A.1. Modified Bessel functions [056Q]

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A.1. Modified Bessel functions

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 $K_{\nu}$ .

Lemma A.1.

Given $\nu\in\mathbb{R}$ , then the following integral formulae hold for each $y>0$ ,

(A.7)		$\displaystyle I_{\nu}(y)$	$\displaystyle=\frac{1}{\pi}\int_{0}^{\pi}e^{y\cos\theta}\cos(\nu\theta)d\theta-\frac{\sin(\nu\pi)}{\pi}\int_{0}^{\infty}e^{-y\cosh t-\nu t}dt,$
(A.8)		$\displaystyle K_{\nu}(y)$	$\displaystyle=\int_{0}^{\infty}e^{-y\cosh t}\cosh(\nu t)dt.$

Proof.

First, we prove the integral formula for $I_{\nu}$ . The idea of the proof was originally inspired by Hankel’s representation formula for the reciprocal gamma function. In fact, let $\mathcal{L}\subset\mathbb{C}$ be a contour winding around the negative $Ox$ -axis. In our particular case, $\mathcal{L}=\mathcal{L}_{1}+\mathcal{L}_{2}+\mathcal{L}_{3}$ , where $\mathcal{L}_{1}$ and $\mathcal{L}_{3}$ are two rays parallel to $Ox$ and $\mathcal{L}_{2}$ is an arc of the unit circle centered at the origin (See Figure A.1). So Hankel’s representation formula gives that

(A.9)

\frac{1}{\Gamma(k+\nu+1)}=\frac{1}{2\pi\sqrt{-1}}\int_{\mathcal{L}}e^{w}w^{-(k+\nu+1)}dw,\ w\in\mathbb{C}.

By the power series definition of $I_{\nu}$ ,

(A.10)	$\displaystyle I_{\nu}(y)$	$\displaystyle=$	$\displaystyle\sum\limits_{k=0}^{\infty}\frac{1}{\Gamma(k+1)\Gamma(k+\nu+1)}\Big(\frac{y}{2}\Big)^{2k+\nu}$
		$\displaystyle=$	$\displaystyle(\frac{y}{2})^{\nu}\frac{1}{2\pi\sqrt{-1}}\int_{\mathcal{L}}e^{w}w^{-\nu-1}\sum\limits_{k=0}^{\infty}\frac{(\frac{y^{2}}{4w})^{k}}{k!}dw$
		$\displaystyle=$	$\displaystyle(\frac{y}{2})^{\nu}\frac{1}{2\pi\sqrt{-1}}\int_{\mathcal{L}}e^{w+\frac{y^{2}}{4w}}w^{-\nu-1}dw.$

For every $y>0$ , we make change of variables for each $w\in\mathbb{C}$ ,

(A.11)

w=\frac{y\cdot e^{\zeta}}{2}=\frac{ye^{t}}{2}\cdot e^{\sqrt{-1}\theta},\ 0<t<\infty,\ 0\leq\theta\leq 2\pi.

Letting $\mathcal{L}_{1}$ and $\mathcal{L}_{3}$ tend to each other, then in terms of the variables $(t,\theta)$ ,

(A.12)

\int_{\mathcal{L}}e^{w+\frac{y^{2}}{4w}}w^{-\nu-1}dw=\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.

The integral formula for $K_{\nu}$ follows easily from the above integral representation for $I_{\nu}$ and the definition

(A.13)

K_{\nu}(y)=\frac{\pi(I_{-\nu}(y)-I_{\nu}(y))}{2\sin(\nu\pi)}.

Figure A.1. The contour

\mathcal{L}=\mathcal{L}_{1}+\mathcal{L}_{2}+\mathcal{L}_{3}

for the integral (A.9)

∎