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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

$$\frac{\pi}{A_{N}}=\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^{N/2}}, N\in\Bbb{N}$$

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    $\begingroup$ For sure a coincidence. It’s hard to prove a negative, but I would be extremely surprised if there turned out to be any natural connection between these two formulas. $\endgroup$ – Winther Jul 27 '18 at 18:05
  • $\begingroup$ You can also think of the value $\frac{\pi}{A_{2n}}$ is the expected number of random points selected from a unit circle when you finally select a point in (a fixed) regular $2N$-gon. Not sure that makes any sense, but the $\frac{1}{A_{2n}}$ reading seems unlikely to give you what you want. $\endgroup$ – Thomas Andrews Jul 27 '18 at 18:30
  • $\begingroup$ Thank you for your suggestion! That was just my wild guess though:) $\endgroup$ – Mythomorphic Jul 27 '18 at 18:31
  • $\begingroup$ Why restrict formula to positive integer values of $N$ where $N>1$? Half integers $>1$ work as well e.g. $N=\frac{3}{2}$ gives the area of a triangle. $\endgroup$ – James Arathoon Jul 30 '18 at 3:11
  • $\begingroup$ True. I should have included the half integers so as to include polygons with odd number of sides. $\endgroup$ – Mythomorphic Jul 30 '18 at 3:31
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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.

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  • $\begingroup$ Does your answer relate the integral to the area? $\endgroup$ – Mythomorphic Aug 12 '18 at 3:09
  • $\begingroup$ @Mythomorphic Thanks. I think that now it relates. $\endgroup$ – Yuri Negometyanov Aug 12 '18 at 14:13
  • $\begingroup$ When we solve this integral using the residue theorem and the standard wedge contour only one of the poles is inside the contour. math.stackexchange.com/questions/247866/… Even if you had all the residues inside the contour I still see no connection (each term is not the area of a triangle). $\endgroup$ – Winther Aug 12 '18 at 15:10
  • $\begingroup$ @Winther The standard contour is the semicircle $\endgroup$ – Yuri Negometyanov Aug 12 '18 at 15:16
  • $\begingroup$ That does not work for odd $N$. The function is not even so you cannot related $\int_0^\infty$ to $\int_{-\infty}^\infty$ and apply a semi-circle contour. $\endgroup$ – Winther Aug 12 '18 at 15:17

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