In the proof the constant $C(\nu)$ may vary from line to line. First we prove Item (1). To start with, we prove the upper bound estimate for the solution $K_{\nu}(y)$. Notice that $\cosh(t)\geq 1+\frac{t^{2}}{2}$ for every $t\geq 0$, then

(5.50)		$\displaystyle K_{\nu}(y)$	$\displaystyle=$	$\displaystyle\int_{0}^{\infty}e^{-y\cosh t}\cosh(\nu t)dt$
			$\displaystyle\leq$	$\displaystyle\int_{0}^{\infty}e^{-y(1+\frac{t^{2}}{2})}\cosh(\nu t)dt$
			$\displaystyle=$	$\displaystyle\frac{e^{-y}}{2}\Big(\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}+\nu t}dt+\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}-\nu t}dt\Big).$

Now we prove that, for $y\geq 1$ and $\nu\in\mathbb{R}$,

(5.51)

$$\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}+\nu t}dt\leq C(\nu)\cdot\frac{1}{\sqrt{y}}.$$

It is by straightforward computation that

(5.52)		$\displaystyle\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}+\nu t}dt$	$\displaystyle=$	$\displaystyle\int_{0}^{\infty}e^{-(\sqrt{\frac{y}{2}}t-\frac{\nu}{2}\sqrt{\frac{2}{y}})^{2}+\frac{\nu^{2}}{2y}}dt$
			$\displaystyle=$	$\displaystyle\sqrt{\frac{2}{y}}\cdot e^{\frac{\nu^{2}}{2y}}\int_{-\frac{\nu}{2}\sqrt{\frac{2}{y}}}^{\infty}e^{-\tau^{2}}d\tau,$

where $\tau=\sqrt{\frac{y}{2}}t-\frac{\nu}{2}\sqrt{\frac{2}{y}}$. Notice that

(5.53)

$$\int_{-\frac{\nu}{2}\sqrt{\frac{2}{y}}}^{\infty}e^{-\tau^{2}}d\tau\leq\int_{-\infty}^{\infty}e^{-\tau^{2}}d\tau=\sqrt{\pi}.$$

Moreover, the assumption $y\geq 1$ implies $e^{\frac{\nu^{2}}{2y}}\leq e^{\frac{\nu^{2}}{2}}$, so it holds that

(5.54)

$$\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}+\nu t}dt\leq C(\nu)\cdot\frac{1}{\sqrt{y}}.$$

Similarly,

(5.55)

$$\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}-\nu t}dt\leq C(\nu)\cdot\frac{1}{\sqrt{y}}.$$

Therefore, we have

(5.56)

$$K_{\nu}(y)\leq C(\nu)\cdot\frac{e^{-y}}{\sqrt{y}},$$

where $C(\nu)>0$ depends only on $\nu$.

Next we prove the lower bound estimate for $K_{\nu}(y)$. The integral representation of $K_{\nu}(y)$ can be written as follows,

(5.57)

$\displaystyle K_{\nu}(y)$

$\displaystyle=\frac{e^{-y}}{2}\Big(\int_{0}^{\infty}e^{-y(\cosh t-1)+\nu t}dt+\int_{0}^{\infty}e^{-y(\cosh t-1)-\nu t}dt\Big).$

We will give lower bound estimates for the above two integrals respectively. It is straightforward that

(5.58)

$$\int_{0}^{\infty}e^{-y(\cosh t-1)+\nu t}dt\geq\int_{0}^{1}e^{-y(\cosh t-1)+\nu t}dt=\int_{0}^{1}e^{-\frac{y\cdot\cosh(\theta_{t})t^{2}}{2}+\nu t}dt$$

for some $0\leq\theta_{t}\leq 1$, which implies that

(5.59)

$$\int_{0}^{\infty}e^{-y(\cosh t-1)+\nu t}dt\geq\int_{0}^{1}e^{-2yt^{2}+\nu t}dt.$$

The calculations in the last step imply that for $y\geq 1$,

(5.60)

$$\frac{C^{-1}(\nu)}{\sqrt{y}}\leq\int_{0}^{1}e^{-2yt^{2}+\nu t}\leq\frac{C(\nu)}{\sqrt{y}}.$$

Therefore,

(5.61)

$$\int_{0}^{1}e^{-y(\cosh t-1)+\nu t}dt\geq\frac{C^{-1}(\nu)}{\sqrt{y}}.$$

By the same calculations,

(5.62)

$$\int_{0}^{1}e^{-y(\cosh t-1)-\nu t}dt\geq\frac{C^{-1}(\nu)}{\sqrt{y}}.$$

This completes the proof of (5.47).

To see (5.48) we first assume $y\geq 1$. We use the integral representation

(5.63)

$$I_{\nu}(y)=\frac{1}{\pi}\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta-\frac{\sin(\nu\pi)}{\pi}\int_{0}^{\infty}e^{-y\cosh t-\nu t}dt.$$

To estimate the second term, we use the integral estimate

(5.64)

$$\int_{0}^{\infty}e^{-y\cosh t-\nu t}dt\leq e^{-y}\int_{0}^{\infty}e^{-\frac{yt^{2}}{2}-\nu t}dt\leq C(\nu)\cdot\frac{e^{-y}}{\sqrt{y}}.$$

Next, we estimate the first term of $I_{\nu}(y)$. Since for every $\theta\in[0,\frac{\pi}{3}]$,

(5.65)

$$\cos\theta\leq 1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{24}\leq 1-\frac{\theta^{2}}{4},$$

then

(5.66)

$\displaystyle\Big|\frac{1}{\pi}\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta\Big|$

$\displaystyle\leq$

$\displaystyle\frac{1}{\pi}\int_{0}^{\frac{\pi}{3}}e^{y\cos\theta}d\theta+\frac{1}{\pi}\int_{\frac{\pi}{3}}^{\pi}e^{y\cos\theta}d\theta$

Estimating the right hand side separately, we get

$\displaystyle\Big|\frac{1}{\pi}\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta\Big|$

$\displaystyle\leq$

$\displaystyle\frac{e^{y}}{\pi}\int_{0}^{\frac{\pi}{3}}e^{-\frac{y\cdot\theta^{2}}{4}}d\theta+\frac{2e^{\frac{y}{2}}}{3}\leq\frac{2e^{y}}{\sqrt{\pi}\cdot\sqrt{y}}+\frac{2e^{\frac{y}{2}}}{3}\leq\frac{10e^{y}}{\sqrt{y}}.$

Therefore,

(5.67)

$$I_{\nu}(y)\leq\frac{10e^{y}}{\sqrt{y}}+\frac{C(\nu)\cdot e^{-y}}{\sqrt{y}}\leq\frac{C(\nu)\cdot e^{y}}{\sqrt{y}}.$$

Now we assume $y\in(0,1]$. Since $I_{\nu}$ is smooth, we only need to analyze the behavior of $I_{\nu}(y)$ as $y\to 0$. By the definition of $I_{\nu}(y)$ we see if $\nu\geq 0$ or $\nu$ is a negative integer, $\lim\limits_{y\rightarrow 0}I_{\nu}(y)=0$. For any $\nu<0$, we have

(5.68)

$$\lim\limits_{y\to 0}I_{\nu}(y)\Big/\frac{(\frac{y}{2})^{\nu}}{\Gamma(\nu+1)}=1.$$

Therefore, for any $y\in(0,1]$,

(5.69)

$$I_{\nu}(y)\leq C(\nu)\cdot y^{\nu}.$$

Now we prove Item (2). First we observe that by the definition of $I_{\nu}$ using power series, when $\nu\in(-1,0)$, $I_{\nu}(y)$ is positive for all $y\in(0,\infty)$. So the lower bound of $I_{\nu}$ for $y\in(0,1]$ follows just as before. Now we assume $y\geq 1$. To get the lower bound on $I_{\nu}$, it suffices to get the lower bound on the first term of (5.63). Suppose $\nu\neq 0$, denote $\eta_{\nu}=\min(\pi,\frac{\pi}{3|\nu|})$, then we divide the integral into two parts

(5.70)

$$\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta=\int_{0}^{\eta_{\nu}}e^{y\cos\theta}\cos(\nu\theta)d\theta+\int_{\eta_{\nu}}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta.$$

Since $\cos\theta\geq 1-\frac{\theta^{2}}{2}$ we get

(5.71)

$$\int_{0}^{\eta_{\nu}}e^{y\cos\theta}\cos(\nu\theta)d\theta\geq\frac{1}{2}e^{y}\int_{0}^{\eta_{\nu}}e^{-\frac{\theta^{2}}{2}y}d\theta\geq C(\nu)\frac{e^{y}}{\sqrt{y}},$$

and for the second term we have

(5.72)

$$\Big|\int_{\eta_{\nu}}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta\Big|\leq\int_{\eta_{\nu}}^{\pi}e^{y\cos\theta}d\theta\leq(\pi-\eta_{\nu})e^{\cos(\eta_{\nu})y}.$$

So we get

(5.73)

$$I_{\nu}(y)\geq C^{-1}(\nu)\frac{e^{y}}{\sqrt{y}}.$$

For $\nu=0$ the argument is similar. This completes the proof of Item (1).

∎