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multi-dimensional integral of modified Vandermonde determinant
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I'm looking for suggestions on how one might try to compute the following $(N-1)$-dimensional integral:
$$I_N = frac1(2pi)^N-1(N-1)! intcdotsint \
beginvmatrix
1 & 1 & cdots & 1 & 1 \
e^ialpha_1 & e^ialpha_2 & cdots & e^ialpha_N-1 & e^-i(alpha_1+dots+alpha_N-1) \
e^i2alpha_1 & e^i2alpha_2 & cdots & e^i2alpha_N-1 & e^-i2(alpha_1+dots+alpha_N-1) \
vdots & vdots & ddots & vdots & vdots \
e^i(N-1)alpha_1 & e^i(N-1)alpha_2 & cdots & e^i(N-1)alpha_N-1 & e^-i(N-1)(alpha_1+dots+alpha_N-1) \
endvmatrix^2
dalpha_1cdots dalpha_N-1$$
over the region given by the following conditions:
$$alpha_N-1inleft(-pi, -pi+frac2piNright) text and alpha_1,dots,alpha_N-2,-(alpha_1+dots+alpha_N-1)inleft(-pi+frac2piN, piright)$$
More generally, the integration region could be described as: one of the variables $alpha_i$, $1leq ileq N-1$ must be inside the interval $left(-pi, -pi+frac2piNright)$, and the rest of the $alpha_i$, together with $-alpha_1-dots-alpha_N-1$ must be outside this interval $mod 2pi$. $alpha_N-1$ is just a particular choice and the value of $I_N$ doesn't change if $alpha_N-1$ is replaced by some other $alpha_i$; it also doesn't change if the interval
$left(-pi, -pi+frac2piNright)$ gets replaced with $left(pi-frac2piN, piright)$
Regarding the shape of the integration region, in the $N=3$ case it is a hexagon, in the $N=4$ case it's an octahedron, and in general, the region for $I_N$ is an $(N-1)$-dimensional polytope with $2N$ sides. The normalization factor in front of the integral is just to ensure that $0leq I_Nleq 1$
The integrand is the square of the absolute value of the determinant, and can also be expressed as
$$prod_1leq j<kleq N-1 |e^ialpha_j-e^ialpha_k|^2 prod_1leq jleq N-1 |e^ialpha_j-e^-i(alpha_1+dots+alpha_N-1)|^2$$
or equivalently
$$2^N(N-1) prod_1leq j<kleq N-1 sin^2fracalpha_j-alpha_k2 prod_1leq jleq N-1 sin^2fracalpha_j+alpha_1+dots+alpha_N-12 $$
Using Wolfram Mathematica, I have explicitly computed the values of $I_N$ when $N=2,3,4$
beginalign
I_2 & =1 \[8pt]
I_3 & =frac712+frac113pi^2 \[8pt]
I_4 & =frac512+frac1001360pi^2+frac4232675pi^3
endalign
and I've noticed that if $Ngeq 4, I_N$ can be split into two equal iterated integrals, for which only the first two variables $alpha_1$ and $alpha_2$ have non-constant integration limits:
$$ frac1(2pi)^N-1(N-1)! displaystyle int_-pi^-pi+frac2piN int_-pi+frac2piN^pi dots int_-pi+frac2piN^pi left( int_-frac2piN-alpha_3-ldots-alpha_N-1^pi int_-pi+frac2piN^pi-frac2piN-alpha_2-ldots-alpha_N-1 |dots|^2 dalpha_1 dalpha_2 right) dalpha_3 dots dalpha_N-2 dalpha_N-1 = $$
$$ = frac1(2pi)^N-1(N-1)! displaystyle int_-pi^-pi+frac2piN int_-pi+frac2piN^pi dots int_-pi+frac2piN^pi left( int_-pi+frac2piN^-frac2piN-alpha_3-ldots-alpha_N-1 int_-pi-alpha_2-ldots-alpha_N-1^pi |dots|^2 dalpha_1 dalpha_2 right) dalpha_3 dots dalpha_N-2 dalpha_N-1 = frac12I_N $$
but so far, I haven't been able to figure out a method for computing either of these two integrals for arbitrary $N$.
multivariable-calculus definite-integrals determinant multiple-integral random-matrices
asked Jul 22 at 14:48
Amadocta
163
add a comment |Â
up vote
3
down vote
favorite
I'm looking for suggestions on how one might try to compute the following $(N-1)$-dimensional integral:
$$I_N = frac1(2pi)^N-1(N-1)! intcdotsint \
beginvmatrix
1 & 1 & cdots & 1 & 1 \
e^ialpha_1 & e^ialpha_2 & cdots & e^ialpha_N-1 & e^-i(alpha_1+dots+alpha_N-1) \
e^i2alpha_1 & e^i2alpha_2 & cdots & e^i2alpha_N-1 & e^-i2(alpha_1+dots+alpha_N-1) \
vdots & vdots & ddots & vdots & vdots \
e^i(N-1)alpha_1 & e^i(N-1)alpha_2 & cdots & e^i(N-1)alpha_N-1 & e^-i(N-1)(alpha_1+dots+alpha_N-1) \
endvmatrix^2
dalpha_1cdots dalpha_N-1$$
over the region given by the following conditions:
$$alpha_N-1inleft(-pi, -pi+frac2piNright) text and alpha_1,dots,alpha_N-2,-(alpha_1+dots+alpha_N-1)inleft(-pi+frac2piN, piright)$$
More generally, the integration region could be described as: one of the variables $alpha_i$, $1leq ileq N-1$ must be inside the interval $left(-pi, -pi+frac2piNright)$, and the rest of the $alpha_i$, together with $-alpha_1-dots-alpha_N-1$ must be outside this interval $mod 2pi$. $alpha_N-1$ is just a particular choice and the value of $I_N$ doesn't change if $alpha_N-1$ is replaced by some other $alpha_i$; it also doesn't change if the interval
$left(-pi, -pi+frac2piNright)$ gets replaced with $left(pi-frac2piN, piright)$
Regarding the shape of the integration region, in the $N=3$ case it is a hexagon, in the $N=4$ case it's an octahedron, and in general, the region for $I_N$ is an $(N-1)$-dimensional polytope with $2N$ sides. The normalization factor in front of the integral is just to ensure that $0leq I_Nleq 1$
The integrand is the square of the absolute value of the determinant, and can also be expressed as
$$prod_1leq j<kleq N-1 |e^ialpha_j-e^ialpha_k|^2 prod_1leq jleq N-1 |e^ialpha_j-e^-i(alpha_1+dots+alpha_N-1)|^2$$
or equivalently
$$2^N(N-1) prod_1leq j<kleq N-1 sin^2fracalpha_j-alpha_k2 prod_1leq jleq N-1 sin^2fracalpha_j+alpha_1+dots+alpha_N-12 $$
Using Wolfram Mathematica, I have explicitly computed the values of $I_N$ when $N=2,3,4$
beginalign
I_2 & =1 \[8pt]
I_3 & =frac712+frac113pi^2 \[8pt]
I_4 & =frac512+frac1001360pi^2+frac4232675pi^3
endalign
and I've noticed that if $Ngeq 4, I_N$ can be split into two equal iterated integrals, for which only the first two variables $alpha_1$ and $alpha_2$ have non-constant integration limits:
$$ frac1(2pi)^N-1(N-1)! displaystyle int_-pi^-pi+frac2piN int_-pi+frac2piN^pi dots int_-pi+frac2piN^pi left( int_-frac2piN-alpha_3-ldots-alpha_N-1^pi int_-pi+frac2piN^pi-frac2piN-alpha_2-ldots-alpha_N-1 |dots|^2 dalpha_1 dalpha_2 right) dalpha_3 dots dalpha_N-2 dalpha_N-1 = $$
$$ = frac1(2pi)^N-1(N-1)! displaystyle int_-pi^-pi+frac2piN int_-pi+frac2piN^pi dots int_-pi+frac2piN^pi left( int_-pi+frac2piN^-frac2piN-alpha_3-ldots-alpha_N-1 int_-pi-alpha_2-ldots-alpha_N-1^pi |dots|^2 dalpha_1 dalpha_2 right) dalpha_3 dots dalpha_N-2 dalpha_N-1 = frac12I_N $$
but so far, I haven't been able to figure out a method for computing either of these two integrals for arbitrary $N$.
multivariable-calculus definite-integrals determinant multiple-integral random-matrices
asked Jul 22 at 14:48
Amadocta
163
Are you sure your integrals $I_2,I_3,I_4$ are correct? I get a different result. For example, for $N=2$, the determinant is $e^-ialpha_1-e^+ialpha_1$, and therefore $I_2=frac14piint_-pi^0 4sin^2(alpha_1)mathrm dalpha_1= 1/2neq1$.
– AccidentalFourierTransform
Jul 22 at 19:12
1
Sorry, I made a typo in the formula for $I_N$: it was supposed to be $(N-1)!$ instead of $N!$. Now, the values for $I_2, I_3, I_4$ are correct.
– Amadocta
Jul 22 at 20:11
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm looking for suggestions on how one might try to compute the following $(N-1)$-dimensional integral:
$$I_N = frac1(2pi)^N-1(N-1)! intcdotsint \
beginvmatrix
1 & 1 & cdots & 1 & 1 \
e^ialpha_1 & e^ialpha_2 & cdots & e^ialpha_N-1 & e^-i(alpha_1+dots+alpha_N-1) \
e^i2alpha_1 & e^i2alpha_2 & cdots & e^i2alpha_N-1 & e^-i2(alpha_1+dots+alpha_N-1) \
vdots & vdots & ddots & vdots & vdots \
e^i(N-1)alpha_1 & e^i(N-1)alpha_2 & cdots & e^i(N-1)alpha_N-1 & e^-i(N-1)(alpha_1+dots+alpha_N-1) \
endvmatrix^2
dalpha_1cdots dalpha_N-1$$
over the region given by the following conditions:
$$alpha_N-1inleft(-pi, -pi+frac2piNright) text and alpha_1,dots,alpha_N-2,-(alpha_1+dots+alpha_N-1)inleft(-pi+frac2piN, piright)$$
More generally, the integration region could be described as: one of the variables $alpha_i$, $1leq ileq N-1$ must be inside the interval $left(-pi, -pi+frac2piNright)$, and the rest of the $alpha_i$, together with $-alpha_1-dots-alpha_N-1$ must be outside this interval $mod 2pi$. $alpha_N-1$ is just a particular choice and the value of $I_N$ doesn't change if $alpha_N-1$ is replaced by some other $alpha_i$; it also doesn't change if the interval
$left(-pi, -pi+frac2piNright)$ gets replaced with $left(pi-frac2piN, piright)$
Regarding the shape of the integration region, in the $N=3$ case it is a hexagon, in the $N=4$ case it's an octahedron, and in general, the region for $I_N$ is an $(N-1)$-dimensional polytope with $2N$ sides. The normalization factor in front of the integral is just to ensure that $0leq I_Nleq 1$
The integrand is the square of the absolute value of the determinant, and can also be expressed as
$$prod_1leq j<kleq N-1 |e^ialpha_j-e^ialpha_k|^2 prod_1leq jleq N-1 |e^ialpha_j-e^-i(alpha_1+dots+alpha_N-1)|^2$$
or equivalently
$$2^N(N-1) prod_1leq j<kleq N-1 sin^2fracalpha_j-alpha_k2 prod_1leq jleq N-1 sin^2fracalpha_j+alpha_1+dots+alpha_N-12 $$
Using Wolfram Mathematica, I have explicitly computed the values of $I_N$ when $N=2,3,4$
beginalign
I_2 & =1 \[8pt]
I_3 & =frac712+frac113pi^2 \[8pt]
I_4 & =frac512+frac1001360pi^2+frac4232675pi^3
endalign
and I've noticed that if $Ngeq 4, I_N$ can be split into two equal iterated integrals, for which only the first two variables $alpha_1$ and $alpha_2$ have non-constant integration limits:
$$ frac1(2pi)^N-1(N-1)! displaystyle int_-pi^-pi+frac2piN int_-pi+frac2piN^pi dots int_-pi+frac2piN^pi left( int_-frac2piN-alpha_3-ldots-alpha_N-1^pi int_-pi+frac2piN^pi-frac2piN-alpha_2-ldots-alpha_N-1 |dots|^2 dalpha_1 dalpha_2 right) dalpha_3 dots dalpha_N-2 dalpha_N-1 = $$
$$ = frac1(2pi)^N-1(N-1)! displaystyle int_-pi^-pi+frac2piN int_-pi+frac2piN^pi dots int_-pi+frac2piN^pi left( int_-pi+frac2piN^-frac2piN-alpha_3-ldots-alpha_N-1 int_-pi-alpha_2-ldots-alpha_N-1^pi |dots|^2 dalpha_1 dalpha_2 right) dalpha_3 dots dalpha_N-2 dalpha_N-1 = frac12I_N $$
but so far, I haven't been able to figure out a method for computing either of these two integrals for arbitrary $N$.
multivariable-calculus definite-integrals determinant multiple-integral random-matrices
asked Jul 22 at 14:48
Amadocta
163
I'm looking for suggestions on how one might try to compute the following $(N-1)$-dimensional integral:
$$I_N = frac1(2pi)^N-1(N-1)! intcdotsint \
beginvmatrix
1 & 1 & cdots & 1 & 1 \
e^ialpha_1 & e^ialpha_2 & cdots & e^ialpha_N-1 & e^-i(alpha_1+dots+alpha_N-1) \
e^i2alpha_1 & e^i2alpha_2 & cdots & e^i2alpha_N-1 & e^-i2(alpha_1+dots+alpha_N-1) \
vdots & vdots & ddots & vdots & vdots \
e^i(N-1)alpha_1 & e^i(N-1)alpha_2 & cdots & e^i(N-1)alpha_N-1 & e^-i(N-1)(alpha_1+dots+alpha_N-1) \
endvmatrix^2
dalpha_1cdots dalpha_N-1$$
over the region given by the following conditions:
$$alpha_N-1inleft(-pi, -pi+frac2piNright) text and alpha_1,dots,alpha_N-2,-(alpha_1+dots+alpha_N-1)inleft(-pi+frac2piN, piright)$$
More generally, the integration region could be described as: one of the variables $alpha_i$, $1leq ileq N-1$ must be inside the interval $left(-pi, -pi+frac2piNright)$, and the rest of the $alpha_i$, together with $-alpha_1-dots-alpha_N-1$ must be outside this interval $mod 2pi$. $alpha_N-1$ is just a particular choice and the value of $I_N$ doesn't change if $alpha_N-1$ is replaced by some other $alpha_i$; it also doesn't change if the interval
$left(-pi, -pi+frac2piNright)$ gets replaced with $left(pi-frac2piN, piright)$
Regarding the shape of the integration region, in the $N=3$ case it is a hexagon, in the $N=4$ case it's an octahedron, and in general, the region for $I_N$ is an $(N-1)$-dimensional polytope with $2N$ sides. The normalization factor in front of the integral is just to ensure that $0leq I_Nleq 1$
The integrand is the square of the absolute value of the determinant, and can also be expressed as
$$prod_1leq j<kleq N-1 |e^ialpha_j-e^ialpha_k|^2 prod_1leq jleq N-1 |e^ialpha_j-e^-i(alpha_1+dots+alpha_N-1)|^2$$
or equivalently
$$2^N(N-1) prod_1leq j<kleq N-1 sin^2fracalpha_j-alpha_k2 prod_1leq jleq N-1 sin^2fracalpha_j+alpha_1+dots+alpha_N-12 $$
Using Wolfram Mathematica, I have explicitly computed the values of $I_N$ when $N=2,3,4$
beginalign
I_2 & =1 \[8pt]
I_3 & =frac712+frac113pi^2 \[8pt]
I_4 & =frac512+frac1001360pi^2+frac4232675pi^3
endalign
and I've noticed that if $Ngeq 4, I_N$ can be split into two equal iterated integrals, for which only the first two variables $alpha_1$ and $alpha_2$ have non-constant integration limits:
$$ frac1(2pi)^N-1(N-1)! displaystyle int_-pi^-pi+frac2piN int_-pi+frac2piN^pi dots int_-pi+frac2piN^pi left( int_-frac2piN-alpha_3-ldots-alpha_N-1^pi int_-pi+frac2piN^pi-frac2piN-alpha_2-ldots-alpha_N-1 |dots|^2 dalpha_1 dalpha_2 right) dalpha_3 dots dalpha_N-2 dalpha_N-1 = $$
$$ = frac1(2pi)^N-1(N-1)! displaystyle int_-pi^-pi+frac2piN int_-pi+frac2piN^pi dots int_-pi+frac2piN^pi left( int_-pi+frac2piN^-frac2piN-alpha_3-ldots-alpha_N-1 int_-pi-alpha_2-ldots-alpha_N-1^pi |dots|^2 dalpha_1 dalpha_2 right) dalpha_3 dots dalpha_N-2 dalpha_N-1 = frac12I_N $$
but so far, I haven't been able to figure out a method for computing either of these two integrals for arbitrary $N$.
multivariable-calculus definite-integrals determinant multiple-integral random-matrices
asked Jul 22 at 14:48
Amadocta
163
edited Aug 5 at 11:16
asked Jul 22 at 14:48
Amadocta
163
asked Jul 22 at 14:48
Amadocta
163
asked Jul 22 at 14:48
Amadocta
163
163
Are you sure your integrals $I_2,I_3,I_4$ are correct? I get a different result. For example, for $N=2$, the determinant is $e^-ialpha_1-e^+ialpha_1$, and therefore $I_2=frac14piint_-pi^0 4sin^2(alpha_1)mathrm dalpha_1= 1/2neq1$.
– AccidentalFourierTransform
Jul 22 at 19:12
1
Sorry, I made a typo in the formula for $I_N$: it was supposed to be $(N-1)!$ instead of $N!$. Now, the values for $I_2, I_3, I_4$ are correct.
– Amadocta
Jul 22 at 20:11
add a comment |Â
Are you sure your integrals $I_2,I_3,I_4$ are correct? I get a different result. For example, for $N=2$, the determinant is $e^-ialpha_1-e^+ialpha_1$, and therefore $I_2=frac14piint_-pi^0 4sin^2(alpha_1)mathrm dalpha_1= 1/2neq1$.
– AccidentalFourierTransform
Jul 22 at 19:12
1
Sorry, I made a typo in the formula for $I_N$: it was supposed to be $(N-1)!$ instead of $N!$. Now, the values for $I_2, I_3, I_4$ are correct.
– Amadocta
Jul 22 at 20:11
Are you sure your integrals $I_2,I_3,I_4$ are correct? I get a different result. For example, for $N=2$, the determinant is $e^-ialpha_1-e^+ialpha_1$, and therefore $I_2=frac14piint_-pi^0 4sin^2(alpha_1)mathrm dalpha_1= 1/2neq1$.
– AccidentalFourierTransform
Jul 22 at 19:12
Are you sure your integrals $I_2,I_3,I_4$ are correct? I get a different result. For example, for $N=2$, the determinant is $e^-ialpha_1-e^+ialpha_1$, and therefore $I_2=frac14piint_-pi^0 4sin^2(alpha_1)mathrm dalpha_1= 1/2neq1$.
– AccidentalFourierTransform
Jul 22 at 19:12
1
1
Sorry, I made a typo in the formula for $I_N$: it was supposed to be $(N-1)!$ instead of $N!$. Now, the values for $I_2, I_3, I_4$ are correct.
– Amadocta
Jul 22 at 20:11
Sorry, I made a typo in the formula for $I_N$: it was supposed to be $(N-1)!$ instead of $N!$. Now, the values for $I_2, I_3, I_4$ are correct.
– Amadocta
Jul 22 at 20:11
add a comment |Â
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Are you sure your integrals $I_2,I_3,I_4$ are correct? I get a different result. For example, for $N=2$, the determinant is $e^-ialpha_1-e^+ialpha_1$, and therefore $I_2=frac14piint_-pi^0 4sin^2(alpha_1)mathrm dalpha_1= 1/2neq1$.
– AccidentalFourierTransform
Jul 22 at 19:12
1
Sorry, I made a typo in the formula for $I_N$: it was supposed to be $(N-1)!$ instead of $N!$. Now, the values for $I_2, I_3, I_4$ are correct.
– Amadocta
Jul 22 at 20:11