phi_l (x;\mu,\sigma)=\frac{1}{\sqrt{2 \pi } \sigma }\int_0^{\infty } \alpha  L^{\alpha } x^{-\alpha -1} e^{-\frac{(\alpha -\mu )^2}{2 \sigma ^2}} \, \mathrm{d}\alpha$$
and
$$\phi_g (x;a,b)=\frac{b^{-\frac{a}{b}} }{\Gamma \left(\frac{a}{b}\right)}\int_0^{\infty }  e^{-\frac{\alpha }{b}} L^{\alpha } x^{-\alpha -1} \alpha ^{a/b}\, \mathrm{d}\alpha$$
Background: these correspond to the density of a Pareto distribution $\alpha  L^{\alpha } x^{-\alpha -1} $ with its tail exponent $\alpha$: 1) Lognormally distributed in the first case, and  2) following a Gamma distribution in the second case.
With gratitude.
$\textbf{Later Comment}$: It turned out that the first integral was not the Lognormally distributed exponent (I copied the wrong equation), but I leave here for its calculus/integration interest, rather than distributional importance. –
 [/INST] Let's start with $\phi_l$:
\begin{align}
\phi_l(x;\mu,\sigma) &= \frac{L}{\sqrt{2\pi \sigma^2}}\int_0^\infty \alpha \left(\frac{L}{x}\right)^{\alpha - 1} e^{-\tfrac{(\alpha - \mu)^2}{2\sigma^2}}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}} \frac{d}{dy}\bigg|_{y = L/x} \int_0^\infty y^\alpha e^{-\tfrac{(\alpha - \mu)^2}{2\sigma^2}}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}} \frac{d}{dy}\bigg|_{y = L/x} \int_0^\infty e^{-\tfrac{(\alpha - \mu)^2}{2\sigma^2} + \alpha \ln y}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}}\frac{d}{dy}\bigg|_{y = L/x} \int_0^\infty e^{-\tfrac{(\alpha - \mu - \sigma^2\ln y)^2 + \mu^2 - (\mu + \sigma^2 \ln y)^2}{2\sigma^2}}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}} \frac{d}{dy}\bigg|_{y = L/x} e^{-\tfrac{\mu^2 - (\mu + \sigma^2 \ln y)^2}{2\sigma^2}} \int_0^\infty e^{-\tfrac{(\alpha - \mu - \sigma^2\ln y)^2}{2\sigma^2}}\, d\alpha\\
&= \frac{L}{\sqrt{2\pi \sigma^2}} \frac{d}{dy}\bigg|_{y = L/x} e^{\mu\ln y\, + \frac{\sigma^2}{2}\ln^2 y} \sqrt{2\pi \sigma^2}\operatorname{erfc}\left(-\frac{\mu + \sigma^2\ln y}{\sqrt{2\sigma^2}}\right)\\
&= x\exp\left\{-\mu\ln \frac{x}{L} + \frac{\sigma^2}{2}\ln^2 \frac{x}{L}\right\}\left(\mu - \sigma^2 \ln \frac{x}{L}\right) \operatorname{erfc}\left(-\frac{\mu + \sigma^2 \ln y}{\sqrt{2\sigma^2}}\right)\\
&\quad + \frac{\sigma x}{\sqrt{2}}\exp\left\{-\mu\ln \frac{x}{L} + \frac{\sigma^2}{2}\ln^2 \frac{x}{L}-\frac{(\mu - \sigma^2 \ln \frac{x}{L})^2}{2\sigma^2}\right\}.
\end{align}
Now with $\phi_g$:
\begin{align}
\phi_g(x;a,b) &= \frac{b^{-\tfrac{a}{b}}}{x\Gamma\left(\frac{a}{b}\right)} \int_0^\infty \alpha^{\tfrac{a}{b}} e^{-\left(\frac{1}{b} + \ln \frac{x}{L}\right)\alpha}\, d\alpha\\
&= \frac{b^{-\frac{a}{b}}}{x\Gamma\left(\frac{a}{b}\right)} \left(\frac{1}{b} + \ln \frac{x}{L}\right)^{-\frac{a}{b} - 1}\int_0^\infty u^{\frac{a}{b}} e^{-u}\, du, \quad u = \left(\frac{1}{b} + \ln \frac{x}{L}\right)\alpha\\
&= \frac{b^{-\frac{a}{b}}}{x\Gamma\left(\frac{a}{b}\right)} \left(\frac{1}{b} + \ln \frac{x}{L}\right)^{-\frac{a}{b} - 1} \Gamma\left(\frac{a}{b} + 1\right)\\
&= \frac{b^{-\frac{a}{b}}}{x\Gamma\left(\frac{a}{b}\right)} \left(\frac{1}{b} + \ln \frac{x}{L}\right)^{-\frac{a}{b} - 1} \frac{a}{b}\Gamma\left(\frac{a}{b}\right)\\
&= \frac{a}{x}b^{-\frac{a}{b} - 1} \left(\frac{1}{b} + \ln \frac{x}{L}\right)^{-\frac{a}{b} - 1}\\
&= \frac{a}{x}\left(1 + b\ln \frac{x}{L}\right)^{-\frac{a}{b} - 1}.
\end{align}
</s>