Inequalities for the modified Bessel function of the second kind and the kernel of the Kr\"{a}tzel integral transformation

TL;DR

This paper derives new inequalities for the modified Bessel function of the second kind by establishing bounds for the kernel of the Krätzel integral transformation, connecting special functions and integral transforms.

Contribution

It introduces novel inequalities for $K_ u$ based on the Krätzel kernel, expanding the understanding of these functions and their bounds.

Findings

New inequalities for $K_ u$ in terms of gamma functions

Bounds derived for the Krätzel integral transformation kernel

Special cases recover known inequalities

Abstract

We obtain new inequalities for the modified Bessel function of the second kind $K_\nu$ in terms of the gamma function. These bounds follow as special cases of inequalities that we derive for the kernel of the Kr\"{a}tzel integral transformation.

Equations

$$\frac{1}{\sqrt{x+\frac{1}{2}}}<\frac{\Gamma (x+\frac{1}{2})}{\Gamma (x+1)}<\sqrt{\frac{2}{\pi }}{\mathrm{e}}^x{K}_0(x).$$

$$K_{\nu }(x)=\frac{\sqrt{\pi }(\frac{1}{2}x{)}^{\nu }}{\Gamma (\nu +\frac{1}{2})}{\int }_1^{\infty }{\mathrm{e}}^{-xt}(t^2-1{)}^{\nu -\frac{1}{2}}\mathrm{d}t,x>0.$$

$$\lambda_{\nu}^{(n)}(x)=\frac{(2\pi)^{(n-1)/2}\sqrt{n}\big{(}\frac{x}{n}\big{)}^{n\nu}}{\Gamma(\nu+1-\frac{1}{n})}\int_{1}^{\infty}(t^{n}-1)^{\nu-\frac{1}{n}}\mathrm{e}^{-xt}\,\mathrm{d}t,\quad\text{$\nu>\frac{1}{n}-1$, $n=1,2,\ldots$,}$$

$${\mathcal{L}}_{\nu }^{(n)}\{f\}(z)={\int }_0^{\infty }{\lambda }_{\nu }^{(n)}(zt)f(t)\mathrm{d}t,\mathrm{Re}x>0.$$

$$\lambda_{\nu}^{(2)}(x)=2\bigg{(}\frac{x}{2}\bigg{)}^{\nu}K_{\nu}(x).$$

$$\lambda_{\nu}^{(n)}(x)\geq(2\pi)^{(n-1)/2}\frac{\sqrt{n}}{n-1}\bigg{(}\frac{x}{n}\bigg{)}^{n\nu}\frac{\Gamma(\frac{x}{n-1}+\frac{1}{n}-\nu)}{\Gamma(\frac{x}{n-1}+1)}\mathrm{e}^{-x}$$

$$K_{\nu }(x)\ge \sqrt{\frac{\pi }{2}}\frac{x^{\nu }\Gamma (x+\frac{1}{2}-\nu )}{\Gamma (x+1)}{\mathrm{e}}^{-x}$$

$$\lambda_{\nu}^{(n)}(x)=\frac{(2\pi)^{(n-1)/2}\sqrt{n}\big{(}\frac{x}{n}\big{)}^{n\nu}}{\Gamma(\nu+1-\frac{1}{n})}\int_{1}^{\infty}(t^{n}-1)^{\nu-\frac{1}{n}}\mathrm{e}^{-xt}\,\mathrm{d}t.$$

$$\lambda_{\nu}^{(n)}(x)=\frac{(2\pi)^{(n-1)/2}\sqrt{n}\big{(}\frac{x}{n}\big{)}^{n\nu}}{\Gamma(\nu+1-\frac{1}{n})}\frac{2}{n-1}\mathrm{e}^{-x}\int_{0}^{\infty}\bigg{(}\bigg{(}\frac{2u}{n-1}+1\bigg{)}^{n}-1\bigg{)}^{\nu-\frac{1}{n}}\mathrm{e}^{-\frac{2x}{n-1}u}\,\mathrm{d}u.$$

$$\displaystyle\frac{n}{n-1}\big{(}\mathrm{e}^{2u}-1\big{)}$$

$$>\sum _{k=1}^n\frac{(2u{)}^k}{k!}\frac{n}{n-1}\times \frac{n-1}{n-1}\times \cdots \times \frac{n-k+1}{n-1}$$

$$=\sum _{k=1}^n\frac{(2u{)}^k}{k!}\frac{n!}{(n-k)!}\cdot \frac{1}{(n-1{)}^k}$$

$$\displaystyle=\sum_{k=1}^{n}\binom{n}{k}\bigg{(}\frac{2u}{n-1}\bigg{)}^{k}=\bigg{(}1+\frac{2u}{n-1}\bigg{)}^{n}-1.$$

$${\lambda }_{\nu }^{(n)}(x)$$

$$\displaystyle=\frac{(2\pi)^{(n-1)/2}\sqrt{n}\big{(}\frac{x}{n}\big{)}^{n\nu}}{\Gamma(\nu+1-\frac{1}{n})}\frac{2}{n-1}\bigg{(}\frac{n}{n-1}\bigg{)}^{\nu-\frac{1}{n}}\mathrm{e}^{-x}\int_{0}^{\infty}\mathrm{e}^{-(\frac{2x}{n-1}+\frac{2}{n}-2\nu)u}(1-\mathrm{e}^{-2u})^{\nu-\frac{1}{n}}\,\mathrm{d}u.$$

$${\int }_0^{\infty }{\mathrm{e}}^{-(\frac{2x}{n-1}+\frac{2}{n}-2\nu )u}(1-{\mathrm{e}}^{-2u}{)}^{\nu -\frac{1}{n}}\mathrm{d}u$$

$$=\frac{1}{2}B(\nu +1-\frac{1}{n},\frac{x}{n-1}+\frac{1}{n}-\nu )$$

$$=\frac{\Gamma (\nu +1-\frac{1}{n})\Gamma (\frac{x}{n-1}+\frac{1}{n}-\nu )}{2\Gamma (\frac{x}{n-1}+1)},$$

$$\bigg{(}\frac{x}{x+\frac{1}{2}-\nu}\bigg{)}^{\nu+\frac{1}{2}}<\sqrt{\frac{2}{\pi}}\mathrm{e}^{x}K_{\nu}(x)<1.$$

$$1-\frac{\frac{1}{2}(\frac{1}{4}-{\nu }^2)}{x+\frac{1}{2}(\frac{1}{4}-{\nu }^2)}<\sqrt{\frac{2x}{\pi }}{\mathrm{e}}^x{K}_{\nu }(x)<1-\frac{\frac{1}{2}(\frac{1}{4}-{\nu }^2)}{x+\frac{1}{4}(\frac{9}{4}-{\nu }^2)},x>0,0\le \nu <\frac{1}{2}.$$

Full text

Keywords: Modified Bessel function, Krätzel integral transformation, inequality

AMS 2010 Subject Classification: Primary 33C10

1 Introduction

The modified Bessel function of the second kind $K_{\nu}$ is an important and widely used special function. There exists a substantial literature concerning inequalities for the modified Bessel function of the second; see, for example, [7] and [1] and references therein. In a recent work, [3] derived the following simple lower bound for the function $K_{0}$:

[TABLE]

In this note, we generalise this inequality to the modified Bessel function $K_{\nu}$ for $\nu\geq 0$. In deriving our inequality, we follow the approach of [3] by exploiting the following integral representation of the modified Bessel function of the second kind ([8], formula 10.32.8):

[TABLE]

This integral representation of the modified Bessel function $K_{\nu}$ closely resembles the kernel

[TABLE]

of the Krätzel integral transformation [6] (see also [2] and references therein for further properties) defined by

[TABLE]

Indeed, a simple manipulation yields the relation

[TABLE]

Due to the similarity between the representations (1.2) and (1.3), our approach to bounding the kernel $\lambda_{\nu}^{(n)}$ is no more difficult than bounding the modified Bessel function $K_{\nu}$. In this note, we exploit the representation (1.3) to derive inequalities for the kernel $\lambda_{\nu}^{(n)}$ and then use (1.4) to immediately deduce inequalities for the modified Bessel function $K_{\nu}$. These inequalities are a natural generalisation of the inequality (1.1).

2 Results and proofs

The following is the main result of this note.

Theorem 2.1**.**

(i). Let $x>0$. Then, for $0\leq\nu\leq\frac{1}{n}$, $n=1,2,\ldots$, we have

[TABLE]

with equality if and only if $\nu=\frac{1}{n}$. If $\nu>\frac{1}{n}$, the strict inequality is reversed and holds for all $x>(n-1)(\nu-\frac{1}{n})$.

(ii). Let $x>0$. Then, for $0\leq\nu\leq\frac{1}{2}$, we have

[TABLE]

with equality if and only if $\nu=\frac{1}{2}$. If $\nu>\frac{1}{2}$, the strict inequality is reversed and holds for all $x>\nu-\frac{1}{2}$.

Proof.

We first note that part (ii) follows immediately from setting $n=2$ in part (i), due to the relation (1.4). In order to establish part (i), we recall the integral representation of the kernel $\lambda_{\nu}^{(n)}$:

[TABLE]

Setting $t=\frac{2}{n-1}u+1$ gives

[TABLE]

We now suppose that $0\leq\nu<\frac{1}{n}$ and prove that under this condition inequality (2.5) is strict. For $u>0$, we have

[TABLE]

Applying this inequality to (2.7) yields

[TABLE]

Making the change of variables $y=\mathrm{e}^{-2u}$ gives

[TABLE]

where $B(a,b)$ is the beta function, and we used the standard formula $B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$. This completes the proof that inequality (2.5) holds for $0\leq\nu<\frac{1}{n}$. When $\nu>\frac{1}{n}$ inequality (2.8) is reversed and so inequality (2.5) is also reversed. Note that when $\nu>\frac{1}{n}$ the integral in inequality (2.8) only exists if $x>(n-1)(\nu-\frac{1}{n})$. Finally, we note that we have equality when $\nu=\frac{1}{n}$, because (2.8) becomes an equality in this case. ∎

Corollary 2.2**.**

Let $0\leq\nu<\frac{1}{2}$. Then for all $x>0$,

[TABLE]

Proof.

The upper bound holds because $K_{\nu}(x)<K_{1/2}(x)=\sqrt{\frac{\pi}{2x}}\mathrm{e}^{-x}$ for all $\nu<\frac{1}{2}$ (see [5]). The lower bound follows from Theorem 2.1 and an application of the inequality $\frac{\Gamma(x+a)}{\Gamma(x+1)}>\frac{1}{(x+a)^{1-a}}$ for $0<a<1$ (see [4]). ∎

Remark 2.3**.**

The following bounds for $K_{\nu}(x)$ were obtained by [7]:

[TABLE]

Despite taking a relatively simple form, numerical experiments show that, for $0\leq\nu<\frac{1}{2}$, the bounds of [7] and Theorem 2.1, part (ii) are remarkably accurate for all but very small $x$, for which the modified Bessel function $K_{\nu}(x)$ has a singularity as $x\downarrow 0$. The bound (2.6) outperforms the lower bound of [7] for very small $x$, as it is $O(x^{\nu})$ as $x\downarrow 0$, as opposed to $O(x^{1/2})$ which is the case for that bound of [7]. However, the bound of [7] performs better for large $x$.

Acknowledgements

The author is supported by a Dame Kathleen Ollerenshaw Research Fellowship.