Let $\nu\in\mathbb{R}$, we consider the following modified Bessel equation

(A.1)

$$y^{2}\cdot\frac{d^{2}\mathcal{B}(y)}{dy^{2}}+y\cdot\frac{d\mathcal{B}(y)}{dy}-(y^{2}+\nu^{2})\cdot\mathcal{B}(y)=0,\ y\geq 0.$$

First, for any $\nu\in\mathbb{R}$, we define

(A.2)

$\displaystyle I_{\nu}(y)\equiv\sum\limits_{k=0}^{\infty}\frac{1}{\Gamma(k+1)\Gamma(k+\nu+1)}\Big(\frac{y}{2}\Big)^{2k+\nu}.$

In the special case $\nu=-\ell$ with $\ell\in\mathbb{Z}_{+}$, then the above definition can be also explained as

(A.3)

$$I_{\nu}(y)=\sum\limits_{k=\ell}^{\infty}\frac{1}{\Gamma(k+1)\Gamma(k-\ell+1)}\Big(\frac{y}{2}\Big)^{2k-\ell}.$$

Immediately, for any positive integer $\ell\in\mathbb{Z}_{+}$, we have

(A.4)

$$I_{-\ell}(z)=I_{\ell}(z).$$

Next we define $K_{\nu}(z)$ as follows,

(A.5)

$\displaystyle K_{\nu}(y)\equiv\begin{cases}\frac{\pi}{2\sin(\nu\pi)}\cdot(I_{-\nu}(y)-I_{\nu}(y)),&\nu\not\in\mathbb{Z},\\ \lim\limits_{\begin{subarray}{c}\nu^{\prime}\to\nu\\ \nu^{\prime}\not\in\mathbb{Z}\end{subarray}}K_{\nu^{\prime}}(y),&\nu\in\mathbb{Z}.\end{cases}$

One can check that $I_{\nu}(y)$ and $K_{\nu}(y)$ are two linearly independent solutions to (A.1). In the literature, $I_{\nu}$ and $K_{\nu}$ are usually called modified Bessel functions.

In our context, mainly we are interested in the solutions $I_{\nu}$ and $K_{\nu}$ with an index $\nu=\frac{1}{n}$ and $n\geq 2$. The simples case is $n=2$ such that both $I_{\frac{1}{2}}(y)$ and $K_{\frac{1}{2}}(y)$ have explicit formulae:

(A.6)

$$I_{\frac{1}{2}}(y)=\sqrt{\frac{2}{\pi y}}\sinh(y),\ K_{\frac{1}{2}}(y)=\sqrt{\frac{\pi}{2y}}e^{-y}.$$

The main part of this subsection is to prove the following useful integral representations for $I_{\nu}$ and $K_{\nu}$.

Now we summarize some results regarding the confluent hypergeometric functions which are used in Section 5. Given $\alpha,\beta\in\mathbb{R}$ such that $\alpha>\beta$ and $\alpha$ is not a negative integer, we consider the following confluent hypergeometric equation

(A.14)

$$y\cdot\frac{d^{2}\mathcal{J}(y)}{dy^{2}}+(\fa-y)\cdot\frac{d\mathcal{J}(y)}{dy}-\fb\cdot\mathcal{J}(y)=0.$$

Let

(A.15)

$$\Ku(\beta,\alpha,y)\equiv\sum\limits_{k=0}^{\infty}\frac{(\beta)_{k}}{(\alpha)_{k}}\cdot\frac{y^{k}}{k!},$$

where we define the notation $(x)_{k}\equiv\prod\limits_{m=1}^{k}(x+m-1)$ and $(x)_{0}=1$. So the power series $\Ku(\fb,\fa,z)$ is always well-defined for all $\fb\in\mathbb{C}$, $z\in\mathbb{C}$ and $\fa\in\mathbb{C}\setminus\{0,-1,-2,\ldots\}$. Moreover, for any fixed $z\in\mathbb{C}$, the function $\Ku$ is entire in $\fb$ and meromorphic in $\fa$ with simple poles at negative integers.

It is by straightforward calculations that the function $\Ku(\fb,\fa,y)$ is a solution to (A.14). In the literature, $\Ku$ is called Kummer’s (confluent hypergeometric) function. Moreover, when $y>0$, one can directly check that the function $\widehat{\Ku}(\fb,\fa,y)\equiv y^{1-\fa}\cdot\Ku(1+\fb-\fa,2-\fa,y)$, which is linearly independent of $\Ku(\fb,\fa,y)$, also solves (A.14). Therefore, the general solution of (A.14) for $y>0$ is

(A.16)

$$\mathcal{J}(y)=C\cdot\Ku(\fb,\fa,y)+C^{*}\cdot y^{1-\fa}\cdot\Ku(1+\fb-\fa,2-\fa,y).$$

The power series definition of $\Ku(\fb,\fa,y)$ immediately gives the following integral representation formula which is well known in the literature. We include a short proof just for the convenience of the readers.

Given $\fb>0$ and $y>0$, we define the function

(A.21)

$$\mathcal{U}(\fb,\fa,y)\equiv\frac{1}{\Gamma(\fb)}\int_{0}^{\infty}e^{-yt}t^{\fb-1}(1+t)^{\fa-\fb-1}dt.$$

Quick computations show that for each $\fb>0$, the function $\mathcal{U}(\fb,\fa,y)$ is a solution to the confluent hypergeometric equation (A.14) on the positive real axis $\mathbb{R}_{+}$. Now let $\fb>0$ and $\fa\in\mathbb{R}\setminus\{0,-1,-2,-3,\ldots\}$, thanks to (A.16), the function $\mathcal{U}(\fb,\fa,y)$ can be written in terms of Kummer’s function $\Ku$. Evaluating those functions and their derivatives at $y=0$, one can easily obtain

(A.22)

$$\mathcal{U}(\fb,\fa,y)=\frac{\Gamma(1-\fa)}{\Gamma(1+\fb-\fa)}\cdot\Ku(\fb,\fa,y)+\frac{\Gamma(\fa-1)}{\Gamma(\fb)}\cdot y^{1-\fa}\cdot\Ku(1+\fb-\fa,2-\fa,y).$$

Notice that, the above relation is well-defined for each $y\geq 0$ and non-integral $\alpha$. Moreover, if $\alpha\to n+1\in\mathbb{Z}_{+}$, then the right hand side of (A.22) will tend to a definite limit. The function $\mathcal{U}(\fb,\fa,y)$ is usually called Tricomi’s (confluent hypergeometric) function. In our context, we are also interested in the case $y<0$. It can be directly verified that, if $y<0$, the function

(A.23)

$$\Tri(\fb,\fa,y)\equiv e^{y}\cdot\mathcal{U}(\fa-\fb,\fa,-y)$$

solves equation (A.14). Moreover, it immediately follows from the integral representation of $\mathcal{U}$ that for any $y<0$,

(A.24)

$$\Tri(\beta,\alpha,y)=\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt.$$

In summary, if $y<0$, the equation (A.14) has two linearly independent solutions $\Ku(\fb,\fa,y)$ and $\Tri(\fb,\fa,y)$.

The asymptotic behavior of $\Ku(\fb,\fa,y)$, $\mathcal{U}(\fb,\fa,y)$ and $\Tri(\fb,\fa,y)$ can be easily seen from the above integral formulae. In fact, we have the following

Next we introduce some recurrence formulae for Kummer’s function.

With the above recurrence formula, we can extend the domain of indices in Lemma A.3 for Kummer’s function.

Next we give another integral representation for Kummer’s function $\Ku(\fb,\fa,y)$ in the case $y\leq 0$, which has a crucial role in Section 5.

The following corollary shows the asymptotic behavior of $I_{\nu}(y)$ and $K_{\nu}(y)$ as $y\to+\infty$.