Integrate: $int_0^1 ||vec x||^{-m}mathrm{d}vec{x}$












2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 19:25







becko

















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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $[0,1]^n$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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



















1












$begingroup$

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:



$$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$






share|cite|improve this answer











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019159%2fintegrate-int-01-vec-x-m-mathrmd-vecx%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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 $[0,1]^n$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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
















    1












    $begingroup$

    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 $[0,1]^n$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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














    1












    1








    1





    $begingroup$

    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 $[0,1]^n$.






    share|cite|improve this answer









    $endgroup$



    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 $[0,1]^n$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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


















    • $begingroup$
      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
    $endgroup$
    – Calvin Khor
    Dec 6 '18 at 18:32




    $begingroup$
    @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











    1












    $begingroup$

    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:



    $$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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:



      $$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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:



        $$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$






        share|cite|improve this answer











        $endgroup$



        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:



        $$1-frac{2umathrm e^{-u^2}}{sqrt{pi} mathrm{erf} (u)} = alpha $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 7 '18 at 21:47

























        answered Dec 7 '18 at 21:25









        beckobecko

        2,35431942




        2,35431942






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019159%2fintegrate-int-01-vec-x-m-mathrmd-vecx%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more