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We are interested in the following integration: $$ \int_a^b \left[ \left(1 - \frac{a}{x} \right)\left(\frac{b}{x}-1 \right) \right]^{1/2} dx. $$

I try to use substitution with: $$u = \left(1-\tfrac{a}{x}\right)\left(\tfrac{b}{x} -1\right) = (a+b)x^{-1} - 1 - (ab)x^{-2}.$$ $$ du = \left( (-a-b)x^{-2} + (2ab)(x^{-3}) \right)dx $$

or
$$ dx = du \left( \left( (-a-b)x^{-2} + (2ab)(x^{-3}) \right)^{-1} \right)$$

Now the integration becomes something ugly. I try integration by parts on the ugly part, and it becomes even uglier. Obviously, I don't see some easier ways. I would appreciate some advices or the solution!

Edit: It may be easier to see the problem as

$$ \int_a^b \left[ (a-x)(x-b)x^{-2} \right]^{1/2} dx .$$

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    $\begingroup$ Mind if I ask where this came from? If $a<b$, the answer is supposedly $\frac\pi2(\sqrt b-\sqrt a)^2$, though I really cannot see how to arrive at it... :/ $\endgroup$ Commented Sep 4, 2019 at 1:18
  • $\begingroup$ There is no details given. This is an exercise question in the 'warming up' section of a stats course. I tried many ways, nothing worked so far. $\endgroup$ Commented Sep 4, 2019 at 1:21

3 Answers 3

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Rescale the integration range with the following variable $u = \frac{x-a}{b-a}$ and the shorthand $q=\frac{b}{a}-1$ to simplify the integral

$$ I = \int_a^b \left[ \left(1 - \frac{a}{x} \right)\left(\frac{b}{x}-1 \right) \right]^{1/2} dx=aq^2\int_0^1 \frac{\sqrt{u(1-u)}}{1+qu} du$$

Then, let $u=\sin^2 t$

\begin{align} I &= 2a\int_0^{\pi/2} \frac{q^2\sin^2 t\cos^2 t}{1+q\sin^2 t}dt\\ &= 2a\int_0^{\pi/2}\left( 1+q\cos^2 t-\frac{1+q}{1+q\sin^2 t}\right) dt \end{align}

where

$$\int_0^{\pi/2}\left( 1+q\cos^2 t\right) dt=\frac{\pi}{2}\left(1+\frac{q}{2}\right)$$

$$\int_0^{\pi/2} \frac{1+q}{1+q\sin^2 t}dt=\int_0^{\infty} \frac{(1+q) d(\tan t)}{1+(1+q)\tan^2 t}=\frac{\pi}{2}\sqrt{1+q}$$

Thus

$$I= \frac{\pi}{2}(b+a-2\sqrt{ab})$$

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Let $0<a<b$.

By rewriting a bit we get

$$I(a,b)=\int_a^b\frac{\sqrt{(x-a)(b-x)}}x~\mathrm dx$$

Completing the square gives us

$$I(a,b)=\int_a^b\frac{\sqrt{\frac14(b-a)^2-\left(x-\frac{b+a}2\right)^2}}x~\mathrm dx$$

Substitute $\displaystyle\frac{b-a}2\sin(\theta)=x-\frac{b+a}2$ to get

$$I(a,b)=\frac{(b-a)^2}2\int_{-\pi/2}^{\pi/2}\frac{\cos^2(\theta)}{(b-a)\sin(\theta)+b+a}~\mathrm d\theta$$

By manipulating symmetry,

$$I(a,b)=\frac{(b-a)^2}4\int_0^{2\pi}\frac{\cos^2(\theta)}{(b-a)\sin(\theta)+b+a}~\mathrm d\theta$$

from which one may conclude

$$I(a,b)=\frac\pi2(\sqrt b-\sqrt a)^2$$

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  • $\begingroup$ Thank you! What is the name of $\displaystyle\frac{b-a}2\sin(\theta)=x-\frac{b+a}2$ substitution? I would like to understand it. $\endgroup$ Commented Sep 4, 2019 at 1:45
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    $\begingroup$ :p let $a>0\dots$ $\endgroup$ Commented Sep 4, 2019 at 1:50
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    $\begingroup$ Can you explain your final conclusion in more detail, I don't quite follow it? $\endgroup$
    – B. Mehta
    Commented Sep 4, 2019 at 2:37
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    $\begingroup$ Short and elegant ! $\endgroup$ Commented Sep 4, 2019 at 5:01
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    $\begingroup$ @B.Mehta see here for example. $\endgroup$ Commented Sep 4, 2019 at 16:54
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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.
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