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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 $n\geq 2$

$$I(n)=\int_{\Delta^{(n)}}\mathrm{d}x_1\dots \mathrm{d}x_n \, \mathrm{det}\big[\exp \big(-\frac{(x_i-x_j)^2}{2}\big)\big]_{i,j\in [1,n]} $$

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

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

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

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  • 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 $I(n)=\int_{\mathbb{R}_+^n}\mathrm{d}x_1\dots \mathrm{d}x_n \, \delta(\sum_{i=1}^n x_i-1)\mathrm{det}\big[\exp \big(-\frac{(x_i-x_j)^2}{2}\big)\big]_{i,j\in [1,n]}$ 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
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Let $$I_1(a,b,c,x) = \int e^{-ax^2-bx-c}\,\mathrm dx = \dfrac1{2\sqrt a}\left(\sqrt π \,e^{b^2/(4a)-c}\,\mathrm{erf}\left(\frac{2ax+b}{2\sqrt a}\right)\right) + constant\tag1$$ (see also Wolfram Alpha), $$I_{0}(a,b,x) = \int e^{-a(x-b)^2}\,\mathrm dx = I_1(a,-2ab,ab^2,x) =\sqrt{\dfrac{\pi}{4a}}\,\mathrm{erf}\left(\sqrt a(x-b)\right) + constant,\tag2$$ $$I_{20}(a,b,x) = \int \mathrm{erf}(a(x-b))\,\mathrm dx = \dfrac1{\sqrt\pi a}e^{-a^2(x-b)^2} + (x-b)\, \mathrm{erf}(a(x-b)) + constant\tag3$$ (see also Wolfram Alpha), $$I_{21}(a,b,x) = \int(x-b) \mathrm{erf}(a(x-b))\,\mathrm dx = {\small \dfrac{\sqrt\pi(1-2a^2 (b^2-2bx+x^2))\,\mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4\sqrt\pi a^2}},\tag4$$ (see also Wolfram Alpha).

Then $$J(1) = \int\limits_0^1 \,\mathrm dx_1 = 1.$$ 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 $$d_2=\begin{vmatrix} 1 & e^{-(x-y)^2}\\ e^{-(x-y)^2} & 1\\ \end{vmatrix} =1-e^{-2(x-y)^2},$$ $$J(2) = \int\limits_0^1\left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}\right)\,\mathrm dx =3-(2I_0(2,0,x)+I_0(8,1/2,x))\bigg|_0^1 =3-\left(\sqrt{\dfrac\pi2}\,\mathrm{erf}(\sqrt2x)+\dfrac14\sqrt{\dfrac\pi2}\,\mathrm{erf}\left(\sqrt2(2x-1)\right)\right)\bigg|_0^1 = 3-\dfrac34\sqrt{2\pi}\,\mathrm{erf}(\sqrt2) \approx 1.20557$$ (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 $$d_3=\begin{vmatrix} 1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \\ e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\\ e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\\ \end{vmatrix} = 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-\left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}\right),$$

$$J(3) = 3\int\limits_0^1\int\limits_0^{1-x}\left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}\right)\,\mathrm dy\,\mathrm dx$$ $$ + \int\limits_0^1\int\limits_0^{1-x_1}\bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}\bigg)\,\mathrm dx_2\,\mathrm dx_1 $$ $$ = 3\int\limits_0^1\int\limits_0^{1-x}\left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}\right)\,\mathrm dy\,\mathrm dx$$ $$ + \int\limits_0^1\int\limits_0^{1-x}\bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}\bigg)\,\mathrm dx_2\,\mathrm dx_1 $$ $$ = 3\int\limits_0^1\int\limits_0^{1-x}\left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}\right)\,\mathrm dy\,\mathrm dx$$ $$+\int\limits_0^1\int\limits_0^{1-x}\bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}\bigg)\,\mathrm dy\,\mathrm dx $$ $$ = 3J_{31} + J_{32} \approx 0.0929729 + 0.00144917 = 0.09442207.$$ 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.

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  • $\begingroup$ 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

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