For simplicity, we will calculate the asymptotic behavior in $y$. For fixed $h$ and $j$, as $y\rightarrow\infty$, it is straightforward that

(4.66)

$$\begin{split}t_{0}&=y+\frac{h}{2y}+O(y^{-2})\\ s_{0}&=\frac{h}{2y}+O(y^{-2})\end{split}$$

which implies that

(4.67)

$$\begin{split}F(t_{0})&=y^{2}+h\log y+O(y^{-1})\\ U(s_{0})&=-h+h\log h-h\log(2y)+O(y^{-1}).\end{split}$$

First, we prove the asymptotics for $\mathcal{F}$. As in the proof of Lemma 4.6, we get

	$\displaystyle\int_{0}^{\infty}e^{F(t)}dt$	$\displaystyle=e^{F(t_{0})}\int_{0}^{2t_{0}}e^{F(t)-F(t_{0})}dt+\int_{2t_{0}}^{\infty}e^{F(t)}dt$
(4.68)			$\displaystyle=t_{0}e^{F(t_{0})}\int_{-1}^{1}e^{-t_{0}^{2}\epsilon^{2}+h(\log(1+\epsilon)-\epsilon)}d\epsilon+\int_{2t_{0}}^{\infty}e^{F(t)}dt.$

Notice that

(4.69)

$$\lim_{t_{0}\rightarrow\infty}t_{0}\int_{-1}^{1}e^{-t_{0}^{2}\epsilon^{2}+h(\log(1+\epsilon)-\epsilon)}d\epsilon=\sqrt{\pi},$$

and

(4.70)

$$\lim_{t_{0}\rightarrow\infty}e^{-F(t_{0})}\int_{2t_{0}}^{\infty}e^{F(t)}dt=0.$$

Moreover, by (4.66), $\lim\limits_{y\to+\infty}t_{0}(y)\to\infty$. It follows that

(4.71)

$$\lim\limits_{z\to\infty}\frac{\mathcal{F}(z)}{\sqrt{\pi}e^{-\frac{jz^{2}}{2}+F(t_{0}(z))}}=1.$$

Combining the above limit and (4.67), the proof of (4.64) is complete.

In the case $j\in\mathbb{Z}_{+}$ and $h>0$, we will prove the asymptotic behavior of $\mathcal{U}$ and we write

	$\displaystyle\int_{0}^{\infty}e^{U(t)}dt$	$\displaystyle=s_{0}e^{U(s_{0})}\int_{-1}^{\infty}e^{-s_{0}^{2}\epsilon^{2}+h(\log(1+\epsilon)-\epsilon)}d\epsilon$
(4.72)			$\displaystyle=s_{0}e^{U(s_{0})}\int_{-1}^{\infty}e^{-s_{0}^{2}\epsilon^{2}}\cdot(1+\epsilon)^{h}e^{-h\epsilon}d\epsilon.$

We claim that

(4.73)

$$\lim\limits_{s_{0}\to 0}\int_{-1}^{\infty}(e^{-s_{0}^{2}\epsilon^{2}}-1)\cdot(1+\epsilon)^{h}e^{-h\epsilon}d\epsilon=0.$$

In fact, it is straightforward that for any $s_{0}>0$,

(4.74)

$$-1\leq e^{-s_{0}^{2}\epsilon^{2}}-1\leq 0$$

and for any fixed $h>0$,

(4.75)

$$\int_{-1}^{\infty}(1+\epsilon)^{h}e^{-h\epsilon}d\epsilon<\infty.$$

Applying the dominated convergence theorem,

(4.76)

$$\lim\limits_{s_{0}\to 0}\int_{-1}^{\infty}(e^{-s_{0}^{2}\epsilon^{2}}-1)\cdot(1+\epsilon)^{h}e^{-h\epsilon}d\epsilon=0.$$

This completes the proof the the claim. Next, by the definition of the gamma function,

(4.77)

$$\int_{-1}^{\infty}(1+\epsilon)^{h}e^{-h\epsilon}d\epsilon=e^{h}\int_{0}^{\infty}e^{-hs}s^{h}ds=e^{h}h^{-h-1}\Gamma(h+1).$$

Therefore,

(4.78)

$$\lim\limits_{s_{0}\to 0}\int_{-1}^{\infty}e^{-s_{0}^{2}\epsilon^{2}+h(\log(1+\epsilon)-\epsilon)}d\epsilon=e^{h}h^{-h-1}\Gamma(h+1).$$

Since $\lim\limits_{y\to+\infty}s_{0}=0$ and $U$ yields to the asymptotic property (4.67), eventually we obtain (4.65). ∎