Find the area of the region enclosed by $\frac{\sin x}{\sin y}=\frac{\sin x+\sin y}{\sin(x+y)}$ and the $x$-axis (comes from a probability question)

Question

This question resisted attacks at MSE, so I am posting it here.

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Here is the graph of $\dfrac{\sin x}{\sin y}=\dfrac{\sin x+\sin y}{\sin(x+y)}$.

Find the area of the region enclosed by the curve and the $x$-axis, from $x=0$ to $x=\pi$.

Context, and why I suspect the answer is $\frac{\pi^2}{8}$

This question arose when I asked myself: "A triangle's vertices are three uniformly random points on a circle. The side lengths are, in random order, $a,b,c$. The triangle inequality tells us that $P(a+b<c)=0$. But what is $P\left(a+b<\left(\frac{a}{b}\right)c\right)$, given that $\frac{a}{b}>1$ ?"

A simulation of $10^7$ random triangles yielded an estimated probability of $0.49998$, suggesting that the probability is $\frac12$, which would imply that the area in this question is $\frac{\pi^2}{8}$. The connection between the probability and the area is explained in the MSE question.

I decided to phrase this question in terms of area, but we can attack it in terms of area or probability.

Progress so far

I used Wolfram (changing $\sin y$ to $y$, and changing $\sin x$ to $a$) to help me change $\frac{\sin x}{\sin y}=\frac{\sin x+\sin y}{\sin(x+y)}$ to:

$$y=\arcsin\left(\frac13\left(\sin (2x)-\sin x+f(x)-\frac{g(x)}{f(x)}\right)\right)dx$$ where $$f(x)=\left(\frac12\left(h(x)+\sqrt{4g(x)^3+h(x)^2}\right)\right)^\frac13$$ $$g(x)=3\sin^2 x-(\sin 2x-\sin x)^2$$ $$h(x)=-16\left(\cos x\right)\sin^5 x+10\left(\cos x\right)\sin^3x+24\sin^5 x+10\sin^3 x$$

In this answer at the MSE question, user @Masd reports that $\int_0^\pi y \, \mathrm dx$ matches $\frac{\pi^2}{8}$ to $11$ decimal places.

${}$

Answer 1

Here is my computation. First of all, I had no idea whatsoever how to integrate implicit trigonometric functions, so I decided to switch to the integration with respect to the sides $x,y,z$ of the triangle $XYZ$. Recall that the conditions for the domain is $xz\le y^2+xy$, $x>y$ plus the triangle inequality $z>x-y$. This triangle inequality is compatible with the first condition if and only if $x< (\sqrt 2+1)y$. So the admissible domain is given by $$ y< x< (\sqrt 2+1)y,\qquad x-y<z<y+\frac {y^2} x\,. $$ Now we have to integrate over the points on the circle, i.e., for the case when the circumradius is $1$ and with respect to the angular variables. We relax the circumradius to $r>0$, place $Z$ on the positive semi-axis and the circle center at the origin (so $Z=r$), and take $Y=re^{2i\varphi}, X=re^{2i\psi}$, so $$ x=2r\sin\varphi,\qquad y=2r\sin\psi,\qquad z=2r\sin(\varphi-\psi) $$ (the last one up to a sign) where $\varphi,\psi$ run independently over $[0,\pi]$. The condition $x>y$ reduces the $d\varphi\,d\psi$ measure to $\frac{\pi^2}2$ and each admissible triple $x,y,z$ is realized twice (up to the flip about the horizontal axes). We can use any measure on $(0,+\infty)$ we want when integrating over $r$, so we will choose $\frac{dr}{r}$.

Now it is time to compute the Jacobian $J$ of the mapping $(r,\varphi,\psi)\mapsto (x,y,z)$. We have $$ J=8\left|\begin{matrix} \sin\varphi & r\cos\varphi & 0 \\ \sin\psi & 0 & r\cos\psi \\ \sin(\varphi-\psi) &r\cos(\varphi-\psi) & -r\cos(\varphi-\psi) \end{matrix}\right| \\ =8r^2[\cos\varphi\cos\psi\sin(\varphi-\psi)-\sin\varphi\cos\psi\cos(\varphi-\psi)+\cos\varphi\sin\psi\cos(\varphi-\psi)] \\ =8r^2[\cos\varphi\cos\psi\sin(\varphi-\psi)- \sin(\varphi-\psi)\cos(\varphi-\psi)] \\ =8r^2\sin(\varphi-\psi)\sin\varphi\sin\psi=\frac{xyz}r\,, $$ i.e. $$ \frac{dr}r\,d\varphi\,d\psi=\frac{dx\,dy\,dz}{xyz}\,, $$ which is pretty neat.

Now let $\Omega$ be the set of all good triangles and $\omega$ be the set of good triangles with $r=1$. Then the back of the envelope computation is the following: $$ \left[\int_0^\infty\frac{dr}{r}\right]\int_\omega d\varphi\,d\psi= 2\int_{\Omega}\frac{dr}{r}\,d\varphi\,d\psi =2\int_{\Omega}\frac{dx\,dy\,dz}{xyz} \\ = 2\int_0^\infty\frac{dy}{y}\left[\int_{y}^{(\sqrt 2+1)y}\frac{dx}x\left(\int_{x-y}^{y+\frac{y^2}x}\frac{dz}z\right)\right] $$ Substituting $x=ty$, $t\in(1,\sqrt 2+1)$, we evaluate the inner integrals to $$ \int_1^{\sqrt 2+1}\log\frac{t+1}{t(t-1)}\frac{dt}t\,, $$ which is independent of $y$, so we finally get $$2\left[\int_0^\infty\frac{dy}y\right] \int_1^{\sqrt 2+1}\log\frac{t+1}{t(t-1)}\frac{dt}t\,, $$ Now it remains to cancel $\int_0^\infty\frac{ds}s$ to get the claimed identity.

The little (but somewhat irritating) problem is that the integral $\int_0^\infty \frac{ds}s$ diverges. However, as Landau (the physicist) used to say, "A chicken is not a bird and a logarithm is not infinity". So we will fix that with a little bit more elaborate argument. ${}{}$

We shall impose one extra condition on the admissible triangles, namely, we will demand that $y$ is comparable to $r$. We always trivially have $r\ge \frac y2$, so we'll fix a big $A>0$ and demand that $r<Ay$. Notice that it is a condition on the shape only, not on the size. Denoting by $\omega_A$ the set of admissible triangles with circumradius $1$ and by $\Omega_A$ the set of all admissible triangles and choosing a huge finite $R>0$, we can write, as before, $$ \left[\int_1^R\frac{dr}r\right]\int_{\omega_A}d\varphi\,d\psi=2\int_{\Omega_A\cap\{1<r<R\}}\frac{dx\,dy\,dz}{xyz}\,. $$ Now notice that $$ \Omega_A\cap\{2<y<\frac RA\}\subset \Omega_A\cap\{1<r<R\}\subset\Omega_A\cap\{\frac 1A<y<2R\}\,, $$ so instead of one identity, we write two inequalities and conclude that $\log R\int_{\omega_A}d\varphi\,d\psi$ is squeezed between $\log\frac{R}{2A}$ times $2\int_1^{\sqrt 2+1}\left(\int_{E(t,A)}\frac{ds}{s}\right)\frac{dt}t$ and $\log(2AR)$ times the same quantity, where we put $x=ty, z=sy$ and defined $E(t,A)$ to be the set of $s$ for which the triangle with the sides $1,t,s$ is admissible.

Now, dividing by $\log R$, letting $R\to+\infty$ and using the squeeze theorem, we conclude that $$ \int_{\omega_A}d\varphi\,d\psi= 2\int_1^{\sqrt 2+1}\left(\int_{E(t,A)}\frac{ds}{s}\right)\frac{dt}t $$ for every fixed $A$. But when $A\to+\infty$, the set $\omega_A$ expands to $\omega$ and for each $t$, the set $E(t,A)$ expands to $[t-1,1+\frac 1t]$, so the monotone convergence theorem finishes the story.

That's it. I tried to find a simple change of variables that would avoid using non-trivial identities for $\mathrm{Li}_2$ to get the final $\frac{\pi^2}8$ answer, but failed so far. Still, we have the result, so, I hope, somebody will eventually provide a simpler proof. The value $\frac 12$ of the probability is too nice to be there without a clear reason. :-)

Answer 2

$\newcommand{\li}{\operatorname{Li}_2}$The following expression of the area (say $A$) was proposed in fedja's comment: \begin{equation*} A=I:=\int_1^{1+\sqrt2}\frac{dt}t\,\ln\frac{t+1}{t(t-1)}. \tag{10}\label{10} \end{equation*}

It was noted in the comment by GH from MO that Mathematica evaluates the integral $I$ as \begin{equation*} B:=\frac{\pi^2}{4}-\frac12\,\ln(1+\sqrt2)^2+\li(1-\sqrt2)-\li(\sqrt2-1). \tag{20}\label{20} \end{equation*} GH from MO also noted that, according to Lima, \begin{equation*} B=\frac{\pi^2}{8}. \end{equation*}

The purpose of this answer is to provide a simple "human" proof of the Mathematica evaluation that \begin{equation*} I=B. \tag{30}\label{30} \end{equation*}

Toward this end, write \begin{equation*} I=I_1-I_2, \tag{40}\label{40} \end{equation*} where \begin{equation*} I_1:=\int_1^{1+\sqrt2}\frac{dt}t\,\ln\frac{t+1}{t},\quad I_2:=\int_1^{1+\sqrt2}\frac{dt}t\,\ln(t-1). \end{equation*} Next, \begin{align*} I_1&=\int_1^{1+\sqrt2}\frac{dt}t\,\ln(1+1/t) \\ &=\int_1^{1+\sqrt2}\frac{dt}t\,\sum_{k\ge1}\frac{(-1)^{k-1}}{k}\,t^{-k} \\ &=\sum_{k\ge1}\frac{(-1)^{k-1}}{k}\int_1^{1+\sqrt2} dt\,t^{-k-1} \\ &=\sum_{k\ge1}\frac{(-1)^{k-1}}{k^2}\,(1-(\sqrt2-1)^k) \\ &=\sum_{k\ge1}\frac{(-1)^{k-1}}{k^2}+\sum_{k\ge1}\frac{(1-\sqrt2)^k}{k^2} \\ &=\frac{\pi^2}{12}+\li(1-\sqrt2). \tag{50}\label{50} \end{align*} Somewhat similarly, \begin{align*} I_2&=\int_{\sqrt2-1}^1\frac{du}u\,\ln(1/u-1) \\ & =\int_{\sqrt2-1}^1\frac{du}u\,\ln(1-u)-\int_{\sqrt2-1}^1\frac{du}u\,\ln u \\ & =-\sum_{k\ge1}\frac{1}{k}\,\int_{\sqrt2-1}^1 du\,u^{k-1}+\frac12\ln(\sqrt2-1)^2 \\ & =-\sum_{k\ge1}\frac{1}{k^2}\,(1-(\sqrt2-1)^k)+\frac12\ln(\sqrt2-1)^2 \\ & =-\frac{\pi^2}6+\li(\sqrt2-1)+\frac12\ln(\sqrt2-1)^2. \tag{60}\label{60} \end{align*}

Now \eqref{30} follows immediately from \eqref{40}, \eqref{50}, and \eqref{60}. $\quad\Box$

Answer 3

An approach bypassing polylogarithms is as follows:

\begin{align}\int_1^{1+\sqrt2}\log\frac{t+1}{t(t-1)}\frac{dt}t&\stackrel{ibp}=\int_1^{1+\sqrt2}\frac{t^2+2t-1}{t(t^2-1)}\log t\,dt\\&\stackrel{t=e^u}=\int_0^{\operatorname{arsinh}1}\frac{\sinh u+1}{\sinh u}u\,du\\&\stackrel{v=\sinh u}=\frac12\operatorname{arsinh}^21+\int_0^1\frac{\operatorname{arsinh}v}{v\sqrt{1+v^2}}\,dv\\&=\frac{\pi^2}8\end{align} as shown here. This is directly related to the Legendre chi function at $\nu=2$.

Answer 4

Using the book Lewin: Polylogarithms and associated functions (Elsevier, 1981) as a reference, we can give a streamlined proof of the identity $$I:=\int_1^{\sqrt 2+1}\log\left(\frac{t+1}{t(t-1)}\right)\frac{dt}{t}=\frac{\pi^2}{8}.$$ In the "Added" section, I will make this proof self-contained without any reference to the book.

By a change of variable $t=1/x$, we see that $$I=\int_{\sqrt{2}-1}^1\log\left(\frac{x(1+x)}{1-x}\right)\frac{dx}{x}.$$ It is also clear that $$I_1:=\int_{\sqrt{2}-1}^1\log(x)\frac{dx}{x}=\left[\frac{\log^2(x)}{2}\right]_{\sqrt{2}-1}^1=-\frac{\log^2(\sqrt{2}-1)}{2}.$$ On the other hand, by (1.66) of the book, $$I_2:=\int_{\sqrt{2}-1}^1\log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}=2\chi_2(1)-2\chi_2(\sqrt{2}-1).$$ Now it is classical that $$\chi_2(1)=\sum_{k=0}^\infty\frac{1}{(2k+1)^2}=\frac{\pi^2}{8},$$ while (1.68) of the book tells us that $$\chi_2(\sqrt{2}-1)=\frac{\pi^2}{16}-\frac{\log^2(\sqrt{2}-1)}{4}.$$ Putting everything together, $$I=I_1+I_2=-\frac{\log^2(\sqrt{2}-1)}{2}+\frac{\pi^2}{4}-2\left(\frac{\pi^2}{16}-\frac{\log^2(\sqrt{2}-1)}{4}\right)=\frac{\pi^2}{8}.$$ I should add that (1.66) follows immediately from the definition of $\chi_2$, while (1.68) takes a few lines to prove. Both were known to Euler and Legendre.

Added. We can make the calculation of $I_2$ (and hence $I$) self-contained as follows. The key point is the identity $$\int_y^1\log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}-\int_0^\frac{1-y}{1+y}\log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}=\log(y)\log\left(\frac{1-y}{1+y}\right),\qquad y\in(0,1).$$ Indeed, the $y$-derivatives of the two sides are the same, while both sides tend to zero as $y\to 0+$ or $y\to 1-$. Plugging $y:=\sqrt{2}-1$ into this identity, we infer that $$\int_{\sqrt{2}-1}^1\log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}-\int_0^{\sqrt{2}-1}\log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}=\log^2(\sqrt{2}-1).\tag{1}$$ On the other hand, $$\int_{\sqrt{2}-1}^1\log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}+\int_0^{\sqrt{2}-1}\log\left(\frac{1+x}{1-x}\right)\frac{dx}{x}=\frac{\pi^2}{4},\tag{2}$$ as follows readily from $$\log\left(\frac{1+x}{1-x}\right)=2\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1},\qquad x\in(0,1).$$ Averaging $(1)$ and $(2)$, we infer that $$I_2=\frac{\log^2(\sqrt{2}-1)}{2}+\frac{\pi^2}{8},$$ and then $$I=I_1+I_2=\frac{\pi^2}{8}.$$

Answer 5

This is an attempt to work directly with the equation $\frac{\sin x}{\sin y}=\frac{\sin x+\sin y}{\sin(x+y)}$ from the question. As $\cos x$ is injective on $[0,\pi]$, it might be sensible to try to express the area in terms of $\cos x$. To this end, we compute the algebraic relation between $a:=\cos x$ and $b:=\cos y$. Using the addition theorem for $\sin(x+y)$ and eliminating $\sin x$ and $\sin y$ via $\sin^2+\cos^2=1$, we arrive at \begin{equation} (a-b)(a+b)^2 + 2(1-a^2)(1-b^2)=0. \end{equation} The $(a,b)$-curve given by this equation admits a rational parametrization: \begin{align*} a(t) &= \frac{t^3 + t^2 + t - 1}{2t^2}\\ b(t) &= \frac{t^3 + t^2 - t + 1}{2t} \end{align*} The area we are looking for is \begin{equation} I=\int_{0}^\pi y\,dx=\int_{0}^\pi\arccos(\cos y)\,dx. \end{equation} If $x$ runs from $0$ to $\pi$, then $a=\cos x$ runs from $1$ down to $-1$, and this is achieved if $t$ runs from $1$ down to $\sqrt{2}-1$.

From $a(t)=\cos x$ we get $a'(t)=-\sin x\frac{dx}{dt}=-\sqrt{1-a(t)^2}\frac{dx}{dt}$, hence \begin{equation} I=\int_{\sqrt{2}-1}^1\frac{a'(t)}{\sqrt{1-a(t)^2}}\arccos b(t)\,dt. \end{equation} There are several possibilities to continue from there (see my and @Zacky's comments to the OP's question). For instance, using \begin{equation} \arccos b=2\arctan\frac{\sqrt{1-b^2}}{1+b}=2\int_0^{\frac{\sqrt{1-b^2}}{1+b}}\frac{1}{1+z^2}dz=2\int_0^1\frac{\sqrt{1-b^2}}{1+b+(1-b)s^2}ds \end{equation} and $1-b(t)^2=t\cdot(1-a(t)^2)$ we get \begin{equation} I=2\int_{\sqrt{2}-1}^1\int_0^1\frac{ta'(t)}{1+b(t)+(1-b(t))s^2}ds\,dt. \end{equation} Written out explicitly this is \begin{equation} I=2\int_{\sqrt{2}-1}^1\int_0^1\frac{t^3 - t + 2}{(1-s^2)t(t^3 + t^2 + \frac{1+3s^2}{1-s^2}t + 1)}ds\,dt. \end{equation} It is tempting to switch the order of integration and first integrate along $t$ by a partial fraction decomposition. The problem is that the roots of the cubic in the denominator are non-constant algebraic functions in $s$. On the other hand, the simple answer of the finite result should allow for some simple trick here ?!?