10

How can I evaluate this integral? $$\iint\limits_{0\le x\le y\le1}\!\sqrt{1+x^2-y^2}\,{\rm d}x\,{\rm d}y$$ I know from Wolfram Alpha that the answer is $\frac13\ln2+\frac16\approx0.39772$. However, doing this explicitly seems like a nightmare. Is there any nice way of doing this?

jimjim
  • 9,855
Akiva Weinberger
  • 25,412
  • My first thought would be to look at the Taylor series for $\sqrt(1+z)$. Also, why is this called "the fisher and the fish?" – Mike Battaglia Jul 23 '24 at 02:49
  • 1
    @MikeBattaglia I originally had phrased it in terms of a probability question: a fisher is somewhere on the edge of a square-shaped lake, and a fish is somewhere inside of it. What is the probability that the fisher is closer to the center than the fish is? Using the above integral, the answer is $\frac{1-\ln2}3$. While drafting the question I figured it was cleaner just to discuss the integral on its own, but forgot to edit the title to match. (I have now edited the title.) – Akiva Weinberger Jul 23 '24 at 02:51
  • 1
    Is your integral equal to $\int_0^1 \int_0^y \sqrt{1+x^2-y^2} , dx , dy ,, ?$ –  Jul 23 '24 at 02:56
  • @LuckyChouhan Yes. – Akiva Weinberger Jul 23 '24 at 02:56
  • 1
    This feels like it begs for a substitution of the type $x := \rho \cosh \alpha$ and $y := \rho \sinh \alpha$, but for some reason I just can't make the bounds of integration work. – PrincessEev Jul 23 '24 at 03:28
  • @PrincessEev Since $x\le y$, you might want to switch $x$ and $y$ there (right as I wrote this comment, an answer was posted using this haha). – Varun Vejalla Jul 23 '24 at 03:45
  • @PrincessEev the reason the bounds weren't working out was because $y$ is larger than $x$ on the domain and so needs to be assigned the $\cosh$ in the transformation as a result. – Ninad Munshi Jul 23 '24 at 03:46
  • Two valid submissions posted within three or four minutes of each other, using different methods. Which one to give the green checkmark…? – Akiva Weinberger Jul 23 '24 at 03:53
  • The checkmark is meant to be for the answer that personally helped or informed you the most, since you're the one asking the question. – PrincessEev Jul 23 '24 at 04:40

3 Answers3

9

Following @PrincessEev's suggestion, use the following change of variables:

$$\begin{cases}x = s\sinh t \\ y = s\cosh t \end{cases} \implies J = \begin{vmatrix}\sinh t & s\cosh t \\ \cosh t & s\sinh t\end{vmatrix} = s|\sinh^2t-\cosh^2t| = s$$

which leaves us with the integral

$$\int_0^\infty \int_0^{\operatorname{sech}t} s\sqrt{1-s^2}\:ds\:dt = \frac{1}{3}\int_0^\infty 1-\tanh^3t\:dt$$

$$\frac{1}{3}\int_0^\infty \left(\frac{1}{1+\tanh t}+\tanh t\right)\:d(\tanh t) = \frac{1}{3}\left[\log(1+\tanh t)+\frac{1}{2}\tanh^2t\right]_0^\infty = \boxed{\frac{\log 2}{3} +\frac{1}{6}}$$

Ninad Munshi
  • 37,891
  • Wow! Given that the integral has both $+$ and $-$ inside of it, do you think there might also be a way to do this with circular (i.e. regular) trig substitution (in addition to your hyperbolic trig sub method)? If I remember right, the integral of $\sqrt{a^2+x^2}$ is often done with hyperbolic trig sub while $\sqrt{a^2-x^2}$ is often done with circular trig sub, and this problem is like a hybrid of both. – Akiva Weinberger Jul 23 '24 at 03:51
  • 2
    @AkivaWeinberger those methods don't apply here, because after the two variable substitution, this was chain rule to get an $(\cdots)^\frac{3}{2}$ Instead of some inverse trig function. The only thing that mattered was the relationship between the $x$ and the $y$ variables, not the overall sign within the radicand. Because they were squared and had an opposite sign, hyperbolic substitution was the way to go. Multivariable and single variable substitutions have very different considerations. – Ninad Munshi Jul 23 '24 at 03:57
  • 1
    Well done. You just beat me to it saved me minutes of typing! – Ted Shifrin Jul 23 '24 at 04:06
  • Can you explain the bounds? – Akiva Weinberger Jul 23 '24 at 04:21
  • @AkivaWeinberger let's start with the easier ones. $s=\operatorname{sech}t$ What have you tried? What does this correspond to in the original $xy$ coordinates? – Ninad Munshi Jul 23 '24 at 04:24
  • @TedShifrin dozens of minutes! There will be no writer's cramp or texter's thumb today! – Ninad Munshi Jul 23 '24 at 04:25
  • Ha! I’m not as slow and feeble as you think! – Ted Shifrin Jul 23 '24 at 04:28
  • Ahhh of course I should've switched them around, no wonder it wasn't working lol – PrincessEev Jul 23 '24 at 04:40
  • @NinadMunshi Right, I see. At $s=\operatorname{sech}t$, we have $x=\tanh t$ and $y=1$, which parametrizes the right edge of the triangle ${(x,y):0\le x\le y\le 1}$. At $s=0$ we get the point $(0,0)$ in the opposite corner, and for $s$ in between we get the rest of the triangle. – Akiva Weinberger Jul 23 '24 at 04:45
  • *top edge, not right – Akiva Weinberger Jul 23 '24 at 05:50
  • In fact, if I understand right, this immediately generalizes to $$\iint\limits_{0\le mx\le y\le1}!\sqrt{1+x^2-y^2},{\rm d}x,{\rm d}y=\frac13\ln(1+m)+\frac16m^2$$ for $m\ge1$. – Akiva Weinberger Jul 23 '24 at 05:56
  • @AkivaWeinberger I don't know if I buy that formula or how you would have derived it. For one thing, in the limit $m\to\infty$ this integral should approach $0$ but your formula approaches $\infty$. – Ninad Munshi Jul 23 '24 at 06:16
  • @NinadMunshi Oh, whoops, silly error. $$\iint\limits_{0\le mx\le y\le1}!\sqrt{1+x^2-y^2},{\rm d}x,{\rm d}y=\frac13\ln\left(1+\frac1m\right)+\frac1{6m^2}.$$ – Akiva Weinberger Jul 23 '24 at 11:04
  • To elaborate, this is by restricting $t$ to $[0,\coth m]$. – Akiva Weinberger Jul 25 '24 at 01:45
  • I found a new method. See my answer below. – Akiva Weinberger Jul 30 '24 at 02:11
5

Use the integral $$\int\sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}2\ln(x+\sqrt{x^2+a^2})+C,$$ \begin{align} &\iint\limits_{0\le x\le y\le1}\!\sqrt{1+x^2-y^2}\,{\rm d}x\,{\rm d}y\\ =&\int_0^1 \int_0^y \sqrt{1+x^2-y^2}\,{\rm d}x\,{\rm d}y\\ =&\int_0^1\left(\int_0^y \sqrt{\left(\sqrt{1-y^2}\right)^2+x^2}\,{\rm d}x\right)\,{\rm d}y\\ =&\int_0^1\left(\frac{x}{2}\sqrt{1+x^2-y^2}+\frac{1-y^2}2\ln(x+\sqrt{1+x^2-y^2})\bigg|_0^y\right)\,{\rm d}y\\ =&\int_0^1\left[\frac{y}2+\frac{1-y^2}2\ln(1+y)-\frac{1-y^2}4\ln(1-y^2)\right]\,{\rm d}y\\ =&\frac{(2+y^3-3y)\ln(1-y)+(2-y^3+3y)\ln(1+y)+2y^2}{12}\Bigg|_0^1\\ =&\frac{2 \ln\left(2\right)+1}{6}. \end{align}

Riemann
  • 11,801
3

I found a new method. Namely, applying the substitution $s=\frac xy$ sets us up perfectly for the substitution $t=y^2$. \begin{align} \iint\limits_{0\le x\le y\le1}\!\sqrt{1+x^2-y^2}\,{\rm d}x\,{\rm d}y&=\int_0^1\int_0^1 y\sqrt{1+s^2y^2-y^2}\,{\rm d}s\,{\rm d}y\\ &=\frac12\int_0^1\int_0^1\sqrt{1+s^2t-t}\,{\rm d}s\,{\rm d}t\\ &=\frac12\int_0^1\int_0^1\sqrt{1+t(s^2-1)}\,{\rm d}t\,{\rm d}s\\ &=\frac12\int_0^1\frac23\left[\frac{\big(1+(s^2-1)t\big){}^{3/2}}{s^2-1}\right]_{t=0}^{t=1}\,{\rm d}s\\ &=\frac13\int_0^1\frac{s^3-1}{s^2-1}\,{\rm d}s\\ &=\frac13\int_0^1s+\frac1{s+1}\,{\rm d}s\\ &=\frac13\left(\frac12+\ln2\right)=\frac13\ln2+\frac16 \end{align}

Akiva Weinberger
  • 25,412
  • 2
    In theory you could do the substitutions simultaneously, using$${\rm d}s,{\rm d}t=\begin{vmatrix}\frac1y&-\frac x{y^2}\0&2y\end{vmatrix}{\rm d}x,{\rm d}y=2,{\rm d}x,{\rm d}y,$$but that's harder to motivate. – Akiva Weinberger Jul 30 '24 at 02:25