By the definition of $\mathcal{F}$ and $\mathcal{U}$, it suffices to prove

(4.55)

$$\int_{0}^{\infty}e^{F(t)}dt\leq(1+\sqrt{\pi})e^{F(t_{0})},$$

and

(4.56)

$$\int_{0}^{\infty}e^{U(t)}dt\leq(1+\sqrt{\pi})e^{U(s_{0})}.$$

We only prove the first inequality and the second can be proved in exactly the same way. In fact, the second can be proved exactly the same way. Denote $a\equiv 2y$. For $\epsilon\in(-1,1]$ we have

	$\displaystyle F(t_{0}(1+\epsilon))-F(t_{0})$	$\displaystyle=-\epsilon(\epsilon+2)t_{0}^{2}+\epsilon at_{0}+h\log(1+\epsilon)$
		$\displaystyle=-\epsilon^{2}t_{0}^{2}+h(\log(1+\epsilon)-\epsilon)$
(4.57)			$\displaystyle\leq-\epsilon^{2}t_{0}^{2}-h(\frac{\epsilon^{2}}{2}-\frac{\epsilon^{3}}{3})\leq-\epsilon^{2}(t_{0}^{2}+\frac{h}{6}).$

The above computations imply that under the transformation $t=t_{0}(1+\epsilon)$,

	$\displaystyle\int_{0}^{2t_{0}}\exp(F(t))dt$	$\displaystyle\leq\int_{0}^{2t_{0}}\exp\Big(F(t_{0})-\epsilon^{2}(t_{0}^{2}+\frac{h}{6})\Big)dt$
		$\displaystyle=t_{0}\exp(F(t_{0}))\int_{-1}^{1}\exp\Big(-(t_{0}^{2}+\frac{h}{6})\epsilon^{2}\Big)d\epsilon$
		$\displaystyle=\frac{t_{0}}{\sqrt{t_{0}^{2}+\frac{h}{6}}}\exp(F(t_{0}))\int_{-\sqrt{t_{0}^{2}+\frac{h}{6}}}^{\sqrt{t_{0}^{2}+\frac{h}{6}}}\exp(-\tau^{2})d\tau$
(4.58)			$\displaystyle\leq\sqrt{\pi}\exp(F(t_{0})).$

In addition, let $t>2t_{0}$, then

(4.59)

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

and hence

(4.60)

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

It can be directly computed that

(4.61)

$$F^{\prime}(2t_{0})=-4t_{0}+a+\frac{h}{2t_{0}}=-\frac{4t_{0}^{2}+h}{2t_{0}}<0,$$

then

(4.62)

$\displaystyle\int_{2t_{0}}^{\infty}\exp(F(t))dt\leq\frac{2t_{0}}{4t_{0}^{2}+h}\exp(F(2t_{0}))\leq\frac{2t_{0}}{4t_{0}^{2}+h}\cdot\frac{\exp(F(t_{0}))}{\exp(t_{0}^{2}+\frac{h}{6})}\leq\exp(F(t_{0})).$

Combining the above calculations,

(4.63)

$$\int_{0}^{\infty}\exp(F(t))dt\leq(1+\sqrt{\pi})\exp(F(t_{0})).$$

∎