Interpretation of an equality: Area of regular polygon and the integral $int_0^{infty}!...












32












$begingroup$


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



$$A_{2N}=(2N)left(frac12cdot 1^2sin{frac{2pi}{2N}}right)=picdotfrac{sin(pi/N)}{pi/N}$$



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



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






integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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
















32












$begingroup$


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



$$A_{2N}=(2N)left(frac12cdot 1^2sin{frac{2pi}{2N}}right)=picdotfrac{sin(pi/N)}{pi/N}$$



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



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






integration







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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














32












32








32


13



$begingroup$


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



$$A_{2N}=(2N)left(frac12cdot 1^2sin{frac{2pi}{2N}}right)=picdotfrac{sin(pi/N)}{pi/N}$$



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



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






integration







share|cite|improve this question











$endgroup$




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



$$A_{2N}=(2N)left(frac12cdot 1^2sin{frac{2pi}{2N}}right)=picdotfrac{sin(pi/N)}{pi/N}$$



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



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






integration




integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 20:21









the_fox

2,90021537




2,90021537










asked Jul 27 '18 at 17:00









MythomorphicMythomorphic

5,3421834




5,3421834








  • 2




    $begingroup$
    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














  • 2




    $begingroup$
    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








2




2




$begingroup$
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




$begingroup$
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




$begingroup$
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




$begingroup$
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




$begingroup$
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




$begingroup$
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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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 for the any vertice
begin{align}
&x_k = e^{fracpi N(2k+1)i},\
&mathop{mathrm{Res}}limits_{x_k}dfrac1{1+x^N} = limlimits_{xto x_k}dfrac{x-x_k}{1+x^N} = dfrac{1}{Nx_k^{N-1}} = -dfrac{x_k}N =-e^{frac{2kpi}Ni}frac{cosfracpi N+isinfracpi N}{N}.\[4pt]
end{align}
The residue theorem defines the issue integral as the sum of residues with the coefficient $2pi i.$ Taking in account that summation eliminates the factor $e^{frac{2kpi}Ni},$ easily to see the analogy of the term $sinfracpi N$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 for the any vertice
begin{align}
&x_k = e^{fracpi N(2k+1)i},\
&mathop{mathrm{Res}}limits_{x_k}dfrac1{1+x^N} = limlimits_{xto x_k}dfrac{x-x_k}{1+x^N} = dfrac{1}{Nx_k^{N-1}} = -dfrac{x_k}N =-e^{frac{2kpi}Ni}frac{cosfracpi N+isinfracpi N}{N}.\[4pt]
end{align}
The residue theorem defines the issue integral as the sum of residues with the coefficient $2pi i.$ Taking in account that summation eliminates the factor $e^{frac{2kpi}Ni},$ easily to see the analogy of the term $sinfracpi N$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















2












$begingroup$

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 for the any vertice
begin{align}
&x_k = e^{fracpi N(2k+1)i},\
&mathop{mathrm{Res}}limits_{x_k}dfrac1{1+x^N} = limlimits_{xto x_k}dfrac{x-x_k}{1+x^N} = dfrac{1}{Nx_k^{N-1}} = -dfrac{x_k}N =-e^{frac{2kpi}Ni}frac{cosfracpi N+isinfracpi N}{N}.\[4pt]
end{align}
The residue theorem defines the issue integral as the sum of residues with the coefficient $2pi i.$ Taking in account that summation eliminates the factor $e^{frac{2kpi}Ni},$ easily to see the analogy of the term $sinfracpi N$ 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












2








2





$begingroup$

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 for the any vertice
begin{align}
&x_k = e^{fracpi N(2k+1)i},\
&mathop{mathrm{Res}}limits_{x_k}dfrac1{1+x^N} = limlimits_{xto x_k}dfrac{x-x_k}{1+x^N} = dfrac{1}{Nx_k^{N-1}} = -dfrac{x_k}N =-e^{frac{2kpi}Ni}frac{cosfracpi N+isinfracpi N}{N}.\[4pt]
end{align}
The residue theorem defines the issue integral as the sum of residues with the coefficient $2pi i.$ Taking in account that summation eliminates the factor $e^{frac{2kpi}Ni},$ easily to see the analogy of the term $sinfracpi N$ 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.






share|cite|improve this answer











$endgroup$



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 for the any vertice
begin{align}
&x_k = e^{fracpi N(2k+1)i},\
&mathop{mathrm{Res}}limits_{x_k}dfrac1{1+x^N} = limlimits_{xto x_k}dfrac{x-x_k}{1+x^N} = dfrac{1}{Nx_k^{N-1}} = -dfrac{x_k}N =-e^{frac{2kpi}Ni}frac{cosfracpi N+isinfracpi N}{N}.\[4pt]
end{align}
The residue theorem defines the issue integral as the sum of residues with the coefficient $2pi i.$ Taking in account that summation eliminates the factor $e^{frac{2kpi}Ni},$ easily to see the analogy of the term $sinfracpi N$ 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 23 '18 at 18:41

























answered Aug 10 '18 at 21:39









Yuri NegometyanovYuri Negometyanov

11.9k1729




11.9k1729












  • $begingroup$
    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




















  • $begingroup$
    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


















$begingroup$
Does your answer relate the integral to the area?
$endgroup$
– Mythomorphic
Aug 12 '18 at 3:09




$begingroup$
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




$begingroup$
@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






$begingroup$
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




$begingroup$
@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






$begingroup$
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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