Integrate: $int_0^1 ||vec x||^{-m}mathrm{d}vec{x}$
$begingroup$
Let $m,n$ be two positive integers with $0 < m < n$.
Can we integrate this:
$$I = int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2}$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$lim_{nrightarrowinfty} frac{1}{n} log [Gamma(n/2) I]$$
where it is assumed that the ratio $m/n$ remains fixed. I introduce the factor $Gamma(n/2)$ so that the argument to the log has simple exponential growth.
Related: Integrate: $int_{-infty}^infty exp(-||vec x||^2) ||vec x||^{-m}mathrm{d}vec{x}$.
definite-integrals large-deviation-theory
asked Nov 29 '18 at 20:27
beckobecko
2,35431942
$endgroup$
|
show 6 more comments
$begingroup$
Let $m,n$ be two positive integers with $0 < m < n$.
Can we integrate this:
$$I = int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2}$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$lim_{nrightarrowinfty} frac{1}{n} log [Gamma(n/2) I]$$
where it is assumed that the ratio $m/n$ remains fixed. I introduce the factor $Gamma(n/2)$ so that the argument to the log has simple exponential growth.
Related: Integrate: $int_{-infty}^infty exp(-||vec x||^2) ||vec x||^{-m}mathrm{d}vec{x}$.
definite-integrals large-deviation-theory
asked Nov 29 '18 at 20:27
beckobecko
2,35431942
$endgroup$
$begingroup$
It can certainly be estimated easily in polar coordinates.
$endgroup$
– T. Bongers
Nov 29 '18 at 20:28
$begingroup$
@T.Bongers So no exact analytical formula?
$endgroup$
– becko
Nov 29 '18 at 20:29
$begingroup$
I wouldn't be surprised if there is one... but depending on the application, do you really need one, or is an estimate good enough?
$endgroup$
– T. Bongers
Nov 29 '18 at 20:29
$begingroup$
@T.Bongers I need an analytical formula. But I will then take the limit $m,n rightarrowinfty$ (with a fixed ratio $m/n$). So in reality I only require an asymptotic estimate.
$endgroup$
– becko
Nov 29 '18 at 20:31
$begingroup$
@T.Bongers Specifically I'll be satisfied by a large deviation limit $lim_{nrightarrowinfty} (1/n) ln I$, where $I$ is the above integral and the ratio $m/n$ is fixed.
$endgroup$
– becko
Nov 29 '18 at 21:05
|
show 6 more comments
$begingroup$
Let $m,n$ be two positive integers with $0 < m < n$.
Can we integrate this:
$$I = int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2}$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$lim_{nrightarrowinfty} frac{1}{n} log [Gamma(n/2) I]$$
where it is assumed that the ratio $m/n$ remains fixed. I introduce the factor $Gamma(n/2)$ so that the argument to the log has simple exponential growth.
Related: Integrate: $int_{-infty}^infty exp(-||vec x||^2) ||vec x||^{-m}mathrm{d}vec{x}$.
definite-integrals large-deviation-theory
asked Nov 29 '18 at 20:27
beckobecko
2,35431942
$endgroup$
Let $m,n$ be two positive integers with $0 < m < n$.
Can we integrate this:
$$I = int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2}$$
If a closed analytical expression is not possible, what I really need is to evaluate a large deviation limit of the form:
$$lim_{nrightarrowinfty} frac{1}{n} log [Gamma(n/2) I]$$
where it is assumed that the ratio $m/n$ remains fixed. I introduce the factor $Gamma(n/2)$ so that the argument to the log has simple exponential growth.
Related: Integrate: $int_{-infty}^infty exp(-||vec x||^2) ||vec x||^{-m}mathrm{d}vec{x}$.
definite-integrals large-deviation-theory
definite-integrals large-deviation-theory
asked Nov 29 '18 at 20:27
beckobecko
2,35431942
asked Nov 29 '18 at 20:27
beckobecko
2,35431942
edited Dec 6 '18 at 19:25
becko
asked Nov 29 '18 at 20:27
beckobecko
2,35431942
asked Nov 29 '18 at 20:27
beckobecko
2,35431942
asked Nov 29 '18 at 20:27
beckobecko
2,35431942
2,35431942
$begingroup$
It can certainly be estimated easily in polar coordinates.
$endgroup$
– T. Bongers
Nov 29 '18 at 20:28
$begingroup$
@T.Bongers So no exact analytical formula?
$endgroup$
– becko
Nov 29 '18 at 20:29
$begingroup$
I wouldn't be surprised if there is one... but depending on the application, do you really need one, or is an estimate good enough?
$endgroup$
– T. Bongers
Nov 29 '18 at 20:29
$begingroup$
@T.Bongers I need an analytical formula. But I will then take the limit $m,n rightarrowinfty$ (with a fixed ratio $m/n$). So in reality I only require an asymptotic estimate.
$endgroup$
– becko
Nov 29 '18 at 20:31
$begingroup$
@T.Bongers Specifically I'll be satisfied by a large deviation limit $lim_{nrightarrowinfty} (1/n) ln I$, where $I$ is the above integral and the ratio $m/n$ is fixed.
$endgroup$
– becko
Nov 29 '18 at 21:05
|
show 6 more comments
$begingroup$
It can certainly be estimated easily in polar coordinates.
$endgroup$
– T. Bongers
Nov 29 '18 at 20:28
$begingroup$
@T.Bongers So no exact analytical formula?
$endgroup$
– becko
Nov 29 '18 at 20:29
$begingroup$
I wouldn't be surprised if there is one... but depending on the application, do you really need one, or is an estimate good enough?
$endgroup$
– T. Bongers
Nov 29 '18 at 20:29
$begingroup$
@T.Bongers I need an analytical formula. But I will then take the limit $m,n rightarrowinfty$ (with a fixed ratio $m/n$). So in reality I only require an asymptotic estimate.
$endgroup$
– becko
Nov 29 '18 at 20:31
$begingroup$
@T.Bongers Specifically I'll be satisfied by a large deviation limit $lim_{nrightarrowinfty} (1/n) ln I$, where $I$ is the above integral and the ratio $m/n$ is fixed.
$endgroup$
– becko
Nov 29 '18 at 21:05
$begingroup$
It can certainly be estimated easily in polar coordinates.
$endgroup$
– T. Bongers
Nov 29 '18 at 20:28
$begingroup$
It can certainly be estimated easily in polar coordinates.
$endgroup$
– T. Bongers
Nov 29 '18 at 20:28
$begingroup$
@T.Bongers So no exact analytical formula?
$endgroup$
– becko
Nov 29 '18 at 20:29
$begingroup$
@T.Bongers So no exact analytical formula?
$endgroup$
– becko
Nov 29 '18 at 20:29
$begingroup$
I wouldn't be surprised if there is one... but depending on the application, do you really need one, or is an estimate good enough?
$endgroup$
– T. Bongers
Nov 29 '18 at 20:29
$begingroup$
I wouldn't be surprised if there is one... but depending on the application, do you really need one, or is an estimate good enough?
$endgroup$
– T. Bongers
Nov 29 '18 at 20:29
$begingroup$
@T.Bongers I need an analytical formula. But I will then take the limit $m,n rightarrowinfty$ (with a fixed ratio $m/n$). So in reality I only require an asymptotic estimate.
$endgroup$
– becko
Nov 29 '18 at 20:31
$begingroup$
@T.Bongers I need an analytical formula. But I will then take the limit $m,n rightarrowinfty$ (with a fixed ratio $m/n$). So in reality I only require an asymptotic estimate.
$endgroup$
– becko
Nov 29 '18 at 20:31
$begingroup$
@T.Bongers Specifically I'll be satisfied by a large deviation limit $lim_{nrightarrowinfty} (1/n) ln I$, where $I$ is the above integral and the ratio $m/n$ is fixed.
$endgroup$
– becko
Nov 29 '18 at 21:05
$begingroup$
@T.Bongers Specifically I'll be satisfied by a large deviation limit $lim_{nrightarrowinfty} (1/n) ln I$, where $I$ is the above integral and the ratio $m/n$ is fixed.
$endgroup$
– becko
Nov 29 '18 at 21:05
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
An upper bound for $n>m$,
$$ int_{[0,1]^n} frac1{|x|^m} dx le frac1{2^n}int_{B(0,sqrt2)} frac1{|x|^m} dx = frac {C_n} {2^n} int_0^{sqrt 2} r^{n-1-m} dr = frac{C_n}{2^n(n-m)}sqrt{2}^{n-m} $$
$C_n$ is the area of the unit sphere (coming from the integral in angles),
$ C_n = 2pi^{n/2}/Gamma (n/2)$. I think it should be clear from here how to get an upper bound on $frac {log I}n$. A very similar lower bound is also possible by similarly cutting up $B(0,1)$ into $2^n$ slices(one cut along each hyperplane ${x_i=0})$, one of which is contained in $[0,1]^n$.
answered Dec 6 '18 at 16:03
Calvin KhorCalvin Khor
11.5k21438
$endgroup$
$begingroup$
Your answer made me realize that I need to introduce a factor $Gamma(n/2)$ to get simple exponential growth. Having done that, your technique produces lower and upper bounds that have different rates of exponential growth. Therefore these bounds are not tight enough to produce a large deviation principle. Or am I missing something? Thanks!
$endgroup$
– becko
Dec 6 '18 at 17:12
$begingroup$
@becko No, I don't think you're missing anything. I haven't thought of a better approximation
$endgroup$
– Calvin Khor
Dec 6 '18 at 17:15
$begingroup$
@becko While I don't think this resolves your question, this paper undoubtedly goes far further than the basic computation I made projecteuclid.org/download/pdf_1/euclid.em/1317758103
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:18
1
$begingroup$
Thanks. According to davidhbailey.com/dhbpapers/boxintegrals.pdf, we have that $I(n,m) ~ (n/3)^{-m/2}$ for large $n$ and fixed $m$. Unfortunately they say nothing about the large $n,m$ limit with fixed ratio $m/n$.
$endgroup$
– becko
Dec 6 '18 at 18:27
$begingroup$
@becko I'm very pleased to see divergence theorem useful, my small scribblings didn't bring me to that conclusion. And I would guess that a clever interchange of integrals doesn't help much either. Good luck
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:32
|
show 2 more comments
$begingroup$
According to https://www.davidhbailey.com/dhbpapers/boxintegrals.pdf, Eq. (33), we have the following exact relation:
$$begin{aligned}
I(m,n) &= int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} \
&= frac{2^{1 - n} pi^{n/2}}{Gamma(m/2)} int_0^{infty}
frac{[mathrm{erf}(u)]^n}{u^{n-m+1}} mathrm d u
end{aligned}$$
For large $m,n$ and fixed $alpha=m/n$, we can evaluate the integral by Laplace's method.
$$int_0^{infty} frac{[mathrm {erf} (u)]^N}{u^{N - M + 1}} mathrm d u asymp exp { N [ln mathrm{erf} (u^{ast}) - (1 - alpha) ln u^{ast}] }$$
where $u^*$ maximizes $ln mathrm{erf}^{} (u) - (1 - alpha) ln u$ with respect to $uge0$.
The notation $a_nasymp b_n$ from large deviation theory means that $lim (1/n)ln a_n = lim(1/n)ln b_n$.
Differentiating we find that $u^{ast}$ is the root of the equation:
$$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$
answered Dec 7 '18 at 21:25
beckobecko
2,35431942
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An upper bound for $n>m$,
$$ int_{[0,1]^n} frac1{|x|^m} dx le frac1{2^n}int_{B(0,sqrt2)} frac1{|x|^m} dx = frac {C_n} {2^n} int_0^{sqrt 2} r^{n-1-m} dr = frac{C_n}{2^n(n-m)}sqrt{2}^{n-m} $$
$C_n$ is the area of the unit sphere (coming from the integral in angles),
$ C_n = 2pi^{n/2}/Gamma (n/2)$. I think it should be clear from here how to get an upper bound on $frac {log I}n$. A very similar lower bound is also possible by similarly cutting up $B(0,1)$ into $2^n$ slices(one cut along each hyperplane ${x_i=0})$, one of which is contained in $[0,1]^n$.
answered Dec 6 '18 at 16:03
Calvin KhorCalvin Khor
11.5k21438
$endgroup$
$begingroup$
Your answer made me realize that I need to introduce a factor $Gamma(n/2)$ to get simple exponential growth. Having done that, your technique produces lower and upper bounds that have different rates of exponential growth. Therefore these bounds are not tight enough to produce a large deviation principle. Or am I missing something? Thanks!
$endgroup$
– becko
Dec 6 '18 at 17:12
$begingroup$
@becko No, I don't think you're missing anything. I haven't thought of a better approximation
$endgroup$
– Calvin Khor
Dec 6 '18 at 17:15
$begingroup$
@becko While I don't think this resolves your question, this paper undoubtedly goes far further than the basic computation I made projecteuclid.org/download/pdf_1/euclid.em/1317758103
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:18
1
$begingroup$
Thanks. According to davidhbailey.com/dhbpapers/boxintegrals.pdf, we have that $I(n,m) ~ (n/3)^{-m/2}$ for large $n$ and fixed $m$. Unfortunately they say nothing about the large $n,m$ limit with fixed ratio $m/n$.
$endgroup$
– becko
Dec 6 '18 at 18:27
$begingroup$
@becko I'm very pleased to see divergence theorem useful, my small scribblings didn't bring me to that conclusion. And I would guess that a clever interchange of integrals doesn't help much either. Good luck
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:32
|
show 2 more comments
$begingroup$
An upper bound for $n>m$,
$$ int_{[0,1]^n} frac1{|x|^m} dx le frac1{2^n}int_{B(0,sqrt2)} frac1{|x|^m} dx = frac {C_n} {2^n} int_0^{sqrt 2} r^{n-1-m} dr = frac{C_n}{2^n(n-m)}sqrt{2}^{n-m} $$
$C_n$ is the area of the unit sphere (coming from the integral in angles),
$ C_n = 2pi^{n/2}/Gamma (n/2)$. I think it should be clear from here how to get an upper bound on $frac {log I}n$. A very similar lower bound is also possible by similarly cutting up $B(0,1)$ into $2^n$ slices(one cut along each hyperplane ${x_i=0})$, one of which is contained in $[0,1]^n$.
answered Dec 6 '18 at 16:03
Calvin KhorCalvin Khor
11.5k21438
$endgroup$
$begingroup$
Your answer made me realize that I need to introduce a factor $Gamma(n/2)$ to get simple exponential growth. Having done that, your technique produces lower and upper bounds that have different rates of exponential growth. Therefore these bounds are not tight enough to produce a large deviation principle. Or am I missing something? Thanks!
$endgroup$
– becko
Dec 6 '18 at 17:12
$begingroup$
@becko No, I don't think you're missing anything. I haven't thought of a better approximation
$endgroup$
– Calvin Khor
Dec 6 '18 at 17:15
$begingroup$
@becko While I don't think this resolves your question, this paper undoubtedly goes far further than the basic computation I made projecteuclid.org/download/pdf_1/euclid.em/1317758103
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:18
1
$begingroup$
Thanks. According to davidhbailey.com/dhbpapers/boxintegrals.pdf, we have that $I(n,m) ~ (n/3)^{-m/2}$ for large $n$ and fixed $m$. Unfortunately they say nothing about the large $n,m$ limit with fixed ratio $m/n$.
$endgroup$
– becko
Dec 6 '18 at 18:27
$begingroup$
@becko I'm very pleased to see divergence theorem useful, my small scribblings didn't bring me to that conclusion. And I would guess that a clever interchange of integrals doesn't help much either. Good luck
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:32
|
show 2 more comments
$begingroup$
An upper bound for $n>m$,
$$ int_{[0,1]^n} frac1{|x|^m} dx le frac1{2^n}int_{B(0,sqrt2)} frac1{|x|^m} dx = frac {C_n} {2^n} int_0^{sqrt 2} r^{n-1-m} dr = frac{C_n}{2^n(n-m)}sqrt{2}^{n-m} $$
$C_n$ is the area of the unit sphere (coming from the integral in angles),
$ C_n = 2pi^{n/2}/Gamma (n/2)$. I think it should be clear from here how to get an upper bound on $frac {log I}n$. A very similar lower bound is also possible by similarly cutting up $B(0,1)$ into $2^n$ slices(one cut along each hyperplane ${x_i=0})$, one of which is contained in $[0,1]^n$.
answered Dec 6 '18 at 16:03
Calvin KhorCalvin Khor
11.5k21438
$endgroup$
An upper bound for $n>m$,
$$ int_{[0,1]^n} frac1{|x|^m} dx le frac1{2^n}int_{B(0,sqrt2)} frac1{|x|^m} dx = frac {C_n} {2^n} int_0^{sqrt 2} r^{n-1-m} dr = frac{C_n}{2^n(n-m)}sqrt{2}^{n-m} $$
$C_n$ is the area of the unit sphere (coming from the integral in angles),
$ C_n = 2pi^{n/2}/Gamma (n/2)$. I think it should be clear from here how to get an upper bound on $frac {log I}n$. A very similar lower bound is also possible by similarly cutting up $B(0,1)$ into $2^n$ slices(one cut along each hyperplane ${x_i=0})$, one of which is contained in $[0,1]^n$.
answered Dec 6 '18 at 16:03
Calvin KhorCalvin Khor
11.5k21438
answered Dec 6 '18 at 16:03
Calvin KhorCalvin Khor
11.5k21438
answered Dec 6 '18 at 16:03
Calvin KhorCalvin Khor
11.5k21438
answered Dec 6 '18 at 16:03
Calvin KhorCalvin Khor
11.5k21438
11.5k21438
$begingroup$
Your answer made me realize that I need to introduce a factor $Gamma(n/2)$ to get simple exponential growth. Having done that, your technique produces lower and upper bounds that have different rates of exponential growth. Therefore these bounds are not tight enough to produce a large deviation principle. Or am I missing something? Thanks!
$endgroup$
– becko
Dec 6 '18 at 17:12
$begingroup$
@becko No, I don't think you're missing anything. I haven't thought of a better approximation
$endgroup$
– Calvin Khor
Dec 6 '18 at 17:15
$begingroup$
@becko While I don't think this resolves your question, this paper undoubtedly goes far further than the basic computation I made projecteuclid.org/download/pdf_1/euclid.em/1317758103
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:18
1
$begingroup$
Thanks. According to davidhbailey.com/dhbpapers/boxintegrals.pdf, we have that $I(n,m) ~ (n/3)^{-m/2}$ for large $n$ and fixed $m$. Unfortunately they say nothing about the large $n,m$ limit with fixed ratio $m/n$.
$endgroup$
– becko
Dec 6 '18 at 18:27
$begingroup$
@becko I'm very pleased to see divergence theorem useful, my small scribblings didn't bring me to that conclusion. And I would guess that a clever interchange of integrals doesn't help much either. Good luck
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:32
|
show 2 more comments
$begingroup$
Your answer made me realize that I need to introduce a factor $Gamma(n/2)$ to get simple exponential growth. Having done that, your technique produces lower and upper bounds that have different rates of exponential growth. Therefore these bounds are not tight enough to produce a large deviation principle. Or am I missing something? Thanks!
$endgroup$
– becko
Dec 6 '18 at 17:12
$begingroup$
@becko No, I don't think you're missing anything. I haven't thought of a better approximation
$endgroup$
– Calvin Khor
Dec 6 '18 at 17:15
$begingroup$
@becko While I don't think this resolves your question, this paper undoubtedly goes far further than the basic computation I made projecteuclid.org/download/pdf_1/euclid.em/1317758103
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:18
1
$begingroup$
Thanks. According to davidhbailey.com/dhbpapers/boxintegrals.pdf, we have that $I(n,m) ~ (n/3)^{-m/2}$ for large $n$ and fixed $m$. Unfortunately they say nothing about the large $n,m$ limit with fixed ratio $m/n$.
$endgroup$
– becko
Dec 6 '18 at 18:27
$begingroup$
@becko I'm very pleased to see divergence theorem useful, my small scribblings didn't bring me to that conclusion. And I would guess that a clever interchange of integrals doesn't help much either. Good luck
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:32
$begingroup$
Your answer made me realize that I need to introduce a factor $Gamma(n/2)$ to get simple exponential growth. Having done that, your technique produces lower and upper bounds that have different rates of exponential growth. Therefore these bounds are not tight enough to produce a large deviation principle. Or am I missing something? Thanks!
$endgroup$
– becko
Dec 6 '18 at 17:12
$begingroup$
Your answer made me realize that I need to introduce a factor $Gamma(n/2)$ to get simple exponential growth. Having done that, your technique produces lower and upper bounds that have different rates of exponential growth. Therefore these bounds are not tight enough to produce a large deviation principle. Or am I missing something? Thanks!
$endgroup$
– becko
Dec 6 '18 at 17:12
$begingroup$
@becko No, I don't think you're missing anything. I haven't thought of a better approximation
$endgroup$
– Calvin Khor
Dec 6 '18 at 17:15
$begingroup$
@becko No, I don't think you're missing anything. I haven't thought of a better approximation
$endgroup$
– Calvin Khor
Dec 6 '18 at 17:15
$begingroup$
@becko While I don't think this resolves your question, this paper undoubtedly goes far further than the basic computation I made projecteuclid.org/download/pdf_1/euclid.em/1317758103
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:18
$begingroup$
@becko While I don't think this resolves your question, this paper undoubtedly goes far further than the basic computation I made projecteuclid.org/download/pdf_1/euclid.em/1317758103
$endgroup$
– Calvin Khor
Dec 6 '18 at 18:18
1
1
$begingroup$
Thanks. According to davidhbailey.com/dhbpapers/boxintegrals.pdf, we have that $I(n,m) ~ (n/3)^{-m/2}$ for large $n$ and fixed $m$. Unfortunately they say nothing about the large $n,m$ limit with fixed ratio $m/n$.
$endgroup$
– becko
Dec 6 '18 at 18:27
$begingroup$
Thanks. According to davidhbailey.com/dhbpapers/boxintegrals.pdf, we have that $I(n,m) ~ (n/3)^{-m/2}$ for large $n$ and fixed $m$. Unfortunately they say nothing about the large $n,m$ limit with fixed ratio $m/n$.
$endgroup$
– becko
Dec 6 '18 at 18:27
$begingroup$
@becko I'm very pleased to see divergence theorem useful, my small scribblings didn't bring me to that conclusion. And I would guess that a clever interchange of integrals doesn't help much either. Good luck
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– Calvin Khor
Dec 6 '18 at 18:32
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@becko I'm very pleased to see divergence theorem useful, my small scribblings didn't bring me to that conclusion. And I would guess that a clever interchange of integrals doesn't help much either. Good luck
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– Calvin Khor
Dec 6 '18 at 18:32
|
show 2 more comments
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According to https://www.davidhbailey.com/dhbpapers/boxintegrals.pdf, Eq. (33), we have the following exact relation:
$$begin{aligned}
I(m,n) &= int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} \
&= frac{2^{1 - n} pi^{n/2}}{Gamma(m/2)} int_0^{infty}
frac{[mathrm{erf}(u)]^n}{u^{n-m+1}} mathrm d u
end{aligned}$$
For large $m,n$ and fixed $alpha=m/n$, we can evaluate the integral by Laplace's method.
$$int_0^{infty} frac{[mathrm {erf} (u)]^N}{u^{N - M + 1}} mathrm d u asymp exp { N [ln mathrm{erf} (u^{ast}) - (1 - alpha) ln u^{ast}] }$$
where $u^*$ maximizes $ln mathrm{erf}^{} (u) - (1 - alpha) ln u$ with respect to $uge0$.
The notation $a_nasymp b_n$ from large deviation theory means that $lim (1/n)ln a_n = lim(1/n)ln b_n$.
Differentiating we find that $u^{ast}$ is the root of the equation:
$$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$
answered Dec 7 '18 at 21:25
beckobecko
2,35431942
$endgroup$
add a comment |
$begingroup$
According to https://www.davidhbailey.com/dhbpapers/boxintegrals.pdf, Eq. (33), we have the following exact relation:
$$begin{aligned}
I(m,n) &= int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} \
&= frac{2^{1 - n} pi^{n/2}}{Gamma(m/2)} int_0^{infty}
frac{[mathrm{erf}(u)]^n}{u^{n-m+1}} mathrm d u
end{aligned}$$
For large $m,n$ and fixed $alpha=m/n$, we can evaluate the integral by Laplace's method.
$$int_0^{infty} frac{[mathrm {erf} (u)]^N}{u^{N - M + 1}} mathrm d u asymp exp { N [ln mathrm{erf} (u^{ast}) - (1 - alpha) ln u^{ast}] }$$
where $u^*$ maximizes $ln mathrm{erf}^{} (u) - (1 - alpha) ln u$ with respect to $uge0$.
The notation $a_nasymp b_n$ from large deviation theory means that $lim (1/n)ln a_n = lim(1/n)ln b_n$.
Differentiating we find that $u^{ast}$ is the root of the equation:
$$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$
answered Dec 7 '18 at 21:25
beckobecko
2,35431942
$endgroup$
add a comment |
$begingroup$
According to https://www.davidhbailey.com/dhbpapers/boxintegrals.pdf, Eq. (33), we have the following exact relation:
$$begin{aligned}
I(m,n) &= int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} \
&= frac{2^{1 - n} pi^{n/2}}{Gamma(m/2)} int_0^{infty}
frac{[mathrm{erf}(u)]^n}{u^{n-m+1}} mathrm d u
end{aligned}$$
For large $m,n$ and fixed $alpha=m/n$, we can evaluate the integral by Laplace's method.
$$int_0^{infty} frac{[mathrm {erf} (u)]^N}{u^{N - M + 1}} mathrm d u asymp exp { N [ln mathrm{erf} (u^{ast}) - (1 - alpha) ln u^{ast}] }$$
where $u^*$ maximizes $ln mathrm{erf}^{} (u) - (1 - alpha) ln u$ with respect to $uge0$.
The notation $a_nasymp b_n$ from large deviation theory means that $lim (1/n)ln a_n = lim(1/n)ln b_n$.
Differentiating we find that $u^{ast}$ is the root of the equation:
$$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$
answered Dec 7 '18 at 21:25
beckobecko
2,35431942
$endgroup$
According to https://www.davidhbailey.com/dhbpapers/boxintegrals.pdf, Eq. (33), we have the following exact relation:
$$begin{aligned}
I(m,n) &= int_0^1 mathrm{d}x_1 dots int_0^1 mathrm{d}x_n left(sum_{i=1}^n x_i^2right)^{-m/2} \
&= frac{2^{1 - n} pi^{n/2}}{Gamma(m/2)} int_0^{infty}
frac{[mathrm{erf}(u)]^n}{u^{n-m+1}} mathrm d u
end{aligned}$$
For large $m,n$ and fixed $alpha=m/n$, we can evaluate the integral by Laplace's method.
$$int_0^{infty} frac{[mathrm {erf} (u)]^N}{u^{N - M + 1}} mathrm d u asymp exp { N [ln mathrm{erf} (u^{ast}) - (1 - alpha) ln u^{ast}] }$$
where $u^*$ maximizes $ln mathrm{erf}^{} (u) - (1 - alpha) ln u$ with respect to $uge0$.
The notation $a_nasymp b_n$ from large deviation theory means that $lim (1/n)ln a_n = lim(1/n)ln b_n$.
Differentiating we find that $u^{ast}$ is the root of the equation:
$$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$
answered Dec 7 '18 at 21:25
beckobecko
2,35431942
edited Dec 7 '18 at 21:47
answered Dec 7 '18 at 21:25
beckobecko
2,35431942
answered Dec 7 '18 at 21:25
beckobecko
2,35431942
answered Dec 7 '18 at 21:25
beckobecko
2,35431942
2,35431942
add a comment |
add a comment |
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$begingroup$
It can certainly be estimated easily in polar coordinates.
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– T. Bongers
Nov 29 '18 at 20:28
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@T.Bongers So no exact analytical formula?
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– becko
Nov 29 '18 at 20:29
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I wouldn't be surprised if there is one... but depending on the application, do you really need one, or is an estimate good enough?
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– T. Bongers
Nov 29 '18 at 20:29
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@T.Bongers I need an analytical formula. But I will then take the limit $m,n rightarrowinfty$ (with a fixed ratio $m/n$). So in reality I only require an asymptotic estimate.
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– becko
Nov 29 '18 at 20:31
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@T.Bongers Specifically I'll be satisfied by a large deviation limit $lim_{nrightarrowinfty} (1/n) ln I$, where $I$ is the above integral and the ratio $m/n$ is fixed.
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– becko
Nov 29 '18 at 21:05