Interpretation of an equality: Area of regular polygon and the integral $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^N}$ Recently, I learned the integral from this post:
$$\mathcal I=\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^N}=\frac{\pi/N}{\sin(\pi/N)}$$
This reminds me of the area of a regular polygon. Consider a $2N$-gon with unit "radius". (Distance from centre to vertices)
The area is given by
$$A_{2N}=(2N)\left(\frac12\cdot 1^2\sin{\frac{2\pi}{2N}}\right)=\pi\cdot\frac{\sin(\pi/N)}{\pi/N}$$
Question
Turns out we found the equality
$$\frac{1}{A_{2N}}=\frac{1}{\pi}\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^N}$$
It may look like a coincidence, but is it?
Do we have an interpretation of any kind for this equality?

Thoughts
The term $\frac{1}{A_{2N}}$ is equivalent to the probability of choosing a region with arbitrary boundary and squared unit area from a regular $2N$-gon with unit "radius".
Or, as @Thomas Andrews suggests, we may rewrite the equation as
$$\frac{\pi}{A_{N}}=\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^{N/2}}, N\in\Bbb{N}$$
 A: The partial interpretation is given by the residue theorem.
Residues are calculated in the vertices of the polygon considered in OP. At the same time, only half of the vertices is used, but this does not change the common situation.
Vertices of the polygon are in the circle of radius $1.$ The first vertex has the polar angle $\frac\pi N,$ the angle step between neighbour vertices is double.
Taking in account L'Hospital rule, one can get for the any vertice
\begin{align}
&x_k = e^{\frac\pi N(2k+1)i},\\
&\mathop{\mathrm{Res}}\limits_{x_k}\dfrac1{1+x^N} = \lim\limits_{x\to x_k}\dfrac{x-x_k}{1+x^N} = \dfrac{1}{Nx_k^{N-1}} = -\dfrac{x_k}N =-e^{\frac{2k\pi}Ni}\frac{\cos\frac\pi N+i\sin\frac\pi N}{N}.\\[4pt]
\end{align}
The residue theorem defines the issue integral as the sum of residues with the coefficient $2\pi i.$ Taking in account that summation eliminates the factor $e^{\frac{2k\pi}Ni},$ easily to see the analogy of the term $\sin\frac\pi N$ with the square of any of the triangles, which form the polygon. Also is reasonable the factor $\pi.$
At the same time, the summation via the residue theorem does not add the factor $N$ to the numerator. Vice versa, using of the L'Hospital rule adds this factor to the denominator.
And this detail defines the difference.
