Proposition 15.36.8.704167C1A830 The first moment
\[ M_\alpha =\int _ Yu_\alpha q_\alpha \kappa ^ k \]
is finite exactly when $\alpha <k/(k+1)$. In this range,
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15.36.9
\begin{equation} -M_\alpha =\alpha \int _1^\infty x^{\alpha -1} \left[\alpha x^{\alpha -1}+(1-\alpha x^{\alpha -1})\mathrm e^{1-x}\right]^ k\, \mathrm dx. \end{equation}
For the joint limits of theorem 15.36.5, if $\alpha <k/(k+1)$, then
\[ a_{\lambda _ j,\alpha _ j}\to -M_\alpha ,\qquad U_{\lambda _ j,\alpha _ j}-\lambda _ j^{-1}\log b \to u_\alpha -M_\alpha \]
in $L^1$, in capacity, and locally uniformly away from the pole fiber. If $\alpha \geq k/(k+1)$, then $a_{\lambda _ j,\alpha _ j}\to +\infty $ and the latter translated potentials tend to $+\infty $ locally uniformly off the pole fiber, although $\lambda _ ja_{\lambda _ j,\alpha _ j}\to 0$. In both cases the exact scalar error is
\[ \lambda _ j^{-1}\log (c_{\lambda _ j,\alpha _ j}/b) =a_{\lambda _ j,\alpha _ j}. \]
Proof.
As $t=-s\to \infty $, the radial profile of lemma 15.30.1 satisfies
\[ u_\alpha \sim -t^\alpha ,\qquad Q_\alpha '(s)\sim k(1-\alpha )\alpha ^ k t^{-1-k(1-\alpha )}. \]
Thus the first moment is finite precisely when $(k+1)\alpha <k$. In that case, write $x=1-\psi $. Its cumulative flux is
\[ Q_\alpha =\left[\alpha x^{\alpha -1} +(1-\alpha x^{\alpha -1})\mathrm e^{1-x}\right]^ k. \]
Integration by parts gives (15.36.9); the pole boundary term vanishes since $(k+1)\alpha -k<0$.
Suppose $\alpha <k/(k+1)$ and choose the parameter interval in the preceding proof with $\alpha _+<k/(k+1)$. Its dominating density $g$ satisfies $(-u_{\alpha _+})g\in L^1$. Write $w_ j=w_{\lambda _ j,\alpha _ j}=v_ j+a_ j$. Since $w_ j\geq u_{\alpha _ j}\geq u_{\alpha _+}$, the mass identity and $\mathrm e^ r\geq 1+r$ give $\int _ Yw_ jq_{\alpha _ j}\kappa ^ k\leq 0$. On a fixed annulus $A$ away from $o$, the normalized smooth-volume $L^1$ bound gives
\[ \int _ Aw_ jq_{\alpha _ j}\kappa ^ k \geq a_ j\int _ Aq_{\alpha _ j}\kappa ^ k-C. \]
On the complement, use $w_ j\geq u_{\alpha _+}$ and the integrable lower bound. The annulus has uniformly positive measure, so $0\leq a_ j\leq C_*$. Along any subsequence with $a_ j\to a$, local uniform convergence gives $w_ j\to u_\alpha +a$ off $o$. In the exact identity
\[ \int _ Y\frac{\mathrm e^{\lambda _ jw_ j}-1}{\lambda _ j} q_{\alpha _ j}\kappa ^ k=0, \]
the negative part is dominated by $(-u_{\alpha _+})g$ and the positive part by $C_*\mathrm e^{C_*}g$. Dominated convergence gives $M_\alpha +a=0$. Every subsequence therefore has the asserted scalar limit. Adding these convergent constants to the normalized potentials gives all three claimed topologies.
If $\alpha \geq k/(k+1)$, use instead
\[ \int _ Y\frac{1-\mathrm e^{\lambda _ jv_ j}}{\lambda _ j} q_{\alpha _ j}\kappa ^ k =\frac{1-\mathrm e^{-\lambda _ ja_ j}}{\lambda _ j}\leq a_ j. \]
The integrands are nonnegative and tend off $o$ to $(-u_\alpha )q_\alpha $. Fatou's lemma forces $a_ j\to +\infty $. The preceding theorem gives $\lambda _ ja_ j\to 0$; local uniform convergence of $v_ j$ gives the stated divergence of $w_ j$. This includes sequences approaching the moment boundary from below.
$\square$
Pome
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