Some useful integral and series representations related to these means will be developed next. First, make the substitution \(t = ( y^{p} +1)^{-1}\) with \(dy =- p^{-1} t^{-1-1/p} (1- t)^{1/p -1} \,dt\) so \(c_{p}\) can be put in the form of the beta function integral
$$ \frac{1}{c_{p}} = \frac{1}{p} \int _{0}^{1} t^{1/p -1} ( 1 -t)^{1/p -1} \,dt = \frac{\varGamma (\frac{1}{p})^{2}}{p \varGamma (\frac{2}{p})}. $$
(12)
Returning to the integral for \(M_{p} (a,b)\), let us make the following change of variable:
$$ x^{p} = a^{p} \biggl( \frac{1}{t} -1\biggr) = a^{p} \biggl( \frac{1-t}{t}\biggr) , \qquad dx =- \frac{a}{p} t^{-1-1/p} (1- t)^{1/p -1} \,dt. $$
In this case, the integral for \(M_{p} (a,b)\) takes the following form:
$$ \frac{1}{M_{p} (a,b)} = c_{p} \frac{a}{p} \int _{0}^{1} \frac{t^{1- 1/p} (1-t)^{1/p-1}}{ \frac{a}{t^{1/p}} (a^{p} \frac{1-t}{t} + b ^{p})^{1/p}} \,dt = \frac{c_{p}}{p} \int _{0}^{1} \frac{t^{1/p-1} ( 1 - t)^{1/p-1}}{(a^{p} (1-t) + b^{p} t)^{1/p}} \,dt. $$
(13)
Theorem 2
The function
\(1/ c_{p}\)
increases monotonically for all
\(p \in (0, \infty )\). Moreover, as
\(p \rightarrow \infty \), this function admits the asymptotic expansion
$$ \frac{1}{c_{p}} = 2 - \frac{\pi ^{2}}{2 p^{2}} + \frac{4 \zeta (3)}{p ^{3}} + O \biggl( \frac{1}{p^{4}} \biggr). $$
(14)
Proof
Beginning with (12) and setting \(h (p) =1/ c _{p}\), we have
$$ \log \bigl( h (p)\bigr) = \log \biggl( \frac{\varGamma (\frac{1}{p})^{2}}{p \varGamma ( \frac{2}{p})} \biggr). $$
Differentiating both sides of this with respect to p, it follows that
$$ \frac{h' (p)}{h (p)} = \frac{1}{p^{2}} \biggl( 2 \psi \biggl(\frac{2}{p} \biggr) - 2 \psi \biggl( \frac{1}{p}\biggr) -p\biggr). $$
It suffices to show the quantity in brackets is always positive. To this end, substitute the series form of \(\psi (z)\) to obtain
$$\begin{aligned} &2 \psi \biggl( \frac{2}{p}\biggr) - 2 \psi \biggl( \frac{1}{p} \biggr) -p \\ &\quad = -p + \frac{2}{p} \sum_{n=1}^{\infty } \frac{1}{n (n + \frac{2}{p})} + 2 p - \frac{1}{p} \sum_{n=1}^{\infty } \frac{1}{n (n + \frac{1}{p})} - p \\ &\quad = \frac{1}{p} \sum_{n=1}^{\infty } \frac{1}{ (n + \frac{1}{p})(n + \frac{2}{p})} >0. \end{aligned}$$
Since \(h (p) \rightarrow 0\) as \(p \rightarrow 0^{+}\) and \(h' (p) >0\) on \((0, \infty )\), it follows that the function \(h (p)\) is positive and strictly increasing on \((0, \infty )\). □
The reciprocal of \(M_{p} (a,b)\) can be expanded into an infinite series. This expansion will be seen to have several uses. Let us introduce the following expressions:
$$ (a, k) = a (a+1) \cdots (a+k-1), \qquad (a,0)=1, \qquad (a,-1)=0, \quad a \neq 0. $$
Theorem 3
Given
\(a,b >0\)
and
\(p \in (0, \infty )\), the following expansion holds:
$$ \frac{1}{M_{p} (a,b)} = \frac{1}{ \max (a,b)} \sum_{k=0}^{\infty } \frac{ ( \frac{1}{p} , k)}{(\frac{2}{p} , k) k!} \biggl[ 1 - \biggl( \frac{ \min (a,b)}{\max (a,b)} \biggr)^{p} \biggr]^{k}, $$
(15)
with
\(0! =1\).
Proof
Both sides of (15) are equal to \(1/a\) when \(a=b\). Assume without loss of generality that \(a > b >0\). Set \(\beta = 1 - (b/a)^{p}\) so we have \(0 < \beta <1\) and
$$ a^{p} (1-t) + b^{p} t = a^{p} (1 - \beta t). $$
Expanding the denominator of (13) into power series gives
$$ \frac{1}{[ a^{p} (1-t) + b^{p} t]^{1/p}} = \frac{1}{a} ( 1 - \beta t)^{-1/p} = \frac{1}{a} \sum_{k=0}^{\infty } \biggl( \frac{1}{p} ,k \biggr) \frac{\beta ^{k}}{k!} t^{k}. $$
(16)
The series on the right-hand side of (16) converges when \(\beta \in (0,1)\), and so integration of the series term by term is justified. Substitute (16) into (13) to obtain
$$\begin{aligned} \frac{1}{M_{p} (a,b)} =& \frac{1}{a B (\frac{1}{p}, \frac{1}{p})} \int _{0}^{1} \sum_{k=0}^{\infty } \prod_{m=0}^{k-1} \biggl( \frac{1}{p} + m\biggr) \frac{\beta ^{k}}{k!} t^{k+1/p-1} (1- t)^{1/p-1} \,dt \\ =& \frac{1}{a B( \frac{1}{p}, \frac{1}{p})} \sum_{k=0}^{\infty } \biggl( \frac{1}{p}, k\biggr) \frac{\beta ^{k}}{k!} \int _{0}^{1} t^{k+1/p-1} (1-t)^{1/p-1} \,dt \\ =& \frac{1}{a} \sum_{k=0}^{\infty } \biggl( \frac{1}{p} , k\biggr) \frac{\alpha ^{k}}{k!} \frac{B ( k + \frac{1}{p}, \frac{1}{p})}{B ( \frac{1}{p}, \frac{1}{p})}. \end{aligned}$$
(17)
Expressing the beta functions in (17) in terms of the gamma function, we find that
$$ \frac{B ( k + \frac{1}{p}, \frac{1}{p})}{B ( \frac{1}{p}, \frac{1}{p})} = \frac{\varGamma (k + \frac{1}{p}) \varGamma (\frac{1}{p})}{\varGamma ( \frac{1}{p} ) \varGamma ( k + \frac{2}{p})} = \prod_{m=0}^{k-1} \frac{( \frac{1}{p} +m)}{( \frac{2}{p} + m)} = \frac{( \frac{1}{p}, k)}{( \frac{2}{p}, k)}. $$
(18)
Substituting (18) into (17), the required result (15) is obtained. □
Another expansion which is relevant to \(M_{p} (a,b)\) is given in the following theorem.
Theorem 4
Given
\(a, b>0\)
and
\(p \in (0, \infty )\),
$$ \frac{1}{M_{p} (a,b)} = \biggl( \frac{2}{a^{p} + b^{p}} \biggr)^{1/p} \sum _{k=0} ^{\infty } \frac{( \frac{1}{p}, k)}{k!} \frac{(\frac{1}{p} , 2k)}{( \frac{2}{p} ,2k) } \cdot \biggl( \frac{a^{p} - b^{p}}{a^{p} + b^{p}} \biggr)^{2k}. $$
(19)
Proof
Completing the square, we can write
$$\begin{aligned} \bigl( x^{p} + a^{p} \bigr) \bigl( x^{p} + b^{p}\bigr) =& \biggl( x^{p} - \frac{a^{p} + b^{p}}{2} \biggr)^{2} - \biggl( \frac{a^{p} - b^{p}}{2} \biggr)^{2} \\ =& \biggl(x^{p} + \frac{a^{p} + b ^{p}}{2} \biggr)^{2} \bigl( 1 + \tau (x)\bigr), \end{aligned}$$
(20)
where \(\tau (x)\) is defined to be
$$ \tau (x) = \frac{ \frac{a^{p} - b^{p}}{2}}{x^{p} + \frac{a ^{p} + b^{p}}{2}}. $$
Clearly, \(| \tau (x)| < 1\) and using (20), the following expansion holds:
$$\begin{aligned} \bigl[ \bigl( x^{p} + a^{p} \bigr) \bigl( x^{p} + b^{p} \bigr) \bigr]^{-1/p} =& \biggl(1 + \frac{a^{p} + b ^{p}}{2} \biggr)^{-2/p} \bigl(1- \tau (x)^{2}\bigr)^{-1/p} \\ =& \biggl( x^{p} + \frac{a^{p} + b^{p}}{2} \biggr)^{-2/p} \sum _{k=0}^{\infty } \frac{( \frac{1}{p} ,k )}{k!} \tau (x)^{k}. \end{aligned}$$
(21)
Substituting expansion (21) into the integral (13) for \(M_{p} (a,b)^{-1}\), we obtain
$$\begin{aligned} \frac{1}{M_{p} (a,b)} =& c_{p} \int _{0}^{\infty } \frac{1}{(x^{p} + \frac{a^{p} + b^{p}}{2} )^{2/p}} \sum _{k=0}^{\infty } \frac{( \frac{1}{p} ,k ) }{k!} {\tau (x)^{2k}} dx \\ =& c_{p} \int _{0}^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b ^{p}}{2})^{2/p}} + c_{p} \sum_{k=1}^{\infty } \frac{( \frac{1}{p},k)}{k!} \int _{0}^{\infty } \frac{\tau (x)^{2k}}{(x^{p} + \frac{a^{p} + b^{p}}{2} )^{2/p}} dx \\ =& \biggl( \frac{a^{p} + b^{p}}{2}\biggr)^{-1/p} + c_{p} \sum _{k=1}^{\infty } \frac{( \frac{1}{p} ,k ) }{k!} \int _{0}^{\infty } \frac{ \tau (x)^{2k}}{(x^{p} + \frac{a^{p} + b^{p}}{2} )^{2/p}} dx \\ =& \biggl( \frac{2}{a^{p} + b^{p}}\biggr)^{1/p} + c_{p} \sum _{k=1}^{\infty } \frac{( \frac{1}{p},k)}{k!} \biggl( \frac{a^{p} - b^{p}}{2} \biggr)^{2k} \int _{0} ^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b^{p}}{2} )^{2k +2/p}}. \end{aligned}$$
(22)
Consider the integral apart from (22) and make use of the substitution
$$ x = \biggl( \frac{a^{p} + b^{p}}{2} \biggr)^{1/p} y. $$
The integral in (22) takes the form
$$ \int _{0}^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b^{p}}{2})^{2k+2/p}} = \biggl( \frac{a ^{p} + b^{p}}{2} \biggr)^{-2k-1/p} \int _{0}^{\infty } \frac{dy}{(y ^{p} +1)^{2k+2/p}}. $$
Finally, introduce the change of variable \(y^{p} +1 = t^{-1}\) into the integral so it takes the form
$$ \int _{0}^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b^{p}}{2})^{2k+2/p}} = \biggl( \frac{a ^{p} + b^{p}}{2}\biggr)^{-2k-1/p} \frac{1}{p} B \biggl( 2k + \frac{1}{p}, \frac{1}{p}\biggr). $$
(23)
Multiply (23) by \(c_{p}\) from (12) to obtain
$$ c_{p} \int _{0}^{\infty } \frac{dx}{(x^{p} + \frac{a^{p} + b ^{p}}{2})^{2k+2/p}} = \biggl( \frac{a^{p} + b^{p}}{2}\biggr)^{-2k-1/p} \frac{B (2k + \frac{1}{p}, \frac{1}{p})}{B ( \frac{1}{p}, \frac{1}{p})} = \biggl( \frac{2}{a ^{p} + b^{p}}\biggr)^{2k +1/p} \frac{(\frac{1}{p} ,2k)}{( \frac{2}{p} , 2k)}. $$
Substituting this integral into (22), we arrive at the desired expansion
$$ \frac{1}{M_{p} (a,b)} = \biggl( \frac{2}{a^{p} + b^{p}}\biggr)^{1/p} + \biggl( \frac{2}{a ^{p} + b^{p}}\biggr)^{1/p} \sum_{k=1}^{\infty } \frac{( \frac{1}{p} ,k )}{k!} \frac{( \frac{1}{p} , 2k)}{(\frac{2}{p} ,2k)} \biggl( \frac{a ^{p} - b^{p}}{a^{p} + b^{p}} \biggr)^{2k}. $$
□