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

Question

Consider the graph of $\sin^p x+\sin^p y=\sin^p(x+y)$, where $x$ and $y$ are acute, and $p>1$.

Here are examples with, from left to right, $p=1.05,\space 1.25,\space 2,\space 4,\space 100$.

Find the area of the region enclosed by the curve and the $x$-axis and the $y$-axis, in terms of $p$.

Context, and why I suspect the answer is $\left(1-\frac{1}{p^2}\right)\frac{\pi^2}{6}$

An MSE question by user @Nilotpal Sinha in effect asks, "A triangle's vertices are uniformly random points on a circle. The side lengths, in increasing order, are $a,b,c$. Given $p>1$, what is the probability that line segments of lengths $a^p,\space b^p,\space c^p$ could form a triangle? That is, what is $P(a^p+b^p\ge c^p)$ ?

Simulations (by @Nilotpal Sinha and myself) suggest that, elegantly, $P(a^p+b^p\ge c^p)=\frac{1}{p^2}$, which would imply that the area in this question is $\left(1-\frac{1}{p^2}\right)\frac{\pi^2}{6}$, as explained below.

Connection between the probability question and the area question (optional reading). Let $x,y,z$ be the angles opposite the sides with lengths $a,b,c$ respectively. Assuming the radius of the circle is $1$, the law of sines tells us that $a=2\sin x,\space b=2\sin y,\space c=2\sin (x+y)$. Since we are given $a<b<c$, the sample space is bounded by $x=0,\space y=x,\space x+2y=\pi$. For convenience, we double the sample space by including its reflection across $y=x$. Then $1-P(a^p+b^p\ge c^p)$ is the area of the region in this question, divided by the area of the (doubled) sample space, which is $\frac{\pi^2}{6}$.

Similarity with a recent question

This area question is similar to a recent question, "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". Both involve finding the area of a region defined by an implicit trigonometric function. I suspect this question is also non-trivial.

Answer 1

The conjecture is true, and it can be verified with fedja's method developed for your earlier question. We present a simplified version of the method. The idea is to scale the triangle so that one of the sides becomes $1$, and then use the other two sides as parameters (instead of using the angles as parameters). This change of variables makes the calculation more tractable.

Without loss of generality, the vertices opposite the sides $a$, $b$, $c$ are $e^{2i\beta}$, $e^{-2i\alpha}$, $1$, where $(\alpha,\beta)\in\mathbb{R}^2$ is uniformly distributed in the region $$\alpha,\beta>0\qquad\text{and}\qquad \alpha+\beta<\pi.$$ Then the triangle has angles $\alpha$, $\beta$, $\pi-\alpha-\beta$, and $$a=2\sin\alpha,\qquad b=2\sin\beta,\qquad c=2\sin(\alpha+\beta).$$ Let us consider $$s:=\frac{b}{a}=\frac{\sin\beta}{\sin\alpha},\qquad t:=\frac{c}{a}=\frac{\sin(\alpha+\beta)}{\sin\alpha}.$$ The mapping $(\alpha,\beta)\mapsto(s,t)$ is bijective onto the set of pairs $(s,t)\in\mathbb{R}^2$ satisfying $$s,t>0\qquad\text{and}\qquad s+t>1>|s-t|.$$ Moreover, the Jacobian of the mapping is $st$, whence $$d\alpha\,d\beta=\frac{ds\,dt}{st}.$$ We note for later reference that the set of all pairs $(\alpha,\beta)$ has Lebesgue measure $\pi^2/2$, hence the subset corresponding to the event $c\ge b\ge a$ has Lebesgue measure $\pi^2/12$.

Using the change of variables $(\alpha,\beta)\mapsto(s,t)$, we see that $$P(a^p+b^p\ge c^p\mid c\ge b\ge a)= \frac{12}{\pi^2}\int_{\Omega(p)}\frac{ds\,dt}{st},$$ where $$\Omega(p):=\left\{(s,t)\in\mathbb{R}^2:(1+s^p)^{1/p}\ge t\ge s\ge 1\right\}.$$ Now an application of Fubini's theorem shows that the probability in question equals $$\frac{12}{\pi^2}\int_1^\infty\left(\int_s^{(1+s^p)^{1/p}}\frac{dt}{t}\right)\frac{ds}{s}=\frac{12}{\pi^2}\int_1^{\infty}\frac{\log(1+s^{-p})}{ps}\,ds.$$ On the other hand, $$\log(1+s^{-p})=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k s^{kp}},\qquad s\ge 1,\tag{1}$$ therefore $$P(a^p+b^p\ge c^p\mid c\ge b\ge a)=\frac{12}{\pi^2}\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^2 p^2}=\frac{1}{p^2}.$$

Added. In the last step, we used that $$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^2}=\frac{\pi^2}{12},\tag{2}$$ which is equivalent to the solution of the Basel problem. On the other hand, for $p=1$, the probability in question equals $1$ by the triangle inequality, hence in that case the above calculation yields a proof of $(2)$. This was observed by the OP, and in a slightly different context by Nilotpal Sinha. In fact the triangle inequality allows us to eliminate both $(1)$ and $(2)$ from the above calculation. Namely, by a change of variable we can see that $$\int_1^{\infty}\log(1+s^{-p})\frac{ds}{s}$$ is proportional to $p^{-1}$, hence the probability sought is proportional to $p^{-2}$.