Arithmetic Means for a Class of Functions and the Modified Bessel Functions of the First Kind

Abstract

In the paper, by virtue of the residue theorem in the theory of complex functions, the authors establish several identities between arithmetic means for a class of functions and the modified Bessel functions of the first kind, present several identities between arithmetic means for a class of functions and infinite series, and find several series expressions for the modified Bessel functions of the first kind.

1. Preliminaries

Recall from ([1], Chapter 6) and ([2], Chapter 5) that the classical Euler gamma function can be defined by

$$\mathsf{\Gamma }(w)={\int }_0^{\infty }{t}^{w-1}{e}^{-t}dt,\Re (w)>0$$

or by

$$\mathsf{\Gamma }(w)=\underset{n\to \infty }{lim}\frac{n!n^w}{{\prod }_{k=0}^n(w+k)},w\in \mathbb{C}\setminus \{0,-1,-2,\cdots \}$$

and that the logarithmic derivative

$$\psi (w)={[ln\mathsf{\Gamma }(w)]}^{\prime }=\frac{{\mathsf{\Gamma }}^{\prime }(w)}{\mathsf{\Gamma }(w)}$$

is called the psi or digamma function. One can also find these notions in the papers [3,4], the handbooks [1,2], and closely related references therein.

In most calculus texts, the quantity $\frac{1}{b-a}{\int }_a^{b}f(x)dx$ is called the arithmetic mean for $f(x)$ on the interval $[a,b]$. This is justified by noting that the approximating Riemann sums are the arithmetic means of values of $f(x)$ at points distributed across $[a,b]$. See ([5], p. 368).

Recall from ([1], Chapter 9) and ([2], Chapter 10) that the modified Bessel functions of the first kind $I_{\nu }(w)$ can be defined by

$$I_{\nu }(w)=\sum _{k=0}^{\infty }\frac{1}{k!\mathsf{\Gamma }(\nu +k+1)}{\left(\frac{w}{2}\right)}^{2k+\nu },|argw|<\pi ,\nu \in \mathbb{C}.$$

One can find some inequalities for $I_{\nu }(w)$ in the papers [6,7,8,9,10] and closely related references therein.

2. Motivations

In ([11], Lemma 2.1), by virtue of two approaches, the identity

$$\frac{2}{\pi }{\int }_0^{\pi /2}{\alpha }^{{cos}^2\varphi }{\beta }^{{sin}^2\varphi }d\varphi =\sqrt{\alpha \beta }{I}_0\left(ln\sqrt{\frac{\alpha }{\beta }}\right)$$

(1)

for $\alpha ,\beta >0$ was derived. By virtue of the identity (1) and a double inequality for $I_0(w)$, the double inequality

$$L(\alpha ,\beta )<\frac{2}{\pi }{\int }_0^{\pi /2}{\alpha }^{{cos}^2\varphi }{\beta }^{{sin}^2\varphi }d\varphi <I(\alpha ,\beta )$$

for $\alpha ,\beta >0$ and $\alpha \ne \beta$ was discovered in ([11], Theorem 1.1), where

$$I(\alpha ,\beta )=\frac{1}{e}{\left(\frac{{\beta }^{\beta }}{{\alpha }^{\alpha }}\right)}^{1/(\beta -\alpha )}\mathrm{and}L(\alpha ,\beta )=\frac{\beta -\alpha }{ln\beta -ln\alpha }$$

for $\alpha ,\beta >0$ and $\alpha \ne \beta$ are respectively called the exponential and logarithmic means [5,12,13,14]. Subsequently, there have been several papers, such as [10,15,16,17,18,19,20], dedicated to estimating or bounding the so-called Toader–Qi mean

$$TQ(\alpha ,\beta )=\frac{2}{\pi }{\int }_0^{\pi /2}{\alpha }^{{cos}^2\varphi }{\beta }^{{sin}^2\varphi }d\varphi =\frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{{cos}^2\varphi }{\beta }^{{sin}^2\varphi }d\varphi ,\alpha ,\beta >0.$$

In this paper, we consider the following arithmetic means for a class of functions,

$$Q(\alpha ,\beta ;k,\ell )=\frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{{cos}^k\varphi }{\beta }^{{sin}^{\ell }\varphi }d\varphi ,\alpha ,\beta >0,k,\ell \in \mathbb{N}.$$

It is clear that $Q(\alpha ,\beta ;2,2)$ is just the above Toader–Qi mean $TQ(\alpha ,\beta )$.

3. A Definition and a Lemma

In this paper, one of our main tools is the residue theorem in the theory of complex functions.

Definition 1 

(([21], p. 112) and ([2], p. 19)). Let $f(w)$ have an isolated singularity at $w=w_0$ and let

$$f(w)=\sum _{n=-\infty }^{\infty }{c}_n{(w-w_0)}^n$$

be its Laurent expansion about $w=w_0$. Then the residue of $f(w)$ at $w=w_0$ is defined to be the coefficient $c_{-1}$ and is denoted by $Res[f(w),w_0]=c_{-1}$.

Lemma 1 

(Residue theorem ([21], p. 112) and ([2], p. 19)). Let f be analytic in the domain D except for the isolated singularities $w_1,w_2,\cdots ,w_n$. If γ is a simple closed contour in D and does not pass through any of the points $w_k$ for $1\le k\le n$, then

$$\frac{1}{2\pi i}{\int }_{\gamma }f(w)dw=\sum _{k=1}^{n}Res[f(w),w_k].$$

4. Several Identities and Series for Arithmetic Means

We now state and prove the first identity for the arithmetic mean $Q(\alpha ,\beta ;1,1)$ and the modified Bessel function of the first kind $I_0(w)$ as follows.

Theorem 1.

For $\alpha ,\beta >0$, we have

$$\frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{cos\varphi }{\beta }^{sin\varphi }d\varphi =I_0\left(\sqrt{{(ln\alpha )}^2+{(ln\beta )}^2}\right).$$

(2)

Proof. 

Let $i=\sqrt{-1}$ and $w=e^{i\varphi }$. Then

$$cos\varphi =\frac{w+w^{-1}}{2}\mathrm{and}sin\varphi =\frac{w-w^{-1}}{2i}.$$

(3)

Accordingly, by direct calculation and Lemma 1, we have

$$\begin{array}{cc}\hfill \frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{cos\varphi }{\beta }^{sin\varphi }d\varphi & =\frac{1}{2\pi i}{\oint }_{\left|w\right|=1}\frac{{\alpha }^{(w+w^{-1})/2}{\beta }^{(w-w^{-1})/(2i)}}{w}dw\hfill \\ & =Res\left[\frac{{\alpha }^{(w+w^{-1})/2}{\beta }^{(w-w^{-1})/(2i)}}{w},0\right].\hfill \end{array}$$

(4)

Straightforward computation yields

$$\begin{array}{c}{\alpha }^{(w+w^{-1})/2}{\beta }^{(w-w^{-1})/(2i)}=exp\left(\frac{w+w^{-1}}{2}ln\alpha +\frac{w-w^{-1}}{2i}ln\beta \right)\\ =exp\left(\frac{ln\alpha -iln\beta }{2}w+\frac{ln\alpha +iln\beta }{2}\frac{1}{w}\right)=\sum _{p=0}^{\infty }\frac{1}{p!}{\left(\frac{ln\alpha -iln\beta }{2}w+\frac{ln\alpha +iln\beta }{2}\frac{1}{w}\right)}^p\\ =\sum _{p=0}^{\infty }\frac{1}{p!}\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{p}{q}\right){\left(\frac{ln\alpha -iln\beta }{2}w\right)}^q{\left(\frac{ln\alpha +iln\beta }{2}\frac{1}{w}\right)}^{p-q}\\ =\sum _{p=0}^{\infty }\frac{1}{p!}\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{p}{q}\right){\left(\frac{ln\alpha -iln\beta }{2}\right)}^q{\left(\frac{ln\alpha +iln\beta }{2}\right)}^{p-q}{w}^{2q-p}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{q=0}^{2m}\left(\genfrac{}{}{0pt}{}{2m}{q}\right){\left(\frac{ln\alpha -iln\beta }{2}\right)}^q{\left(\frac{ln\alpha +iln\beta }{2}\right)}^{2m-q}{w}^{2(q-m)}\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{q=0}^{2m+1}\left(\genfrac{}{}{0pt}{}{2m+1}{q}\right){\left(\frac{ln\alpha -iln\beta }{2}\right)}^q{\left(\frac{ln\alpha +iln\beta }{2}\right)}^{2m-q+1}{w}^{2(q-m)-1}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\left(\genfrac{}{}{0pt}{}{2m}{m}\right){\left(\frac{ln\alpha -iln\beta }{2}\right)}^m{\left(\frac{ln\alpha +iln\beta }{2}\right)}^{2m-m}{w}^{2(m-m)}\\ +\sum _{m=0}^{\infty }\frac{1}{(2m)!}\left(\sum _{q=0}^{m-1}+\sum _{q=m+1}^{2m}\right)\left(\genfrac{}{}{0pt}{}{2m}{q}\right){\left(\frac{ln\alpha -iln\beta }{2}\right)}^q{\left(\frac{ln\alpha +iln\beta }{2}\right)}^{2m-q}{w}^{2(q-m)}\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{q=0}^{2m+1}\left(\genfrac{}{}{0pt}{}{2m+1}{q}\right){\left(\frac{ln\alpha -iln\beta }{2}\right)}^q{\left(\frac{ln\alpha +iln\beta }{2}\right)}^{2m-q+1}{w}^{2(q-m)-1}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\left(\genfrac{}{}{0pt}{}{2m}{m}\right){\left[\frac{{(ln\alpha )}^2+{(ln\beta )}^2}{4}\right]}^m\\ +\sum _{m=0}^{\infty }\frac{1}{(2m)!}\left(\sum _{q=0}^{m-1}+\sum _{q=m+1}^{2m}\right)\left(\genfrac{}{}{0pt}{}{2m}{q}\right){\left(\frac{ln\alpha -iln\beta }{2}\right)}^q{\left(\frac{ln\alpha +iln\beta }{2}\right)}^{2m-q}{w}^{2(q-m)}\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{q=0}^{2m+1}\left(\genfrac{}{}{0pt}{}{2m+1}{q}\right){\left(\frac{ln\alpha -iln\beta }{2}\right)}^q{\left(\frac{ln\alpha +iln\beta }{2}\right)}^{2m-q+1}{w}^{2(q-m)-1},\end{array}$$

where any empty sum is taken to be zero. Therefore, by Definition 1, we obtain

$$\begin{array}{c}Res\left[\frac{{\alpha }^{(w+w^{-1})/2}{\beta }^{(w-w^{-1})/(2i)}}{w},0\right]=\sum _{m=0}^{\infty }\frac{1}{(2m)!}\left(\genfrac{}{}{0pt}{}{2m}{m}\right){\left[\frac{{(ln\alpha )}^2+{(ln\beta )}^2}{4}\right]}^m\\ =\sum _{m=0}^{\infty }\frac{1}{{(m!)}^2}{\left[\frac{\sqrt{{(ln\alpha )}^2+{(ln\beta )}^2}}{2}\right]}^{2m}=I_0\left(\sqrt{{(ln\alpha )}^2+{(ln\beta )}^2}\right).\end{array}$$

Substituting this equation into (4) results in the identity (2). The proof of Theorem 1 is thus complete. □

We now state and prove the second and third identities between the arithmetic means $Q(\alpha ,\beta ;1,2)$ and $Q(\alpha ,\beta ;2,1)$ and two infinite series as follows.

Theorem 2.

For $\alpha ,\beta >0$, we have

$$\begin{array}{c}\frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{cos\varphi }{\beta }^{{sin}^2\varphi }d\varphi =\sum _{m=0}^{\infty }\frac{{(ln\beta )}^{2m}}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\left(\genfrac{}{}{0pt}{}{(2n-1)/2}{2m-n}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n}\hfill \\ -\sum _{m=0}^{\infty }\frac{{(ln\beta )}^{2m+1}}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\left(\genfrac{}{}{0pt}{}{(2n-1)/2}{2m-n+1}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n}\end{array}$$

(5)

and

$$\begin{array}{cc}\hfill \frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{{cos}^2\varphi }{\beta }^{sin\varphi }d\varphi & =\sum _{m=0}^{\infty }\frac{{(ln\beta )}^{2m}}{(2m)!}\sum _{n=0}^m\frac{1}{2^{2(m+n)}}\left(\genfrac{}{}{0pt}{}{2m}{2n}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n}\left[\sum _{q=0}^{m+n}{(-1)}^q\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n-q}\right)\right]\hfill \\ & +\frac{1}{4}\sum _{m=0}^{\infty }\frac{{(-1)}^m{(ln\beta )}^{2m+1}}{(2m+1)!}\sum _{n=0}^m{(-1)}^n\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\left(\genfrac{}{}{0pt}{}{[2(2n+1)-1]/2}{m+n+1}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n+1}.\hfill \end{array}$$

(6)

Proof. 

As did in the proof of Theorem 1, we can obtain

$$\frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{cos\varphi }{\beta }^{{sin}^2\varphi }d\varphi =Res\left[\frac{{\alpha }^{(w+w^{-1})/2}{\beta }^{{[(w-w^{-1})/(2i)]}^2}}{w},0\right].$$

Direct computation yields

$$\begin{array}{c}{\alpha }^{(w+w^{-1})/2}{\beta }^{{[(w-w^{-1})/(2i)]}^2}=exp\left[\frac{w+w^{-1}}{2}ln\alpha +{\left(\frac{w-w^{-1}}{2i}\right)}^{2}ln\beta \right]\\ =exp\left[\frac{ln\alpha }{2}\left(w+\frac{1}{w}\right)-\frac{ln\beta }{4}{\left(w-\frac{1}{w}\right)}^2\right]=\sum _{m=0}^{\infty }\frac{1}{m!}{\left[\frac{ln\alpha }{2}\left(w+\frac{1}{w}\right)-\frac{ln\beta }{4}{\left(w-\frac{1}{w}\right)}^2\right]}^m\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left[\frac{ln\alpha }{2}\left(w+\frac{1}{w}\right)\right]}^n{(-1)}^{m-n}{\left[\frac{ln\beta }{4}{\left(w-\frac{1}{w}\right)}^2\right]}^{m-n}\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m{(-1)}^{m-n}\left(\genfrac{}{}{0pt}{}{m}{n}\right)\frac{{(ln\alpha )}^n{(ln\beta )}^{m-n}}{2^{2m-n}}\frac{{\left(1+w^2\right)}^n{\left(w^2-1\right)}^{2(m-n)}}{w^{2m-n}}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^{2m}{(-1)}^{2m-n}\left(\genfrac{}{}{0pt}{}{2m}{n}\right)\frac{{(ln\alpha )}^n{(ln\beta )}^{2m-n}}{2^{4m-n}}\frac{{\left(1+w^2\right)}^n{\left(w^2-1\right)}^{2(2m-n)}}{w^{4m-n}}\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^{2m+1}{(-1)}^{2m+1-n}\left(\genfrac{}{}{0pt}{}{2m+1}{n}\right)\frac{{(ln\alpha )}^n{(ln\beta )}^{2m+1-n}}{2^{2(2m+1)-n}}\frac{{\left(1+w^2\right)}^n{\left(w^2-1\right)}^{2(2m+1-n)}}{w^{2(2m+1)-n}}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n}}{2^{2(2m-n)}}\frac{{\left(1+w^2\right)}^{2n}{\left(w^2-1\right)}^{2(2m-2n)}}{w^{2(2m-n)}}\\ +\sum _{m=0}^{\infty }\frac{-1}{(2m)!}\sum _{n=0}^{m-1}\left(\genfrac{}{}{0pt}{}{2m}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(ln\beta )}^{2m-(2n+1)}}{2^{4m-(2n+1)}}\frac{{\left(1+w^2\right)}^{2n+1}{\left(w^2-1\right)}^{2[2m-(2n+1)]}}{w^{4m-(2n+1)}}\\ +\sum _{m=0}^{\infty }\frac{-1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m+1-2n}}{2^{2(2m+1)-2n}}\frac{{\left(1+w^2\right)}^{2n}{\left(w^2-1\right)}^{2(2m+1-2n)}}{w^{2(2m+1)-2n}}\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(ln\beta )}^{2m+1-(2n+1)}}{2^{2(2m+1)-(2n+1)}}\frac{{\left(1+w^2\right)}^{2n+1}{\left(w^2-1\right)}^{2[2m+1-(2n+1)]}}{w^{2(2m+1)-(2n+1)}}.\end{array}$$

In the second and fourth sums above, all terms are not constant and dependent of the variable w. The first and third sums above can be rewritten as

$$\begin{array}{c}\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n}}{2^{2(2m-n)}}\frac{{\left(1+w^2\right)}^{2n}{\left(w^2-1\right)}^{2(2m-2n)}}{w^{2(2m-n)}}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2(m-n)}}{{(2w)}^{2(2m-n)}}\sum _{p=0}^{2n}\left(\genfrac{}{}{0pt}{}{2n}{p}\right)w^{2p}\sum _{p=0}^{4(m-n)}\left(\genfrac{}{}{0pt}{}{4(m-n)}{p}\right){(-1)}^p{w}^{2p}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2(m-n)}}{{(2w)}^{2(2m-n)}}\sum _{p=0}^{2(2m-n)}\left[\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{4(m-n)}{p-q}\right){(-1)}^{p-q}\right]w^{2p}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2(m-n)}}{2^{2(2m-n)}}\left[\sum _{q=0}^{2m-n}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{4(m-n)}{2m-n-q}\right){(-1)}^{2m-n-q}\right]\\ +\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2(m-n)}}{{(2w)}^{2(2m-n)}}\\ \times \left[\sum _{p=0}^{2m-n-1}+\sum _{p=2m-n+1}^{2(2m-n)}\right]\left[\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{4(m-n)}{p-q}\right){(-1)}^{p-q}\right]w^{2p}\end{array}$$

and

$$\begin{array}{c}\sum _{m=0}^{\infty }\frac{-1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m+1-2n}}{2^{2(2m+1)-2n}}\frac{{\left(1+w^2\right)}^{2n}{\left(w^2-1\right)}^{2(2m+1-2n)}}{w^{2(2m+1)-2n}}\\ =\sum _{m=0}^{\infty }\frac{-1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n+1}}{{(2w)}^{2(2m-n+1)}}\\ \times \sum _{p=0}^{2n}\left(\genfrac{}{}{0pt}{}{2n}{p}\right)w^{2p}\sum _{p=0}^{2(2m-2n+1)}\left(\genfrac{}{}{0pt}{}{2(2m-2n+1)}{p}\right)w^{2p}{(-1)}^p\\ =\sum _{m=0}^{\infty }\frac{-1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n+1}}{{(2w)}^{2(2m-n+1)}}\\ \times \sum _{p=0}^{2(2m-n+1)}\left[\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(2m-2n+1)}{p-q}\right){(-1)}^{p-q}\right]w^{2p}\\ =\sum _{m=0}^{\infty }\frac{-1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n+1}}{2^{2(2m-n+1)}}\\ \times \left[\sum _{q=0}^{2m-n+1}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(2m-2n+1)}{2m-n+1-q}\right){(-1)}^{2m-n+1-q}\right]\\ +\sum _{m=0}^{\infty }\frac{-1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n+1}}{{(2w)}^{2(2m-n+1)}}\\ \times \left[\sum _{p=0}^{2m-n}+\sum _{p=2m-n+2}^{2(2m-n+1)}\right]\left[\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(2m-2n+1)}{p-q}\right){(-1)}^{p-q}\right]w^{2p}.\end{array}$$

Consequently, the constant term in the Laurent expansion of the function ${\alpha }^{(w+w^{-1})/2}{\beta }^{{[(w-w^{-1})/(2i)]}^2}$ at $w=0$ is equal to

$$\begin{array}{c}\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2(m-n)}}{2^{2(2m-n)}}\left[\sum _{q=0}^{2m-n}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{4(m-n)}{2m-n-q}\right){(-1)}^{2m-n-q}\right]\\ +\sum _{m=0}^{\infty }\frac{-1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n+1}}{2^{2(2m-n+1)}}\\ \times \left[\sum _{q=0}^{2m-n+1}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(2m-2n+1)}{2m-n+1-q}\right){(-1)}^{2m-n+1-q}\right]\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m{(-1)}^n\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2(m-n)}}{2^{2(2m-n)}}\sum _{q=0}^{2m-n}{(-1)}^q\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{4(m-n)}{2m-n-q}\right)\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m{(-1)}^n\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n+1}}{2^{2(2m-n+1)}}\sum _{q=0}^{2m-n+1}{(-1)}^q\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(2m-2n+1)}{2m-n+1-q}\right)\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m{(-1)}^n\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2(m-n)}}{2^{2(2m-n)}}{(-4)}^{2m-n}\left(\genfrac{}{}{0pt}{}{(2n-1)/2}{2m-n}\right)\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m{(-1)}^n\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2m-2n+1}}{2^{2(2m-n+1)}}{(-4)}^{2m-n+1}\left(\genfrac{}{}{0pt}{}{(2n-1)/2}{2m-n+1}\right)\\ =\sum _{m=0}^{\infty }\frac{{(ln\beta )}^{2m}}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\left(\genfrac{}{}{0pt}{}{(2n-1)/2}{2m-n}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n}\\ -\sum _{m=0}^{\infty }\frac{{(ln\beta )}^{2m+1}}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\left(\genfrac{}{}{0pt}{}{(2n-1)/2}{2m-n+1}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n},\end{array}$$

where we used the formula

$$\sum _{r=0}^j{(-1)}^r\left(\genfrac{}{}{0pt}{}{q}{r}\right)\left(\genfrac{}{}{0pt}{}{2j-q}{j-r}\right)={(-4)}^j\left(\genfrac{}{}{0pt}{}{(q-1)/2}{j}\right)$$

(7)

in ([22], p. 6, Equation (1.40)) and ([23], p. 63, Equation (3.42)). In a word, it follows that

$$\begin{array}{c}Res\left[\frac{{\alpha }^{(w+w^{-1})/2}{\beta }^{{[(w-w^{-1})/(2i)]}^2}}{w},0\right]=\sum _{m=0}^{\infty }\frac{{(ln\beta )}^{2m}}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\left(\genfrac{}{}{0pt}{}{(2n-1)/2}{2m-n}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n}\hfill \\ \hfill -\sum _{m=0}^{\infty }\frac{{(ln\beta )}^{2m+1}}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\left(\genfrac{}{}{0pt}{}{(2n-1)/2}{2m-n+1}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n}.\end{array}$$

The Equation (5) is thus proved.

As did in the above or in the proof of Theorem 1, we can also obtain

$$\frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{{cos}^2\varphi }{\beta }^{sin\varphi }d\varphi =Res\left[\frac{{\alpha }^{{[(w+w^{-1})/2]}^2}{\beta }^{(w-w^{-1})/(2i)}}{w},0\right].$$

Straightforward calculation shows

$$\begin{array}{c}{\alpha }^{{[(w+w^{-1})/2]}^2}{\beta }^{(w-w^{-1})/(2i)}=exp\left[{\left(\frac{w+w^{-1}}{2}\right)}^{2}ln\alpha +\frac{w-w^{-1}}{2i}ln\beta \right]\\ =exp\left[\frac{ln\alpha }{4}{\left(w+\frac{1}{w}\right)}^2-\frac{iln\beta }{2}\left(w-\frac{1}{w}\right)\right]=\sum _{m=0}^{\infty }\frac{1}{m!}{\left[\frac{ln\alpha }{4}{\left(w+\frac{1}{w}\right)}^2-\frac{iln\beta }{2}\left(w-\frac{1}{w}\right)\right]}^m\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left[\frac{ln\alpha }{4}{\left(w+\frac{1}{w}\right)}^2\right]}^n{(-1)}^{m-n}{\left[\frac{iln\beta }{2}\left(w-\frac{1}{w}\right)\right]}^{m-n}\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m{(-1)}^{m-n}\left(\genfrac{}{}{0pt}{}{m}{n}\right)\frac{{(ln\alpha )}^n{(iln\beta )}^{m-n}}{2^{m+n}}\frac{{\left(1+w^2\right)}^{2n}{\left(w^2-1\right)}^{m-n}}{w^{m+n}}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^{2m}{(-1)}^{2m-n}\left(\genfrac{}{}{0pt}{}{2m}{n}\right)\frac{{(ln\alpha )}^n{(iln\beta )}^{2m-n}}{2^{2m+n}}\frac{{\left(1+w^2\right)}^{2n}{\left(w^2-1\right)}^{2m-n}}{w^{2m+n}}\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^{2m+1}{(-1)}^{2m+1-n}\left(\genfrac{}{}{0pt}{}{2m+1}{n}\right)\frac{{(ln\alpha )}^n{(iln\beta )}^{2m+1-n}}{2^{2m+1+n}}\frac{{\left(1+w^2\right)}^{2n}{\left(w^2-1\right)}^{2m+1-n}}{w^{2m+1+n}}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m{(-1)}^{2m-2n}\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(iln\beta )}^{2m-2n}}{2^{2m+2n}}\frac{{\left(1+w^2\right)}^{4n}{\left(w^2-1\right)}^{2m-2n}}{w^{2m+2n}}\\ +\sum _{m=0}^{\infty }\frac{-1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n}\right)\frac{{(ln\alpha )}^{2n}{(iln\beta )}^{2m+1-2n}}{2^{2m+1+2n}}\frac{{\left(1+w^2\right)}^{4n}{\left(w^2-1\right)}^{2m+1-2n}}{w^{2m+1+2n}}\\ +\sum _{m=0}^{\infty }\frac{-1}{(2m)!}\sum _{n=0}^{m-1}\left(\genfrac{}{}{0pt}{}{2m}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(iln\beta )}^{2m-(2n+1)}}{2^{2m+2n+1}}\frac{{\left(1+w^2\right)}^{2(2n+1)}{\left(w^2-1\right)}^{2m-(2n+1)}}{w^{2m+2n+1}}\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(iln\beta )}^{2(m-n)}}{2^{2(m+n+1)}}\frac{{\left(1+w^2\right)}^{2(2n+1)}{\left(w^2-1\right)}^{2(m-n)}}{w^{2(m+n+1)}}.\end{array}$$

In the second and third sums above, all terms are not constant and dependent of the variable w. The first and fourth sums above can be formulated as

$$\begin{array}{c}\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m{(-1)}^{2m-2n}\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(iln\beta )}^{2m-2n}}{2^{2m+2n}}\frac{{\left(1+w^2\right)}^{4n}{\left(w^2-1\right)}^{2m-2n}}{w^{2m+2n}}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(iln\beta )}^{2(m-n)}}{{(2w)}^{2(m+n)}}\sum _{p=0}^{4n}\left(\genfrac{}{}{0pt}{}{4n}{p}\right)w^{2p}\sum _{p=0}^{2(m-n)}\left(\genfrac{}{}{0pt}{}{2(m-n)}{p}\right){(-1)}^p{w}^{2p}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(iln\beta )}^{2(m-n)}}{{(2w)}^{2(m+n)}}\sum _{p=0}^{2(m+n)}\left[\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{4n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{p-q}\right){(-1)}^{p-q}\right]w^{2p}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(iln\beta )}^{2(m-n)}}{2^{2(m+n)}}\left[\sum _{q=0}^{m+n}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n-q}\right){(-1)}^{m+n-q}\right]\\ +\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(iln\beta )}^{2(m-n)}}{{(2w)}^{2(m+n)}}\\ \times \left[\sum _{p=0}^{m+n-1}+\sum _{p=m+n+1}^{2(m+n)}\right]\left[\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{4n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{p-q}\right){(-1)}^{p-q}\right]w^{2p}\end{array}$$

and

$$\begin{array}{c}\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(iln\beta )}^{2(m-n)}}{2^{2(m+n+1)}}\frac{{\left(1+w^2\right)}^{2(2n+1)}{\left(w^2-1\right)}^{2(m-n)}}{w^{2(m+n+1)}}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(iln\beta )}^{2(m-n)}}{{(2w)}^{2(m+n+1)}}\\ \times \sum _{p=0}^{2(2n+1)}\left(\genfrac{}{}{0pt}{}{2(2n+1)}{p}\right)w^{2p}\sum _{p=0}^{2(m-n)}\left(\genfrac{}{}{0pt}{}{2(m-n)}{p}\right)w^{2p}{(-1)}^{2(m-n)-p}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(iln\beta )}^{2(m-n)}}{{(2w)}^{2(m+n+1)}}\\ \times \sum _{p=0}^{2(m+n+1)}\left[\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{2(2n+1)}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{p-q}\right){(-1)}^{p-q}\right]w^{2p}\\ =\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(iln\beta )}^{2(m-n)}}{2^{2(m+n+1)}}\\ \times \left[\sum _{q=0}^{m+n+1}\left(\genfrac{}{}{0pt}{}{2(2n+1)}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n+1-q}\right){(-1)}^{m+n+1-q}\right]\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(iln\beta )}^{2(m-n)}}{{(2w)}^{2(m+n+1)}}\\ \times \left[\sum _{p=0}^{m+n}+\sum _{p=m+n+2}^{2(m+n+1)}\right]\left[\sum _{q=0}^p\left(\genfrac{}{}{0pt}{}{2(2n+1)}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{p-q}\right){(-1)}^{p-q}\right]w^{2p}.\end{array}$$

Consequently, the constant term in the Laurent expansion of the function ${\alpha }^{{[(w+w^{-1})/2]}^2}{\beta }^{(w-w^{-1})/(2i)}$ at $w=0$ is equal to

$$\begin{array}{c}\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(iln\beta )}^{2(m-n)}}{2^{2(m+n)}}\left[\sum _{q=0}^{m+n}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n-q}\right){(-1)}^{m+n-q}\right]\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(iln\beta )}^{2(m-n)}}{2^{2(m+n+1)}}\\ \times \left[\sum _{q=0}^{m+n+1}\left(\genfrac{}{}{0pt}{}{2(2n+1)}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n+1-q}\right){(-1)}^{m+n+1-q}\right]\\ =\sum _{m=0}^{\infty }\frac{1}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right)\frac{{(ln\alpha )}^{2n}{(ln\beta )}^{2(m-n)}}{2^{2(m+n)}}\left[\sum _{q=0}^{m+n}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n-q}\right){(-1)}^q\right]\\ +\sum _{m=0}^{\infty }\frac{1}{(2m+1)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right)\frac{{(ln\alpha )}^{2n+1}{(ln\beta )}^{2(m-n)}}{2^{2(m+n+1)}}\\ \times \left[\sum _{q=0}^{m+n+1}\left(\genfrac{}{}{0pt}{}{2(2n+1)}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n+1-q}\right){(-1)}^{q-1}\right]\\ =\sum _{m=0}^{\infty }\frac{{(ln\beta )}^{2m}}{(2m)!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{2m}{2n}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n}\frac{1}{2^{2(m+n)}}\left[\sum _{q=0}^{m+n}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n-q}\right){(-1)}^q\right]\\ +\frac{1}{4}\sum _{m=0}^{\infty }\frac{{(-1)}^m{(ln\beta )}^{2m+1}}{(2m+1)!}\sum _{n=0}^m{(-1)}^n\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n+1}\left(\genfrac{}{}{0pt}{}{[2(2n+1)-1]/2}{m+n+1}\right),\end{array}$$

where we used the formula (7). In conclusion, it follows that

$$\begin{array}{l}Res\left[\frac{{\alpha }^{(w+w^{-1})/2}{\beta }^{{[(w-w^{-1})/(2i)]}^2}}{w},0\right]\\ ={\displaystyle \sum _{m=0}^{\infty }}\frac{{(ln\beta )}^{2m}}{(2m)!}{\displaystyle \sum _{n=0}^m}\left(\genfrac{}{}{0pt}{}{2m}{2n}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n}\frac{1}{2^{2(m+n)}}\left[{\displaystyle \sum _{q=0}^{m+n}}\left(\genfrac{}{}{0pt}{}{2n}{q}\right)\left(\genfrac{}{}{0pt}{}{2(m-n)}{m+n-q}\right){(-1)}^q\right]\\ +\frac{1}{4}{\displaystyle \sum _{m=0}^{\infty }}\frac{{(-1)}^m{(ln\beta )}^{2m+1}}{(2m+1)!}{\displaystyle \sum _{n=0}^m}{(-1)}^n\left(\genfrac{}{}{0pt}{}{2m+1}{2n+1}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2n+1}\left(\genfrac{}{}{0pt}{}{[2(2n+1)-1]/2}{m+n+1}\right).\end{array}$$

The Equation (6) is thus proved. The proof of Theorem 2 is complete. □

We now state and prove an identity between the arithmetic mean $Q(\alpha ,\beta ;k,k)$ and an infinite series as follows.

Theorem 3.

For $\alpha ,\beta >0$ and $k\in \mathbb{N}$, we have

$$\frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{{cos}^k\varphi }{\beta }^{{sin}^k\varphi }d\varphi =\sum _{j=0}^{\infty }{(-1)}^{kj}\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}{(-i)}^{kq}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\left(\genfrac{}{}{0pt}{}{(kq-1)/2}{kj}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q.$$

(8)

Proof. 

Employing the transforms in (3) and Lemma 1 reveals

$$\begin{array}{cc}\hfill \frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{{cos}^k\varphi }{\beta }^{{sin}^k\varphi }d\varphi & =\frac{1}{2\pi i}{\oint }_{\left|w\right|=1}\frac{{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^k}}{w}dw\hfill \\ & =Res\left[\frac{{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^k}}{w},0\right].\hfill \end{array}$$

(9)

Easy computation gives

$$\begin{array}{c}{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^k}=exp\left[{\left(\frac{w+w^{-1}}{2}\right)}^{k}ln\alpha +{\left(\frac{w-w^{-1}}{2i}\right)}^{k}ln\beta \right]\\ =exp\left[\frac{{\left(1+w^2\right)}^{k}ln\alpha +{\left(1-w^2\right)}^k{i}^{k}ln\beta }{{(2w)}^k}\right]=\sum _{j=0}^{\infty }\frac{1}{j!}{\left[\frac{{\left(1+w^2\right)}^{k}ln\alpha +{\left(1-w^2\right)}^k{i}^{k}ln\beta }{{(2w)}^k}\right]}^j\\ =\sum _{j=0}^{\infty }\frac{1}{j!}\frac{1}{{(2w)}^{kj}}\sum _{q=0}^j\left(\genfrac{}{}{0pt}{}{j}{q}\right){\left[{\left(1+w^2\right)}^{k}ln\alpha \right]}^q{\left[{\left(1-w^2\right)}^k{i}^{k}ln\beta \right]}^{j-q}\\ =\sum _{j=0}^{\infty }\frac{1}{j!}\frac{1}{{(2w)}^{kj}}\sum _{q=0}^j\left(\genfrac{}{}{0pt}{}{j}{q}\right){(ln\alpha )}^q{(ln\beta )}^{j-q}{i}^{(j-q)k}{\left[{\left(1+w^2\right)}^q{\left(1-w^2\right)}^{j-q}\right]}^k\\ =\sum _{j=0}^{\infty }\sum _{q=0}^{2j}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\frac{{(ln\alpha )}^q{(ln\beta )}^{2j-q}{i}^{(2j-q)k}}{(2j)!}{\left[\frac{{\left(1+w^2\right)}^q{\left(1-w^2\right)}^{2j-q}}{{(4w^2)}^j}\right]}^k\\ +\sum _{j=0}^{\infty }\sum _{q=0}^{2j+1}\left(\genfrac{}{}{0pt}{}{2j+1}{q}\right)\frac{{(ln\alpha )}^q{(ln\beta )}^{2j+1-q}{i}^{(2j+1-q)k}}{(2j+1)!}{\left[\frac{{\left(1+w^2\right)}^q{\left(1-w^2\right)}^{2j+1-q}}{{(2w)}^{2j+1}}\right]}^k.\end{array}$$

It is easy to see that, in the above sum ${\sum }_{j=0}^{\infty }{\sum }_{q=0}^{2j+1}$, there is no any constant term independent of w. On the other hand, because

$$\begin{array}{c}{(1+w)}^q{(1-w)}^{2j-q}=\sum _{r=0}^q\left(\genfrac{}{}{0pt}{}{q}{r}\right)w^r\sum _{s=0}^{2j-q}\left(\genfrac{}{}{0pt}{}{2j-q}{s}\right){(-1)}^s{w}^s=\sum _{p=0}^{2j}\left[\sum _{r=0}^p\left(\genfrac{}{}{0pt}{}{q}{r}\right)\left(\genfrac{}{}{0pt}{}{2j-q}{p-r}\right){(-1)}^{p-r}\right]w^p\\ =\left[\sum _{r=0}^j\left(\genfrac{}{}{0pt}{}{q}{r}\right)\left(\genfrac{}{}{0pt}{}{2j-q}{j-r}\right){(-1)}^{j-r}\right]w^j+\left(\sum _{p=0}^{j-1}+\sum _{p=j+1}^{2j}\right)\left[\sum _{r=0}^p\left(\genfrac{}{}{0pt}{}{q}{r}\right)\left(\genfrac{}{}{0pt}{}{2j-q}{p-r}\right){(-1)}^{p-r}\right]w^p,\end{array}$$

where any empty sum is taken to be zero and $\left(\genfrac{}{}{0pt}{}{r}{s}\right)=0$ for $s>r\ge 0$ and $r,s\in \mathbb{N}$, we have

$$\begin{array}{c}\sum _{j=0}^{\infty }\sum _{q=0}^{2j}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\frac{{(ln\alpha )}^q{(ln\beta )}^{2j-q}{i}^{(2j-q)k}}{(2j)!}{\left[\frac{{\left(1+w^2\right)}^q{\left(1-w^2\right)}^{2j-q}}{{(4w^2)}^j}\right]}^k\\ =\sum _{j=0}^{\infty }\sum _{q=0}^{2j}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\frac{{(ln\alpha )}^q{(ln\beta )}^{2j-q}{i}^{(2j-q)k}}{(2j)!}\frac{{\left(1+w^2\right)}^{kq}{\left(1-w^2\right)}^{k(2j-q)}}{{(4w^2)}^{kj}}\\ =\sum _{j=0}^{\infty }\sum _{q=0}^{2j}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\frac{{(ln\alpha )}^q{(ln\beta )}^{2j-q}{i}^{(2j-q)k}}{(2j)!4^{kj}}\sum _{r=0}^{kj}{(-1)}^{kj-r}\left(\genfrac{}{}{0pt}{}{kq}{r}\right)\left(\genfrac{}{}{0pt}{}{k(2j-q)}{kj-r}\right)\\ +\sum _{j=0}^{\infty }\sum _{q=0}^{2j}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\frac{{(ln\alpha )}^q{(ln\beta )}^{2j-q}{i}^{(2j-q)k}}{(2j)!}\frac{1}{{(4w^2)}^{kj}}\left(\sum _{p=0}^{kj-1}+\sum _{p=kj+1}^{2kj}\right)\left[\sum _{r=0}^p\left(\genfrac{}{}{0pt}{}{kq}{r}\right)\left(\genfrac{}{}{0pt}{}{k(2j-q)}{p-r}\right){(-1)}^{p-r}\right]w^{2p}.\end{array}$$

It is not difficult to see that, in the sum ${\sum }_{p=0}^{kj-1}+{\sum }_{p=kj+1}^{2kj}$, there is no any constant term independent of w. In a word, the constant term independent of w in the Laurent expansion of the function

$${\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^k}$$

at $z=0$ is equal to

$$\sum _{j=0}^{\infty }\sum _{q=0}^{2j}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\frac{{(ln\alpha )}^q{(ln\beta )}^{2j-q}{i}^{(2j-q)k}}{(2j)!4^{kj}}\sum _{r=0}^{kj}{(-1)}^{kj-r}\left(\genfrac{}{}{0pt}{}{kq}{r}\right)\left(\genfrac{}{}{0pt}{}{k(2j-q)}{kj-r}\right).$$

Consequently, by Definition 1, we deduce

$$\begin{array}{c}Res\left[\frac{{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^k}}{w},0\right]\hfill \\ =\sum _{j=0}^{\infty }\frac{1}{4^{kj}}\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}\left(\genfrac{}{}{0pt}{}{2j}{q}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q{(-i)}^{kq}\sum _{r=0}^{kj}{(-1)}^r\left(\genfrac{}{}{0pt}{}{kq}{r}\right)\left(\genfrac{}{}{0pt}{}{k(2j-q)}{kj-r}\right).\hfill \end{array}$$

(10)

Substituting the identity (7) into (10) and simplifying leads to

$$Res\left[\frac{{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^k}}{w},0\right]=\sum _{j=0}^{\infty }{(-1)}^{kj}\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}{(-i)}^{kq}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\left(\genfrac{}{}{0pt}{}{(kq-1)/2}{kj}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q.$$

Substituting this equation into (9) leads to (8). The proof of Theorem 3 is thus complete. □

We now state and prove an identity between the arithmetic mean $Q(\alpha ,\beta ;k,\ell )$ and an infinite series as follows.

Theorem 4.

For $\alpha ,\beta >0$ and $k,\ell \in \mathbb{N}$, we have

$$\begin{array}{c}\frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{{cos}^k\varphi }{\beta }^{{sin}^{\ell }\varphi }d\varphi =\sum _{m=0}^{\infty }{(-1)}^{[kn+\ell (m-n)]/2}\frac{{(-1)}^{\ell (m-n)}+{(-1)}^{kn}}{2}\frac{1}{m!}\hfill \\ \times \sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left(\frac{ln\alpha }{2^k}\right)}^n{\left[\frac{ln\beta }{{(2i)}^{\ell }}\right]}^{m-n}\sum _{r=0}^{[kn+\ell (m-n)]/2}{(-1)}^r\left(\genfrac{}{}{0pt}{}{kn}{r}\right)\left(\genfrac{}{}{0pt}{}{\ell (m-n)}{[kn+\ell (m-n)]/2-r}\right).\hfill \end{array}$$

(11)

Proof. 

Utilizing the transforms in (3) and Lemma 1 results in

$$\begin{array}{cc}\hfill \frac{1}{2\pi }{\int }_0^{2\pi }{\alpha }^{{cos}^k\varphi }{\beta }^{{sin}^{\ell }\varphi }d\varphi & =\frac{1}{2\pi i}{\oint }_{\left|w\right|=1}\frac{{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^{\ell }}}{w}dw\hfill \\ & =Res\left[\frac{{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^{\ell }}}{w},0\right].\hfill \end{array}$$

(12)

Direct computation gives

$$\begin{array}{c}{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^{\ell }}=exp\left[{\left(\frac{w+w^{-1}}{2}\right)}^{k}ln\alpha +{\left(\frac{w-w^{-1}}{2i}\right)}^{\ell }ln\beta \right]\\ =exp\left[{\left(w+\frac{1}{w}\right)}^k\frac{ln\alpha }{2^k}+{\left(w-\frac{1}{w}\right)}^{\ell }\frac{ln\beta }{{(2i)}^{\ell }}\right]=\sum _{m=0}^{\infty }\frac{1}{m!}{\left[{\left(w+\frac{1}{w}\right)}^k\frac{ln\alpha }{2^k}+{\left(w-\frac{1}{w}\right)}^{\ell }\frac{ln\beta }{{(2i)}^{\ell }}\right]}^m\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left[{\left(w+\frac{1}{w}\right)}^k\frac{ln\alpha }{2^k}\right]}^n{\left[{\left(w-\frac{1}{w}\right)}^{\ell }\frac{ln\beta }{{(2i)}^{\ell }}\right]}^{m-n}\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left(\frac{ln\alpha }{2^k}\right)}^n{\left[\frac{ln\beta }{{(2i)}^{\ell }}\right]}^{m-n}{\left(w+\frac{1}{w}\right)}^{kn}{\left(w-\frac{1}{w}\right)}^{\ell (m-n)}\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left(\frac{ln\alpha }{2^k}\right)}^n{\left[\frac{ln\beta }{{(2i)}^{\ell }}\right]}^{m-n}\sum _{p=0}^{kn}\left(\genfrac{}{}{0pt}{}{kn}{p}\right)w^p{\left(\frac{1}{w}\right)}^{kn-p}\sum _{q=0}^{\ell (m-n)}\left(\genfrac{}{}{0pt}{}{\ell (m-n)}{q}\right)w^q{\left(-\frac{1}{w}\right)}^{\ell (m-n)-q}\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left(\frac{ln\alpha }{2^k}\right)}^n{\left[\frac{ln\beta }{{(2i)}^{\ell }}\right]}^{m-n}\sum _{p=0}^{kn}\left(\genfrac{}{}{0pt}{}{kn}{p}\right)w^{2p-kn}\sum _{q=0}^{\ell (m-n)}\left(\genfrac{}{}{0pt}{}{\ell (m-n)}{q}\right){(-1)}^{\ell (m-n)-q}{w}^{2q-\ell (m-n)}\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left(\frac{ln\alpha }{2^k}\right)}^n{\left[\frac{ln\beta }{{(2i)}^{\ell }}\right]}^{m-n}\frac{{(-1)}^{\ell (m-n)}}{w^{kn+\ell (m-n)}}\sum _{p=0}^{kn}\left(\genfrac{}{}{0pt}{}{kn}{p}\right)w^{2p}\sum _{q=0}^{\ell (m-n)}\left(\genfrac{}{}{0pt}{}{\ell (m-n)}{q}\right){(-1)}^q{w}^{2q}\\ =\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left(\frac{ln\alpha }{2^k}\right)}^n{\left[\frac{ln\beta }{{(2i)}^{\ell }}\right]}^{m-n}\frac{{(-1)}^{\ell (m-n)}}{w^{kn+\ell (m-n)}}\sum _{j=0}^{kn+\ell (m-n)}\left[\sum _{r=0}^j{(-1)}^{j-r}\left(\genfrac{}{}{0pt}{}{kn}{r}\right)\left(\genfrac{}{}{0pt}{}{\ell (m-n)}{j-r}\right)\right]w^{2j}.\end{array}$$

Therefore, by Definition 1, we deduce

$$\begin{array}{c}Res\left[\frac{{\alpha }^{{[(w+w^{-1})/2]}^k}{\beta }^{{[(w-w^{-1})/(2i)]}^{\ell }}}{w},0\right]=\sum _{m=0}^{\infty }\frac{1}{m!}\sum _{n=0}^m\left(\genfrac{}{}{0pt}{}{m}{n}\right){\left(\frac{ln\alpha }{2^k}\right)}^n{\left[\frac{ln\beta }{{(2i)}^{\ell }}\right]}^{m-n}\\ \times {(-1)}^{\ell (m-n)}\frac{1+{(-1)}^{kn+\ell (m-n)}}{2}\sum _{r=0}^{[kn+\ell (m-n)]/2}{(-1)}^{[kn+\ell (m-n)]/2-r}\left(\genfrac{}{}{0pt}{}{kn}{r}\right)\left(\genfrac{}{}{0pt}{}{\ell (m-n)}{[kn+\ell (m-n)]/2-r}\right).\end{array}$$

Substituting this result into (12) leads to the identity (11). The proof of Theorem 4 is thus complete. □

5. Two New Series Expressions for $I_0$

Combining the Equation (8) in Theorem 3 with the identities (2) and (1) in sequence, we can find two series expressions for $I_0$ as follows.

Theorem 5.

For $x\in \mathbb{R}$, we have

$$I_0(x)=\sum _{j=0}^{\infty }{(-1)}^j\left(\genfrac{}{}{0pt}{}{-1/2}{j}\right)\frac{x^{2j}}{(2j)!}.$$

(13)

Proof. 

When taking $k=1$, the right hand side of the Equation (8) in Theorem 3 becomes

$$\begin{array}{c}\sum _{j=0}^{\infty }{(-1)}^j\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}{(-i)}^q\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\left(\genfrac{}{}{0pt}{}{(q-1)/2}{j}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q\\ =\sum _{j=0}^{\infty }{(-1)}^j\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}{(-1)}^q{e}^{q\pi i/2}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\left(\genfrac{}{}{0pt}{}{(q-1)/2}{j}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q\\ =\sum _{j=0}^{\infty }{(-1)}^j\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}{(-1)}^q\left(cos\frac{q\pi }{2}+isin\frac{q\pi }{2}\right)\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\left(\genfrac{}{}{0pt}{}{(q-1)/2}{j}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q\\ =\sum _{j=0}^{\infty }{(-1)}^j\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^j{(-1)}^q\left(\genfrac{}{}{0pt}{}{2j}{2q}\right)\left(\genfrac{}{}{0pt}{}{(2q-1)/2}{j}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2q}\\ +i\sum _{j=0}^{\infty }{(-1)}^j\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{j-1}{(-1)}^{q+1}\left(\genfrac{}{}{0pt}{}{2j}{2q+1}\right)\left(\genfrac{}{}{0pt}{}{q}{j}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^{2q+1}\\ =\sum _{q=0}^{\infty }{(-1)}^q\left[\sum _{j=q}^{\infty }{(-1)}^j\frac{{(ln\beta )}^{2j}}{(2j)!}\left(\genfrac{}{}{0pt}{}{2j}{2q}\right)\left(\genfrac{}{}{0pt}{}{(2q-1)/2}{j}\right)\right]{\left(\frac{ln\alpha }{ln\beta }\right)}^{2q}\\ -i\sum _{q=0}^{\infty }{(-1)}^q\left[\sum _{j=q+1}^{\infty }\frac{{(-1)}^j}{(2j)!}\left(\genfrac{}{}{0pt}{}{2j}{2q+1}\right)\left(\genfrac{}{}{0pt}{}{q}{j}\right){(ln\beta )}^{2j}\right]{\left(\frac{ln\alpha }{ln\beta }\right)}^{2q+1}\\ =\sum _{q=0}^{\infty }{(-1)}^q\left[\sum _{j=q}^{\infty }\frac{{(-1)}^j}{(2j)!}\left(\genfrac{}{}{0pt}{}{2j}{2q}\right)\left(\genfrac{}{}{0pt}{}{(2q-1)/2}{j}\right){(ln\beta )}^{2j}\right]{\left(\frac{ln\alpha }{ln\beta }\right)}^{2q},\end{array}$$

where any empty sum is taken to be zero and $\left(\genfrac{}{}{0pt}{}{r}{s}\right)=0$ for $s>r\ge 0$ and $r,s\in \mathbb{N}$. This means that

$$I_0\left(\sqrt{{(ln\alpha )}^2+{(ln\beta )}^2}\right)=\sum _{q=0}^{\infty }{(-1)}^q\left[\sum _{j=q}^{\infty }\frac{{(-1)}^j}{(2j)!}\left(\genfrac{}{}{0pt}{}{2j}{2q}\right)\left(\genfrac{}{}{0pt}{}{(2q-1)/2}{j}\right){(ln\beta )}^{2j}\right]{\left(\frac{ln\alpha }{ln\beta }\right)}^{2q}.$$

Further letting $\alpha \to 1$ and $\beta =e^x$ for $x\in \mathbb{R}$ reduces to (13). The proof of Theorem 5 is complete. □

Theorem 6.

For $x\in \mathbb{R}$, we have

$$I_0(x)=\frac{1}{coshx}\sum _{j=0}^{\infty }\left(\genfrac{}{}{0pt}{}{-1/2}{2j}\right)\frac{{(2x)}^{2j}}{(2j)!}.$$

(14)

Proof. 

When taking $k=2$, the right hand side of the Equation (8) in Theorem 3 becomes

$$\begin{array}{c}\sum _{j=0}^{\infty }{(-1)}^{2j}\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}{(-i)}^{2q}\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\left(\genfrac{}{}{0pt}{}{(2q-1)/2}{2j}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q\\ =\sum _{j=0}^{\infty }\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}{(-1)}^q\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\left(\genfrac{}{}{0pt}{}{(2q-1)/2}{2j}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q.\end{array}$$

This implies that

$$\sqrt{\alpha \beta }{I}_0\left(ln\sqrt{\frac{\alpha }{\beta }}\right)=\sum _{j=0}^{\infty }\frac{{(ln\beta )}^{2j}}{(2j)!}\sum _{q=0}^{2j}{(-1)}^q\left(\genfrac{}{}{0pt}{}{2j}{q}\right)\left(\genfrac{}{}{0pt}{}{(2q-1)/2}{2j}\right){\left(\frac{ln\alpha }{ln\beta }\right)}^q.$$

Further taking $\alpha \to 1$ reveals

$$\sqrt{\beta }{I}_0\left(ln\sqrt{\frac{1}{\beta }}\right)=\sum _{j=0}^{\infty }\frac{{(ln\beta )}^{2j}}{(2j)!}\left(\genfrac{}{}{0pt}{}{-1/2}{2j}\right).$$

Again letting $\beta =e^{\pm 2x}$ results in

$$e^{\pm x}{I}_0(x)=\sum _{j=0}^{\infty }\frac{{(\pm 2x)}^{2j}}{(2j)!}\left(\genfrac{}{}{0pt}{}{-1/2}{2j}\right)$$

which can be rewritten as (14). The proof of Theorem 6 is complete. □