Interpretation of an equality: Area of regular polygon and the integral $int_0^infty! fracmathbbdx1+x^N$

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Recently I learned the integral from this post:

$$mathcal I=int_0^infty! fracmathbbdx1+x^N=fracpi/Nsin(pi/N)$$

This reminds me of the area of regular polygon. Consider a $2N$-gon with unit "radius". (Distance from center to vertices)

The area is given by

$$A_2N=(2N)(frac12cdot 1^2sinfrac2pi2N)=picdotfracsin(pi/N)pi/N$$

Question

Turns out we found the equality

$$frac1A_2N=frac1piint_0^infty! fracmathbbdx1+x^N$$

It may look like a coincidence, but does it?
Do we have an interpretation of any kind for the equality?

---

Thoughts

The term $frac1A_2N$ is equivalent to the probability of choosing a region of 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

$$fracpiA_N=int_0^infty! fracmathbbdx1+x^N/2, NinBbbN$$

integration

share|cite|improve this question

edited Jul 30 at 3:35

asked Jul 27 at 17:00

Mythomorphic

5,0711732

This question has an open bounty worth +100
reputation from Mythomorphic ending ending at 2018-08-13 16:45:23Z">in 2 days.

This question has not received enough attention.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  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.
  – Winther
  Jul 27 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can also think of the value $fracpiA_2n$ 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 $frac1A_2n$ reading seems unlikely to give you what you want.
  – Thomas Andrews
  Jul 27 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your suggestion! That was just my wild guess though:)
  – Mythomorphic
  Jul 27 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why restrict formula to positive integer values of $N$ where $N>1$? Half integers $>1$ work as well e.g. $N=frac32$ gives the area of a triangle.
  – James Arathoon
  Jul 30 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  True. I should have included the half integers so as to include polygons with odd number of sides.
  – Mythomorphic
  Jul 30 at 3:31
  

  
  
  
  

add a comment | 

up vote
25
down vote

favorite

9

Recently I learned the integral from this post:

$$mathcal I=int_0^infty! fracmathbbdx1+x^N=fracpi/Nsin(pi/N)$$

This reminds me of the area of regular polygon. Consider a $2N$-gon with unit "radius". (Distance from center to vertices)

The area is given by

$$A_2N=(2N)(frac12cdot 1^2sinfrac2pi2N)=picdotfracsin(pi/N)pi/N$$

Question

Turns out we found the equality

$$frac1A_2N=frac1piint_0^infty! fracmathbbdx1+x^N$$

It may look like a coincidence, but does it?
Do we have an interpretation of any kind for the equality?

---

Thoughts

The term $frac1A_2N$ is equivalent to the probability of choosing a region of 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

$$fracpiA_N=int_0^infty! fracmathbbdx1+x^N/2, NinBbbN$$

integration

share|cite|improve this question

edited Jul 30 at 3:35

asked Jul 27 at 17:00

Mythomorphic

5,0711732

This question has an open bounty worth +100
reputation from Mythomorphic ending ending at 2018-08-13 16:45:23Z">in 2 days.

This question has not received enough attention.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  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.
  – Winther
  Jul 27 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can also think of the value $fracpiA_2n$ 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 $frac1A_2n$ reading seems unlikely to give you what you want.
  – Thomas Andrews
  Jul 27 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your suggestion! That was just my wild guess though:)
  – Mythomorphic
  Jul 27 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why restrict formula to positive integer values of $N$ where $N>1$? Half integers $>1$ work as well e.g. $N=frac32$ gives the area of a triangle.
  – James Arathoon
  Jul 30 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  True. I should have included the half integers so as to include polygons with odd number of sides.
  – Mythomorphic
  Jul 30 at 3:31
  

  
  
  
  

add a comment | 

up vote
25
down vote

favorite

9

up vote
25
down vote

favorite

9

9

Recently I learned the integral from this post:

$$mathcal I=int_0^infty! fracmathbbdx1+x^N=fracpi/Nsin(pi/N)$$

This reminds me of the area of regular polygon. Consider a $2N$-gon with unit "radius". (Distance from center to vertices)

The area is given by

$$A_2N=(2N)(frac12cdot 1^2sinfrac2pi2N)=picdotfracsin(pi/N)pi/N$$

Question

Turns out we found the equality

$$frac1A_2N=frac1piint_0^infty! fracmathbbdx1+x^N$$

It may look like a coincidence, but does it?
Do we have an interpretation of any kind for the equality?

---

Thoughts

The term $frac1A_2N$ is equivalent to the probability of choosing a region of 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

$$fracpiA_N=int_0^infty! fracmathbbdx1+x^N/2, NinBbbN$$

integration

share|cite|improve this question

edited Jul 30 at 3:35

asked Jul 27 at 17:00

Mythomorphic

5,0711732

Recently I learned the integral from this post:

$$mathcal I=int_0^infty! fracmathbbdx1+x^N=fracpi/Nsin(pi/N)$$

This reminds me of the area of regular polygon. Consider a $2N$-gon with unit "radius". (Distance from center to vertices)

The area is given by

$$A_2N=(2N)(frac12cdot 1^2sinfrac2pi2N)=picdotfracsin(pi/N)pi/N$$

Question

Turns out we found the equality

$$frac1A_2N=frac1piint_0^infty! fracmathbbdx1+x^N$$

It may look like a coincidence, but does it?
Do we have an interpretation of any kind for the equality?

---

Thoughts

The term $frac1A_2N$ is equivalent to the probability of choosing a region of 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

$$fracpiA_N=int_0^infty! fracmathbbdx1+x^N/2, NinBbbN$$

integration

share|cite|improve this question

edited Jul 30 at 3:35

asked Jul 27 at 17:00

Mythomorphic

5,0711732

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 3:35

edited Jul 30 at 3:35

edited Jul 30 at 3:35

asked Jul 27 at 17:00

Mythomorphic

5,0711732

asked Jul 27 at 17:00

Mythomorphic

5,0711732

asked Jul 27 at 17:00

Mythomorphic

5,0711732

5,0711732

This question has an open bounty worth +100
reputation from Mythomorphic ending ending at 2018-08-13 16:45:23Z">in 2 days.

This question has not received enough attention.

This question has an open bounty worth +100
reputation from Mythomorphic ending ending at 2018-08-13 16:45:23Z">in 2 days.

This question has not received enough attention.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  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.
  – Winther
  Jul 27 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can also think of the value $fracpiA_2n$ 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 $frac1A_2n$ reading seems unlikely to give you what you want.
  – Thomas Andrews
  Jul 27 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your suggestion! That was just my wild guess though:)
  – Mythomorphic
  Jul 27 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why restrict formula to positive integer values of $N$ where $N>1$? Half integers $>1$ work as well e.g. $N=frac32$ gives the area of a triangle.
  – James Arathoon
  Jul 30 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  True. I should have included the half integers so as to include polygons with odd number of sides.
  – Mythomorphic
  Jul 30 at 3:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  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.
  – Winther
  Jul 27 at 18:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You can also think of the value $fracpiA_2n$ 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 $frac1A_2n$ reading seems unlikely to give you what you want.
  – Thomas Andrews
  Jul 27 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your suggestion! That was just my wild guess though:)
  – Mythomorphic
  Jul 27 at 18:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why restrict formula to positive integer values of $N$ where $N>1$? Half integers $>1$ work as well e.g. $N=frac32$ gives the area of a triangle.
  – James Arathoon
  Jul 30 at 3:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  True. I should have included the half integers so as to include polygons with odd number of sides.
  – Mythomorphic
  Jul 30 at 3:31
  

  
  
  
  

2

2

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.
– Winther
Jul 27 at 18:05

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.
– Winther
Jul 27 at 18:05

You can also think of the value $fracpiA_2n$ 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 $frac1A_2n$ reading seems unlikely to give you what you want.
– Thomas Andrews
Jul 27 at 18:30

You can also think of the value $fracpiA_2n$ 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 $frac1A_2n$ reading seems unlikely to give you what you want.
– Thomas Andrews
Jul 27 at 18:30

Thank you for your suggestion! That was just my wild guess though:)
– Mythomorphic
Jul 27 at 18:31

Thank you for your suggestion! That was just my wild guess though:)
– Mythomorphic
Jul 27 at 18:31

Why restrict formula to positive integer values of $N$ where $N>1$? Half integers $>1$ work as well e.g. $N=frac32$ gives the area of a triangle.
– James Arathoon
Jul 30 at 3:11

Why restrict formula to positive integer values of $N$ where $N>1$? Half integers $>1$ work as well e.g. $N=frac32$ gives the area of a triangle.
– James Arathoon
Jul 30 at 3:11

True. I should have included the half integers so as to include polygons with odd number of sides.
– Mythomorphic
Jul 30 at 3:31

True. I should have included the half integers so as to include polygons with odd number of sides.
– Mythomorphic
Jul 30 at 3:31

add a comment | 

1 Answer
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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 $fracpi N,$ the angle step between neighbour vertices is double.

Taking in account l'Hospital rule, one can get
beginalign
&x_k = e^fracpi N(2k+1)i,\
&mathopmathrmReslimits_x_kdfrac11+x^N = limlimits_xto x_kdfracx-x_k1+x^N = dfrac1Nx_k^N-1 = dfracx_kN.\[4pt]
endalign
Easy to see that the reason of $N$ in the denominator is the deriving.
At the same time, the residue theorem contains the required constant $pi.$

share|cite|improve this answer

answered 18 hours ago

Yuri Negometyanov

9,2231523

add a comment | 

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1 Answer
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1 Answer
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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 $fracpi N,$ the angle step between neighbour vertices is double.

Taking in account l'Hospital rule, one can get
beginalign
&x_k = e^fracpi N(2k+1)i,\
&mathopmathrmReslimits_x_kdfrac11+x^N = limlimits_xto x_kdfracx-x_k1+x^N = dfrac1Nx_k^N-1 = dfracx_kN.\[4pt]
endalign
Easy to see that the reason of $N$ in the denominator is the deriving.
At the same time, the residue theorem contains the required constant $pi.$

share|cite|improve this answer

answered 18 hours ago

Yuri Negometyanov

9,2231523

add a comment | 

up vote
0
down vote

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 $fracpi N,$ the angle step between neighbour vertices is double.

Taking in account l'Hospital rule, one can get
beginalign
&x_k = e^fracpi N(2k+1)i,\
&mathopmathrmReslimits_x_kdfrac11+x^N = limlimits_xto x_kdfracx-x_k1+x^N = dfrac1Nx_k^N-1 = dfracx_kN.\[4pt]
endalign
Easy to see that the reason of $N$ in the denominator is the deriving.
At the same time, the residue theorem contains the required constant $pi.$

share|cite|improve this answer

answered 18 hours ago

Yuri Negometyanov

9,2231523

add a comment | 

up vote
0
down vote

up vote
0
down vote

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 $fracpi N,$ the angle step between neighbour vertices is double.

Taking in account l'Hospital rule, one can get
beginalign
&x_k = e^fracpi N(2k+1)i,\
&mathopmathrmReslimits_x_kdfrac11+x^N = limlimits_xto x_kdfracx-x_k1+x^N = dfrac1Nx_k^N-1 = dfracx_kN.\[4pt]
endalign
Easy to see that the reason of $N$ in the denominator is the deriving.
At the same time, the residue theorem contains the required constant $pi.$

share|cite|improve this answer

answered 18 hours ago

Yuri Negometyanov

9,2231523

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 $fracpi N,$ the angle step between neighbour vertices is double.

Taking in account l'Hospital rule, one can get
beginalign
&x_k = e^fracpi N(2k+1)i,\
&mathopmathrmReslimits_x_kdfrac11+x^N = limlimits_xto x_kdfracx-x_k1+x^N = dfrac1Nx_k^N-1 = dfracx_kN.\[4pt]
endalign
Easy to see that the reason of $N$ in the denominator is the deriving.
At the same time, the residue theorem contains the required constant $pi.$

share|cite|improve this answer

answered 18 hours ago

Yuri Negometyanov

9,2231523

share|cite|improve this answer

share|cite|improve this answer

answered 18 hours ago

Yuri Negometyanov

9,2231523

answered 18 hours ago

Yuri Negometyanov

9,2231523

answered 18 hours ago

Yuri Negometyanov

9,2231523

9,2231523

add a comment | 

add a comment | 

 

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