5
$\begingroup$

I was trying to solve the following integral: $$ \begin{equation} \int _0^{\alpha }\int _0^{\beta }\min (x,y)dydx \tag{1} \label{eq:1} \end{equation} $$ with $\alpha,\beta > 0$, (generally, $\alpha \ne \beta$).

I've found a question from MSE (how to solve double integral of a min function) where an user suggested the following equivalence: $$ \begin{equation} \int_0^{\beta } \min (x,y) \, dy =\int_0^x \min (x,y) \, dy+\int_x^{\beta } \min(x,y) \, dy \\ =\int_0^x y \, dy+\int_x^{\beta } x \, dy \tag{2} \label{eq:2} \end{equation} $$ The explanation is in link above and I also found it by myself. The issue is that the $(1)$ and the $(2)$ inserted in $(1)$: $$ \int _0^{\alpha }\int _0^{\beta }\min (x,y)dydx \tag{1} $$ $$ \int _0^{\alpha } \bigg( \int_0^x y \, dy+\int_x^{\beta } x \, dy \bigg) dx \tag{3} $$ are giving me different results. In fact, generally, through different cases of $\alpha$ and $\beta$ I set, it seems that, if $\beta \ge \alpha$, the $(1)$ and $(3)$ are equivalent, else they aren't. What am I missing? Have I to suppose some conditions for the $(2)$ is effective? Is there a problem of integral interchange?

$\endgroup$
  • $\begingroup$ It might be useful to sketch pictures of the integration region for the cases (1) $\alpha>\beta$, (2) $\alpha<\beta$, and (3) $\alpha=\beta$. $\endgroup$ – Mark Viola Jul 4 '17 at 15:28
  • $\begingroup$ Yes. I thought so. The area I consider is rectangular and I have to plot the function and see how $\alpha$ and $\beta$ are working. But, is there a way to do it without sketch? I know that $\min(x,y)=\frac{x+y-|y-x|}{2}$ and I solve the problem with that, but I want to know why the reasoning above is not working, if that is true or false and why. $\endgroup$ – Antonio Placentino Jul 4 '17 at 15:36
1
$\begingroup$

By definition

$$\min(x,y)=\begin{cases}x\le y\to x\\x\ge y\to y.\end{cases}$$

Then, assuming $\alpha\le\beta$ you can decompose the domain using

$$I=\int_{x=0}^\alpha\int_{y=0}^\beta\min(x,y)\,dy\,dx=\int_{x=0}^\alpha\int_{y=0}^x y\,dy\,dx+\int_{x=0}^\alpha\int_{y=x}^\beta x\,dy\,dx.$$

The condition is required because $y$ may not exceed $\beta$.

This gives

$$I=\int_0^\alpha\left(\frac{x^2}2+\beta x-x^2\right)dx=\frac{\alpha^2\beta}2-\frac{\alpha^3}6=\alpha^2\frac{3\beta-\alpha}6.$$

And by swapping $\alpha,\beta$, $$\alpha\ge\beta\to I=\beta^2\frac{3\alpha-\beta}6.$$

$\endgroup$
  • $\begingroup$ That "The condition is required because $y$ may not exceed $β$." is what I was looking for, to complete my question. In fact, I haven't seen that, if $\alpha > \beta$, it is wrong to consider the $(2)$ from question box: $$\int_0^{\beta } \min (x,y) \, dy =\int_0^x y \, dy+\int_x^{\beta } x \, dy \tag{2}$$ because it says that $y$ goes from $0$ to $x$ and $x$ can go until $\alpha$ where $y$ is not condiderated ($0<y<\beta<\alpha$). Thanks for that! $\endgroup$ – Antonio Placentino Jul 4 '17 at 17:22
  • $\begingroup$ Moral: Control always the domain, if the calculus bring to a wrong domain of integration, go back and find another way! :) Thank you for your willingness to help. $\endgroup$ – Antonio Placentino Jul 4 '17 at 17:27
3
$\begingroup$

If $\alpha<\beta$, we have

$$\begin{align} \int_0^\alpha \int_0^\beta \min(x,y)\,dy\,dx&=\int_0^\alpha \int_0^\alpha \min(x,y)\,dy\,dx+\int_0^\alpha \int_\alpha^\beta \min(x,y)\,dy\,dx\\\\ &=\int_0^\alpha \left(\int_0^x \min(x,y)\,dy\,dx+\int_x^\alpha \min(x,y)\,dy\,dx\right)+\int_0^\alpha \int_\alpha^\beta \min(x,y)\,dy\,dx\\\\ &=\int_0^\alpha \int_0^x y\,dy\,dx+\int_0^\alpha \int_x^\alpha x\,dy\,dx+\int_0^\alpha \int_\alpha^\beta x\,dy\,dx\\\\ &=\int_0^\alpha \int_0^x y\,dy\,dx+\int_0^\alpha \int_x^\beta x\,dy\,dx \end{align}$$

Can you finish now?

$\endgroup$
  • $\begingroup$ Yes, it is. For $\beta > \alpha$, if we're following the $(3)$ from the question box, we get the expression you generously wrote. I also see that, if $\alpha > \beta$, I cannot do the same transformation of the internal integral from your expression above. Fortunally, thanks to your suggestion, I can find supposition for the interchange of integral and have this for $\alpha > \beta$: $\endgroup$ – Antonio Placentino Jul 4 '17 at 16:53
  • $\begingroup$ $\int _0^{\alpha }\int _0^{\beta }\min (x,y)dydx=\int _0^{\beta }\int _0^{\alpha }\min (x,y)dxdy=\int _{\beta }^{\alpha }\int _0^{\beta }ydydx+\int _0^{\beta }\int _x^{\beta }xdydx+\int _0^{\beta }\int _0^xydydx$ $\endgroup$ – Antonio Placentino Jul 4 '17 at 16:54
  • $\begingroup$ Fubini's Theorem permits the interchange of the order of integration. $\endgroup$ – Mark Viola Jul 4 '17 at 16:54
  • $\begingroup$ Thanks very much for your patience and willingness. I get to the answer thanks to you (and Fubini's Theorem, obviously). Now, I'd like to know why I cannot do the same procedure for $\alpha > \beta$, as I do for $\beta > \alpha$, or why I have to interchange the integral for solving the double integral... I guess, I have to sketch anyway, but if I get a proof of what, it'd be great! $\endgroup$ – Antonio Placentino Jul 4 '17 at 17:00
  • $\begingroup$ Without interchanging the order of integration, we have for $\alpha>\beta$ $$\begin{align} \int_0^\alpha \int_0^\beta \min(x,y)\,dy\,dx&=\int_0^\beta\int_0^\beta \min(x,y)\,dy\,dx+\int_\beta^\alpha \int_0^\beta \min(x,y)\,dy\,dx\\\\ &=\int_0^\beta \left(\int_0^x\min(x,y)\,dy+\int_x^\beta\min(x,y)\,dy\right)\,dx+\int_\beta^\alpha \int_0^\beta y\,dy\,dx\\\\ \end{align}$$ $\endgroup$ – Mark Viola Jul 4 '17 at 17:05
1
$\begingroup$

Probabilistic approach

Without loss of generality assume $\alpha<\beta$. Let us define two independent random variables $X\sim \text{Unif}[0,\alpha]$ and $Y\sim\text{Unif}[0,\beta]$.

The integral you asked for is just: $\alpha\beta \mathbb{E}[\min\{X,Y \}]$

Define $Z=\min\{X,Y\}$. The density function of $Z$ can be obtained via straightforward calculation:

\begin{align} f_Z(z) = \begin{cases} \frac{1}{\alpha}+\frac{1}{\beta}-2\frac{z}{\alpha \beta} & \text{ for } z\in[0,\alpha] \\ 0 & \text{otherwise} \end{cases} \end{align} Thus: \begin{align} \mathbb{E}[\min\{X,Y\}]=\mathbb{E}[Z]=\int_0^\alpha z f_z(z) \mathrm d z = \int^\alpha_0 \frac{z}{\alpha} +\frac{z}{\beta} - \frac{2z^2}{\alpha \beta} \mathrm dz = \frac{\alpha}{2}-\frac{\alpha^2}{6\beta} \end{align} Finally,

\begin{align} \int^\alpha_0 \int^\beta_0 \min\{x,y\} \mathrm d y \mathrm dx = \alpha\beta \mathbb{E}[Z] = \frac{\alpha^2\beta}{2}-\frac{\alpha^3}{6} \end{align}

$\endgroup$
  • $\begingroup$ Thanks for the patience and your willingness. I like that probabilistic approch, can I do the same procedure, if $\alpha > \beta$? $\endgroup$ – Antonio Placentino Jul 4 '17 at 17:02
  • $\begingroup$ Yes, sure! There is no need to do the calculation for $\alpha>\beta$. Just take $\alpha$ to be $\beta$ and vice versa and obtain the result for $\alpha>\beta$. $\endgroup$ – Shashi Jul 4 '17 at 17:06
  • $\begingroup$ Yes, I understand. In fact, the proof you give certainly prevents me from interchange values. This also solve the problem of consider cases of $\alpha > \beta$ and $\beta > \alpha$. Now, I only am trying to understand, why the approch of solving the integral in the question box is not correct for both cases. Thanks for your nice proof. $\endgroup$ – Antonio Placentino Jul 4 '17 at 17:12
  • $\begingroup$ @AntonioPlacentino my answer was meant to show how the integral can be calculated with a different approach. I thought your main question was answered by Mark Viola, right? $\endgroup$ – Shashi Jul 4 '17 at 17:16
  • $\begingroup$ Yes, half of problem. There only was a thing I cannot understand for the integral above and thanks to Yves Daoust, I found it too. I know your approch was different and very interesting and I did appreciate that. Thank you very much! $\endgroup$ – Antonio Placentino Jul 4 '17 at 17:33
0
$\begingroup$

When $\alpha\lt\beta$, we have

x

$$ \begin{align} \int_0^\alpha\int_0^\beta\min(x,y)\,\mathrm{d}y\,\mathrm{d}x &=\overbrace{2\int_0^\alpha\int_0^xy\,\mathrm{d}y\,\mathrm{d}x}^\text{integral over square}+\overbrace{\int_\alpha^\beta\int_0^\alpha y\,\mathrm{d}y\,\mathrm{d}x}^\text{integral over rectangle}\\ &=\frac13\alpha^3+\frac12\alpha^2(\beta-\alpha)\tag{1} \end{align} $$ Using $[\alpha\gt\beta]=\frac{\alpha-\beta+|\alpha-\beta|}{2(\alpha-\beta)}$ and $[\alpha\lt\beta]=\frac{\alpha-\beta-|\alpha-\beta|}{2(\alpha-\beta)}$ and $(1)$, we get $$ \int_0^\alpha\int_0^\beta\min(x,y)\,\mathrm{d}y\,\mathrm{d}x =\frac1{12}\left(|\alpha-\beta|^3-(\alpha+\beta)\left(\alpha^2-4\alpha\beta+\beta^2\right)\right)\tag{2} $$

$\endgroup$
  • $\begingroup$ Thanks! You give a graphical expression of my problem. $\min(x,y)$ can also be rewritten as you suggested. That's a good approach. Thank you for your willingness to help. $\endgroup$ – Antonio Placentino Jul 4 '17 at 20:07
  • $\begingroup$ It is possible to rewrite $\min(x,y)=\frac{x+y-|x-y|}2$, but integrating absolute value expressions often involves many cases. It is usually easier to compute the entire expression assuming $\alpha\lt\beta$ and splice together the integrals using $[\alpha\lt\beta]$ and $[\alpha\gt\beta]$. $\endgroup$ – robjohn Jul 4 '17 at 20:14
  • $\begingroup$ In fact, the approach of splitting integral is the fastest way. You and your colleagues also showed me others ways to approach the problem. Then, I found out I was considering areas out of my problem ($0<y<\beta$ and I was calculating an integral where $y>\beta$ which is out of domain). You all gave me ways of rewriting the integral and I appreciate that. Thanks to you all! $\endgroup$ – Antonio Placentino Jul 4 '17 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.