[Math] Integrating the tail exponent of a Pareto Distribution (difficult integrals).

Question

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. –

Answer 1

The first integral is

$$\frac1{\sqrt{2 \pi} \sigma x} \int_0^{\infty} d\alpha \, \alpha \, e^{\alpha \log{(L/x)}} e^{-(\alpha-\mu)^2/(2 \sigma^2)} $$

This basically has the form

$$C \int_0^{\infty} d\alpha \, \alpha \, e^{-A (\alpha+B)^2} = C \int_B^{\infty} d\alpha \, (\alpha-B) e^{-A \alpha^2}$$

which is an error function plus an exponential term.

The second integral is

$$\frac{b^{-a/b}}{x \Gamma \left ( \frac{a}{b} \right )}\int_0^{\infty} d\alpha \, \alpha^{a/b} \, e^{- \alpha (1/b + \log{(x/L)})} = \frac{a}{b} \frac{b^{-a/b}}{x \left ( \frac1{b} + \log{\left (\frac{L}{x} \right )} \right )^{a/b+1} }$$

because the integral is a Gamma function.