Integral of a determinant over a unit simplex












4












$begingroup$


Recalling the definition of the unit simplex



$$ Delta^{(n)}=lbrace (x_1,dots,x_n)in mathbb{R}_+^{n} ; , sum_{i=1}^n x_i=1 rbrace,$$



I would like to calculate this integral for all integers $ngeq 2$



$$I(n)=int_{Delta^{(n)}}mathrm{d}x_1dots mathrm{d}x_n , mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]} $$



where det is the usual determinant. I tried the triangular change of variable



$$
y_1=x_1 quad qquad \
y_2=x_1+x_2quad ;\
y_3=x_1+x_2+x_3\
dots\
dots\
y_n=sum_{i=1}^n x_i qquad quad
$$



but that does not really help...
Any guess would be welcome !










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
    $endgroup$
    – Song
    Dec 9 '18 at 23:57












  • $begingroup$
    @Song Help me, please, to find my error for $J(2).$
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:59










  • $begingroup$
    @Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
    $endgroup$
    – Alexandre Krajenbrink
    Dec 10 '18 at 10:42


















4












$begingroup$


Recalling the definition of the unit simplex



$$ Delta^{(n)}=lbrace (x_1,dots,x_n)in mathbb{R}_+^{n} ; , sum_{i=1}^n x_i=1 rbrace,$$



I would like to calculate this integral for all integers $ngeq 2$



$$I(n)=int_{Delta^{(n)}}mathrm{d}x_1dots mathrm{d}x_n , mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]} $$



where det is the usual determinant. I tried the triangular change of variable



$$
y_1=x_1 quad qquad \
y_2=x_1+x_2quad ;\
y_3=x_1+x_2+x_3\
dots\
dots\
y_n=sum_{i=1}^n x_i qquad quad
$$



but that does not really help...
Any guess would be welcome !










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
    $endgroup$
    – Song
    Dec 9 '18 at 23:57












  • $begingroup$
    @Song Help me, please, to find my error for $J(2).$
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:59










  • $begingroup$
    @Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
    $endgroup$
    – Alexandre Krajenbrink
    Dec 10 '18 at 10:42
















4












4








4





$begingroup$


Recalling the definition of the unit simplex



$$ Delta^{(n)}=lbrace (x_1,dots,x_n)in mathbb{R}_+^{n} ; , sum_{i=1}^n x_i=1 rbrace,$$



I would like to calculate this integral for all integers $ngeq 2$



$$I(n)=int_{Delta^{(n)}}mathrm{d}x_1dots mathrm{d}x_n , mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]} $$



where det is the usual determinant. I tried the triangular change of variable



$$
y_1=x_1 quad qquad \
y_2=x_1+x_2quad ;\
y_3=x_1+x_2+x_3\
dots\
dots\
y_n=sum_{i=1}^n x_i qquad quad
$$



but that does not really help...
Any guess would be welcome !










share|cite|improve this question











$endgroup$




Recalling the definition of the unit simplex



$$ Delta^{(n)}=lbrace (x_1,dots,x_n)in mathbb{R}_+^{n} ; , sum_{i=1}^n x_i=1 rbrace,$$



I would like to calculate this integral for all integers $ngeq 2$



$$I(n)=int_{Delta^{(n)}}mathrm{d}x_1dots mathrm{d}x_n , mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]} $$



where det is the usual determinant. I tried the triangular change of variable



$$
y_1=x_1 quad qquad \
y_2=x_1+x_2quad ;\
y_3=x_1+x_2+x_3\
dots\
dots\
y_n=sum_{i=1}^n x_i qquad quad
$$



but that does not really help...
Any guess would be welcome !







calculus integration determinant simplex






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 14:33







Alexandre Krajenbrink

















asked Dec 6 '18 at 10:34









Alexandre KrajenbrinkAlexandre Krajenbrink

9215




9215








  • 2




    $begingroup$
    The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
    $endgroup$
    – Song
    Dec 9 '18 at 23:57












  • $begingroup$
    @Song Help me, please, to find my error for $J(2).$
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:59










  • $begingroup$
    @Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
    $endgroup$
    – Alexandre Krajenbrink
    Dec 10 '18 at 10:42
















  • 2




    $begingroup$
    The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
    $endgroup$
    – Song
    Dec 9 '18 at 23:57












  • $begingroup$
    @Song Help me, please, to find my error for $J(2).$
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:59










  • $begingroup$
    @Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
    $endgroup$
    – Alexandre Krajenbrink
    Dec 10 '18 at 10:42










2




2




$begingroup$
The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
$endgroup$
– Song
Dec 9 '18 at 23:57






$begingroup$
The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
$endgroup$
– Song
Dec 9 '18 at 23:57














$begingroup$
@Song Help me, please, to find my error for $J(2).$
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:59




$begingroup$
@Song Help me, please, to find my error for $J(2).$
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:59












$begingroup$
@Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
$endgroup$
– Alexandre Krajenbrink
Dec 10 '18 at 10:42






$begingroup$
@Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
$endgroup$
– Alexandre Krajenbrink
Dec 10 '18 at 10:42












1 Answer
1






active

oldest

votes


















2





+25







$begingroup$

Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).



Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$



At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$

$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$

(see also Wolfram Alpha),



If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$



At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$



$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$



If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$



At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$



And further progress becames too hard.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:57










  • $begingroup$
    @Christoph You are right. Thanks!
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 19:00










  • $begingroup$
    @Christoph First Edition.
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 21:05










  • $begingroup$
    Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
    $endgroup$
    – Christoph
    Dec 10 '18 at 8:54










  • $begingroup$
    @Christoph Thank you, I will check my work and his arguments
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:58











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%2f3028323%2fintegral-of-a-determinant-over-a-unit-simplex%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2





+25







$begingroup$

Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).



Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$



At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$

$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$

(see also Wolfram Alpha),



If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$



At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$



$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$



If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$



At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$



And further progress becames too hard.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:57










  • $begingroup$
    @Christoph You are right. Thanks!
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 19:00










  • $begingroup$
    @Christoph First Edition.
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 21:05










  • $begingroup$
    Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
    $endgroup$
    – Christoph
    Dec 10 '18 at 8:54










  • $begingroup$
    @Christoph Thank you, I will check my work and his arguments
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:58
















2





+25







$begingroup$

Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).



Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$



At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$

$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$

(see also Wolfram Alpha),



If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$



At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$



$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$



If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$



At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$



And further progress becames too hard.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:57










  • $begingroup$
    @Christoph You are right. Thanks!
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 19:00










  • $begingroup$
    @Christoph First Edition.
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 21:05










  • $begingroup$
    Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
    $endgroup$
    – Christoph
    Dec 10 '18 at 8:54










  • $begingroup$
    @Christoph Thank you, I will check my work and his arguments
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:58














2





+25







2





+25



2




+25



$begingroup$

Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).



Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$



At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$

$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$

(see also Wolfram Alpha),



If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$



At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$



$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$



If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$



At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$



And further progress becames too hard.






share|cite|improve this answer











$endgroup$



Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).



Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$



At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$

$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$

(see also Wolfram Alpha),



If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$



At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$



$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$



If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$



At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$



And further progress becames too hard.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 19:05

























answered Dec 9 '18 at 18:38









Yuri NegometyanovYuri Negometyanov

11.6k1728




11.6k1728












  • $begingroup$
    You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:57










  • $begingroup$
    @Christoph You are right. Thanks!
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 19:00










  • $begingroup$
    @Christoph First Edition.
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 21:05










  • $begingroup$
    Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
    $endgroup$
    – Christoph
    Dec 10 '18 at 8:54










  • $begingroup$
    @Christoph Thank you, I will check my work and his arguments
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:58


















  • $begingroup$
    You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
    $endgroup$
    – Christoph
    Dec 9 '18 at 18:57










  • $begingroup$
    @Christoph You are right. Thanks!
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 19:00










  • $begingroup$
    @Christoph First Edition.
    $endgroup$
    – Yuri Negometyanov
    Dec 9 '18 at 21:05










  • $begingroup$
    Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
    $endgroup$
    – Christoph
    Dec 10 '18 at 8:54










  • $begingroup$
    @Christoph Thank you, I will check my work and his arguments
    $endgroup$
    – Yuri Negometyanov
    Dec 10 '18 at 8:58
















$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
$endgroup$
– Christoph
Dec 9 '18 at 18:57




$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
$endgroup$
– Christoph
Dec 9 '18 at 18:57












$begingroup$
@Christoph You are right. Thanks!
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 19:00




$begingroup$
@Christoph You are right. Thanks!
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 19:00












$begingroup$
@Christoph First Edition.
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 21:05




$begingroup$
@Christoph First Edition.
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 21:05












$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
$endgroup$
– Christoph
Dec 10 '18 at 8:54




$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
$endgroup$
– Christoph
Dec 10 '18 at 8:54












$begingroup$
@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58




$begingroup$
@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58


















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%2f3028323%2fintegral-of-a-determinant-over-a-unit-simplex%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

Plaza Victoria

Puebla de Zaragoza

Musa