Our main strategy is to apply Laplace’s method. The basic idea is that the above exponential integrals are concentrated at the critical values $t_{0}$ and $u_{0}$.

First, we prove the uniform estimate for $\Tri(\fb,\fa,y)$. By (5.97),

(5.118)

$$\Tri(\fb,\fa,y)\leq\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{F(t)}dt.$$

Clearly, the upper bound of $\Tri(\fb,\fa,y)$ follows from the upper bound estimate of $\int_{0}^{\infty}e^{F(t)}dt$. Write

(5.119)

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

We will estimate the two terms separately.

To estimate the first term in (5.119), we make a change of variable

(5.120)

$$t=t_{0}\cdot(1+\xi),\ \xi\in(-1,1),$$

then Taylor’s theorem gives that

(5.121)		$\displaystyle F(t)-F(t_{0})$	$\displaystyle=$	$\displaystyle F(t_{0}(1+\xi))-F(t_{0})$
			$\displaystyle=$	$\displaystyle F^{\prime}(t_{0})\cdot t_{0}\cdot\xi+\frac{F^{\prime\prime}(\theta)}{2}\cdot t_{0}^{2}\cdot\xi^{2}$
			$\displaystyle=$	$\displaystyle\frac{F^{\prime\prime}(\theta)}{2}\cdot t_{0}^{2}\cdot\xi^{2},$

where $\theta$ is between $t$ and $t_{0}$. Now we need to estimate the quadratic error term. It is straightforward calculation that

(5.122)

$\displaystyle F^{\prime\prime\prime}(t)$

$\displaystyle=2(\frac{Q}{t^{3}}-\frac{Q}{(t+1)^{3}})>0,$

then $F^{\prime\prime}(t)$ is increasing in $t$. Since $\theta$ is between $t_{0}$ and $t\in[0,2t_{0}]$, the above monotonicity of $F^{\prime\prime}$ implies $F^{\prime\prime}(\theta)\leq F^{\prime\prime}(2t_{0})<0$. So the first term of (5.119) becomes

(5.123)		$\displaystyle\int_{0}^{2t_{0}}e^{F(t)}dt$	$\displaystyle=$	$\displaystyle e^{F(t_{0})}\int_{0}^{2t_{0}}e^{F(t)-F(t_{0})}dt$
			$\displaystyle\leq$	$\displaystyle e^{F(t_{0})}\cdot t_{0}\cdot\int_{-1}^{1}e^{\frac{F^{\prime\prime}(2t_{0})}{2}\cdot t_{0}^{2}\cdot\xi^{2}}d\xi$

By direct computations, $F^{\prime\prime}(2t_{0})=-\frac{(4t_{0}+1)}{4t_{0}^{2}(2t_{0}+1)^{2}}\cdot Q$. So we have,

(5.124)		$\displaystyle\int_{0}^{2t_{0}}e^{F(t)}dt$	$\displaystyle\leq$	$\displaystyle e^{F(t_{0})}\cdot t_{0}\cdot\int_{-1}^{1}e^{-\frac{4t_{0}+1}{8(2t_{0}+1)^{2}}\cdot Q\cdot\xi^{2}}d\xi$
			$\displaystyle\leq$	$\displaystyle C_{n}\cdot\frac{t_{0}(2t_{0}+1)}{\sqrt{4t_{0}+1}\cdot\sqrt{Q}}\cdot e^{F(t_{0})}$
			$\displaystyle\leq$	$\displaystyle C_{n}\cdot Q^{\frac{1}{4}}\cdot e^{F(t_{0})},$

where we used that $t_{0}\leq C_{n}\cdot Q^{1/2}$ (since $y\leq-1$ and $Q\geq 1$). Immediately, we have

(5.125)

$$\Tri(\beta,\alpha,y)\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot\frac{e^{y+F(t_{0})}}{\Gamma(\alpha-\beta)}.$$

Next, we estimate the second term in (5.119). Since we have proved $F^{\prime\prime}(t)<0$, so this implies that $F^{\prime}(t)$ is decreasing and hence $F^{\prime}(t)\leq F^{\prime}(2t_{0})$ for any $t\geq 2t_{0}$. Now Taylor’s theorem gives that

(5.126)

$$F(t)\leq F(2t_{0})+F^{\prime}(2t_{0})\cdot(t-2t_{0}),$$

which implies that

(5.127)

$$\int_{2t_{0}}^{\infty}e^{F(t)}dt\leq e^{F(2t_{0})}\int_{2t_{0}}^{\infty}e^{F^{\prime}(2t_{0})\cdot(t-2t_{0})}dt=\frac{e^{F(2t_{0})}}{-F^{\prime}(2t_{0})}.$$

One can check that $F^{\prime}(2t_{0})=\frac{y(3t_{0}+1)}{2(2t_{0}+1)}<0$ with $0<t_{0}<+\infty$. Since $F^{\prime}(t)<0$ for all $t>t_{0}$, so $F(2t_{0})\leq F(t_{0})$ and hence for $y\leq-1$ we have

(5.128)

$$\int_{2t_{0}}^{\infty}e^{F(t)}dt\leq C_{n}e^{F(t_{0})}.$$

Combining the above, we have

(5.129)

$$\int_{0}^{\infty}e^{F(t)}dt\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot e^{F(t_{0})}.$$

Therefore,

(5.130)

$\displaystyle\Tri(\beta,\alpha,y)$

$\displaystyle\leq C_{n}\cdot Q^{\frac{1}{4}}\cdot\frac{e^{y+F(t_{0})}}{\Gamma(\alpha-\beta)}.$

The lower bound estimate for $\Tri(\fb,\fa,y)$ also follows from Laplace’s method and we just sketch the computations.

	$\displaystyle\Tri(\fb,\fa,y)$	$\displaystyle=\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{F(t)}\cdot\frac{1}{(t+1)^{1+\frac{1}{n}}}dt$
		$\displaystyle\geq\frac{e^{y}}{\Gamma(\alpha-\beta)}\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}\cdot\frac{1}{(t+1)^{1+\frac{1}{n}}}dt$
(5.131)			$\displaystyle\geq\frac{e^{y}}{\Gamma(\alpha-\beta)\cdot(1+2t_{0})^{1+\frac{1}{n}}}\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}dt.$

By the concavity of $F(t)$ and the monotonicity of $F^{\prime\prime}(t)$ in the domain $t_{0}\leq t\leq 2t_{0}$, we have

(5.132)

$$\int_{t_{0}(y)}^{2t_{0}(y)}e^{F(t)}dt\geq e^{F(t_{0})}\int_{t_{0}(y)}^{2t_{0}(y)}e^{\frac{F^{\prime\prime}(t_{0})}{2}(t-t_{0})^{2}}dt\geq C_{n}\cdot e^{F(t_{0})}\frac{t_{0}(t_{0}+1)}{\sqrt{2t_{0}+1}\cdot\sqrt{Q}}$$

It is elementary to see that

(5.133)

$$C_{n}Q^{\frac{1}{2}}(-y)^{-1}\leq t_{0}\leq C_{n}\cdot{Q^{\frac{1}{2}}}$$

Therefore,

(5.134)

$$\Tri(\fb,\fa,y)\geq C_{n}\cdot Q^{-\frac{1}{4}-\frac{1}{2n}}\cdot\frac{e^{y}\cdot(-y)^{-1}}{\Gamma(\alpha-\beta)}\cdot e^{F(t_{0})}.$$

The uniform estimate for $\Ku(\fb,\fa,y)$ stated in (5.117) can be proved in the same way. One just needs to apply Laplace’s method to the integral estimate formula in Lemma 5.8. We can eventually obtain

(5.135)

$$C_{n}^{-1}\cdot Q^{-\frac{1}{4}}\cdot e^{G(u_{0})}\leq\int_{0}^{\infty}e^{G(u)}du\leq C_{n}\cdot e^{G(u_{0})}.$$

We omit the computations here.

∎