Integral results in difference of means $\pi(\frac{a+b}{2} - \sqrt{ab})$ [duplicate]

$$\int_a^b \left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{1/2}dr = \pi\left(\frac{a+b}{2} - \sqrt{ab}\right)$$

What an interesting integral! What strikes me is that the result involves the difference of the arithmetic and geometric mean. Is there an innate geometric explanation that corresponds to this result? And can we generalize this integral to, say, the mean of three or more items?

Some background on where I saw it and how to solve it. This arises in calculating the action variable $I$ for the Kepler problem (Hamiltonian $H=\frac{p_r^2}{2m}+\frac{p_\phi^2}{2mr}-\frac{k}{r}$). The $a,b$ are the minimal and maximal $r$ from the origin set at the focus of an ellipse (i.e. the perihelion/aphelion). See David Tong's Classical Dynamics notes $\S$4.5.4.

I was able to solve the integral by using the third Euler substitution, letting

$$\left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{1/2}=\frac{1}{r}\sqrt{-(r-a)(r-b)}=\frac{1}{r}(r-a)t$$

giving $r=\frac{b+at^2}{1+t^2}$ and

$$2(b-a)\left\{\int_{t(r_2)=0}^{t(r_1)=\infty}\frac{t^2}{(1+t^2)^2}dt - \int_{t(r_2)=0}^{t(r_1)=\infty}\frac{\frac{a}{b}t^2}{(1+t^2)(1+\frac{a}{b}t^2)}dt\right\}$$ $$=\pi\left(\frac{b-a}{2}\right) + \pi\left(a-\sqrt{ab}\right)=\pi\left(\frac{a+b}{2} - \sqrt{ab}\right).$$

Let $n\in\mathbb{N}=\{1,2,\dotsc\}$ and $\boldsymbol{a}=(a_1,a_2,\dotsc,a_n)$ be a positive sequence, that is, $a_k>0$ for $1\le k\le n$. The arithmetic and geometric means $A_n(\boldsymbol{a})$ and $G_n(\boldsymbol{a})$ of the positive sequence $\boldsymbol{a}$ are defined respectively as \begin{equation*} A_n(\boldsymbol{a})=\frac1n\sum_{k=1}^na_k \quad \text{and}\quad G_n(\boldsymbol{a})=\Biggl(\prod_{k=1}^na_k\Biggr)^{1/n}. \end{equation*} For $z\in\mathbb{C}\setminus(-\infty,-\min\{a_k,1\le k\le n\}]$ and $n\ge2$, let $\boldsymbol{e}=(\overbrace{1,1,\dotsc,1}^{n})$ and \begin{equation*} G_n(\boldsymbol{a}+z\boldsymbol{e})=\Biggl[\prod_{k=1}^n(a_k+z)\Biggr]^{1/n}. \end{equation*}

In Theorem 1.1 of the paper [1] below, by virtue of the Cauchy integral formula in the theory of complex functions, the following integral representation was established.

Theorem 1.1. Let $\sigma$ be a permutation of the sequence $\{1,2,\dotsc,n\}$ such that the sequence $\sigma(\boldsymbol{a})=\bigl(a_{\sigma(1)},a_{\sigma(2)},\dotsc,a_{\sigma(n)}\bigr)$ is a rearrangement of $\boldsymbol{a}$ in an ascending order $a_{\sigma(1)}\le a_{\sigma(2)}\le \dotsm \le a_{\sigma(n)}$. Then the principal branch of the geometric mean $G_n(\boldsymbol{a}+z\boldsymbol{e})$ has the integral representation \begin{equation}\label{AG-New-eq1}\tag{1} G_n(\boldsymbol{a}+z\boldsymbol{e})=A_n(\boldsymbol{a})+z-\frac1\pi\sum_{\ell=1}^{n-1}\sin\frac{\ell\pi}n \int_{a_{\sigma(\ell)}}^{a_{\sigma(\ell+1)}} \Biggl|\prod_{k=1}^n(a_k-t)\Biggr|^{1/n} \frac{\textrm{d}\,t}{t+z} \end{equation} for $z\in\mathbb{C}\setminus(-\infty,-\min\{a_k,1\le k\le n\}]$.

Taking $z=0$ in the integral representation \eqref{AG-New-eq1} yields \begin{equation}\label{AG-ineq-int}\tag{2} G_n(\boldsymbol{a})=A_n(\boldsymbol{a})-\frac1\pi\sum_{\ell=1}^{n-1}\sin\frac{\ell\pi}n \int_{a_{\sigma(\ell)}}^{a_{\sigma(\ell+1)}} \Biggl[\prod_{k=1}^n|a_k-t|\Biggr]^{1/n} \frac{\textrm{d}\,t}{t}\le A_n(\boldsymbol{a}). \end{equation} Taking $n=2,3$ in \eqref{AG-ineq-int} gives $$ \frac{a_1+a_2}{2}-\sqrt{a_1a_2}\,=\frac1\pi\int_{a_1}^{a_2} \sqrt{\biggl(1-\frac{a_1}{t}\biggr) \biggl(\frac{a_2}{t}-1\biggr)}\, \textrm{d}\,t\ge0 $$ and $$ \frac{a_1+a_2+a_3}{3}-\sqrt[3]{a_1a_2a_3}\, =\frac{\sqrt{3}\,}{2\pi} \int_{a_1}^{a_3} \sqrt[3]{\biggl| \biggl(1-\frac{a_1}{t}\biggr) \biggl(1-\frac{a_2}{t}\biggr) \biggl(1-\frac{a_3}{t}\biggr)\biggr|}\,\textrm{d}\,t\ge0 $$ for $0<a_1\le a_2\le a_3$.

Weighted version of the integral representation \eqref{AG-New-eq1} can be found in the paper [2] below. We recite the weighted version as follows.

For $n\ge2$, $\boldsymbol{a}=(a_1,a_2,\dotsc,a_n)$, and $\boldsymbol{w}=(w_1,w_2,\dotsc,w_n)$ with $a_k, w_k>0$ and $\sum_{k=1}^nw_k=1$, the weighted arithmetic and geometric means $A_{w,n}(\boldsymbol{a})$ and $G_{w,n}(\boldsymbol{a})$ of $\boldsymbol{a}$ with the positive weight $\boldsymbol{w}$ are defined respectively as \begin{equation} A_{\boldsymbol{w},n}(\boldsymbol{a})=\sum_{k=1}^nw_ka_k \end{equation} and \begin{equation} G_{\boldsymbol{w},n}(\boldsymbol{a})=\prod_{k=1}^na_k^{w_k}. \end{equation} Let us denote $\alpha=\min\{a_k,1\le k\le n\}$. For a complex variable $z\in\mathbb{C}\setminus(-\infty,-\alpha]$, we introduce the complex function \begin{equation}\label{complex-geometric-mean} G_{\boldsymbol{w},n}(\boldsymbol{a}+z)=\prod_{k=1}^n(a_k+z)^{w_k}. \end{equation} In Section 3 of the paper [2] below, with the aid of the Cauchy integral formula in the theory of complex functions, the following integral representation was established.

Theorem 3.1. Let $0<a_k\le a_{k+1}$ for $1\le k\le n-1$ and $z\in\mathbb{C}\setminus(-\infty,-a_1]$. Then the principal branch of the weighted geometric mean $G_{\boldsymbol{w},n}(\boldsymbol{a}+z)$ with a positive weight $\boldsymbol{w}=(w_1,w_2,\dotsc,w_n)$ has the integral representation \begin{equation}\label{AG-New-eq1-weighted}\tag{3} G_{\boldsymbol{w},n}(\boldsymbol{a}+z)=A_{\boldsymbol{w},n}(\boldsymbol{a})+z-\frac1\pi\sum_{\ell=1}^{n-1}\sin\Biggl[\Biggl(\sum_{k=1}^{\ell}w_k\Biggr)\pi\Biggr] \int_{a_\ell}^{a_{\ell+1}} \prod_{k=1}^n|a_k-t|^{w_k} \frac{\textrm{d}\,t}{t+z}. \end{equation} Letting $z=0$ in the integral representation \eqref{AG-New-eq1-weighted} gives \begin{equation}\label{AG-New-eq1-weighted-z=0}\tag{4} G_{\boldsymbol{w},n}(\boldsymbol{a})=A_{\boldsymbol{w},n}(\boldsymbol{a})-\frac1\pi\sum_{\ell=1}^{n-1} \sin\Biggl[\Biggl(\sum_{k=1}^{\ell}w_k\Biggr)\pi\Biggr] \int_{a_\ell}^{a_{\ell+1}} \prod_{k=1}^n|a_k-t|^{w_k} \frac{\textrm{d}\,t}{t}\le A_{\boldsymbol{w},n}(\boldsymbol{a}). \end{equation} Setting $n=2$ in \eqref{AG-New-eq1-weighted-z=0} leads to \begin{equation}\label{AG-New-n=2-weighted-z=0}\tag{5} a_1^{w_1}a_2^{w_2}=w_1a_1+w_2a_2-\frac{\sin(w_1\pi)}\pi \int_{a_1}^{a_2} \biggl(1-\frac{a_1}{t}\biggr)^{w_1} \biggl(\frac{a_2}{t}-1\biggr)^{w_2} \textrm{d}\,t \le w_1a_1+w_2a_2 \end{equation} for $w_1,w_2>0$ such that $w_1+w_2=1$.

There have existed more closely related conclusions published in the following references below.

References

1. Feng Qi, Xiao-Jing Zhang, and Wen-Hui Li, Levy--Khintchine representation of the geometric mean of many positive numbers and applications, Mathematical Inequalities & Applications 17 (2014), no. 2, 719--729; available online at https://doi.org/10.7153/mia-17-53.
2. Feng Qi, Xiao-Jing Zhang, and Wen-Hui Li, An integral representation for the weighted geometric mean and its applications, Acta Mathematica Sinica-English Series 30 (2014), no. 1, 61--68; available online at https://doi.org/10.1007/s10114-013-2547-8.
3. Feng Qi and Bai-Ni Guo, The reciprocal of the weighted geometric mean is a Stieltjes function, Boletin de la Sociedad Matematica Mexicana, Tercera Serie 24 (2018), no. 1, 181--202; available online at https://doi.org/10.1007/s40590-016-0151-5.
4. Feng Qi and Bai-Ni Guo, The reciprocal of the weighted geometric mean of many positive numbers is a Stieltjes function, Quaestiones Mathematicae 41 (2018), no. 5, 653--664; available online at https://doi.org/10.2989/16073606.2017.1396508.
5. Feng Qi and Dongkyu Lim, Integral representations of bivariate complex geometric mean and their applications, Journal of Computational and Applied Mathematics 330 (2018), 41--58; available online at https://doi.org/10.1016/j.cam.2017.08.005.
6. Feng Qi, Bounding the difference and ratio between the weighted arithmetic and geometric means, International Journal of Analysis and Applications 13 (2017), no. 2, 132--135.
7. Feng Qi and Bai-Ni Guo, The reciprocal of the geometric mean of many positive numbers is a Stieltjes transform, Journal of Computational and Applied Mathematics 311 (2017), 165--170; available online at https://doi.org/10.1016/j.cam.2016.07.006.
8. Feng Qi, Xiao-Jing Zhang, and Wen-Hui Li, The harmonic and geometric means are Bernstein functions, Boletin de la Sociedad Matematica Mexicana, Tercera Serie 23 (2017), no. 2, 713--736; available online at https://doi.org/10.1007/s40590-016-0085-y.
9. Feng Qi, Xiao-Jing Zhang, and Wen-Hui Li, An elementary proof of the weighted geometric mean being a Bernstein function, University Politehnica of Bucharest Scientific Bulletin Series A---Applied Mathematics and Physics 77 (2015), no. 1, 35--38.
10. Bai-Ni Guo and Feng Qi, On the degree of the weighted geometric mean as a complete Bernstein function, Afrika Matematika 26 (2015), no. 7, 1253--1262; available online at https://doi.org/10.1007/s13370-014-0279-2.
11. Feng Qi, Xiao-Jing Zhang, and Wen-Hui Li, Levy--Khintchine representations of the weighted geometric mean and the logarithmic mean, Mediterranean Journal of Mathematics 11 (2014), no. 2, 315--327; available online at https://doi.org/10.1007/s00009-013-0311-z.
12. Feng Qi and Bai-Ni Guo, Levy--Khintchine representation of Toader--Qi mean, Mathematical Inequalities & Applications 21 (2018), no. 2, 421--431; available online at https://doi.org/10.7153/mia-2018-21-29.
13. Feng Qi, Viera Cernanova, Xiao-Ting Shi, and Bai-Ni Guo, Some properties of central Delannoy numbers, Journal of Computational and Applied Mathematics 328 (2018), 101--115; available online at https://doi.org/10.1016/j.cam.2017.07.013.

Let $0\le a\le b,$ then \begin{align} &I_1 = \int\limits_a^b \left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{-1/2}\,\mathrm dr = \int\limits_a^b \dfrac{r}{\sqrt{(r-a)(b-r)}}\,\mathrm dr\\ &=\dfrac12\int\limits_a^b \dfrac{2r-(a+b)}{\sqrt{(r-a)(b-r)}}\,\mathrm dr+\dfrac{a+b}2\int\limits_a^b \dfrac{a+b}{\sqrt{(r-a)(b-r)}}\,\mathrm dr\\ &=-\dfrac12\int\limits_a^b \dfrac{\mathrm d((r-a)(b-r))}{\sqrt{(r-a)(b-r)}} + \dfrac{a+b}2\int\limits_a^b \dfrac{\mathrm dr}{\sqrt{\left(\dfrac{b-a}2\right)^2-\left(r-\dfrac{b+a}2\right)^2}}\\ &=-\sqrt{(r-a)(b-r)}\Bigg|_a^b + \dfrac{a+b}2\arcsin\dfrac{2r-(b+a)}{b-a}\Bigg|_a^b = \pi\dfrac{a+b}2,\\ &I_2 = \int\limits_a^b \left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}^{-1/2}\left(1-\left\{ \left(1-\frac{a}{r}\right)\left(\frac{b}{r}-1\right)\right\}\right)\,\mathrm dr\\ & = \int\limits_a^b \dfrac{2r^2-(a+b)r+ab}{\sqrt{(r-a)(b-r)}}\,\dfrac{\mathrm dr}r = \int\limits_a^b \dfrac{2r-(a+b)}{\sqrt{(r-a)(b-r)}}\,\mathrm dr - ab\int\limits_a^b \dfrac{1}{\sqrt{(1-\frac ar)(\frac br-1)}}\,\mathrm d\left(\dfrac1r\right)\\ &=-\sqrt{(r-a)(b-r)}\Bigg|_a^b + ab\int\limits_{1/b}^{1/a}\dfrac{\mathrm dt}{\sqrt{(1-at)(bt-1)}} = \sqrt{ab}\int\limits_{1/b}^{1/a}\dfrac{\mathrm dt}{\sqrt{(\frac1a-t)(\frac1b-t)}}\\ &= \sqrt{ab}\int\limits_{1/b}^{1/a}\dfrac{\mathrm dt}{\sqrt{\left(\frac{\frac1a-\frac1b}2\right)^2 - \left(t-\frac{\frac1a+\frac1b}2\right)^2}} = \sqrt{ab}\arcsin\dfrac{{2t-\left(\frac1a+\frac1b\right)}}{{\frac1a-\frac1b}}\Biggr|_{1/b}^{1/a} = \pi ab.\\ \end{align} This allows to present the issue integral (or the square under the graph of the according function) as the square between two other similar functions, which present AM and GM of $a$ and $b$ (see also Wolphram Alpha plot).