My gut feeling is that there isn't a closed form for n>2. For n=3, upon substituting x[1] -> cos(theta) x[2] -> cos(phi) sin(theta) x[3] -> sin(theta) sin(phi) we have 2pi * integral(FOO, phi=0..pi) where FOO = exp[-(a[1] + a[2]cos(phi)^2 + a[3]sin(phi)^2)/4] * besselI[0, (a[1] - a[2]cos(phi)^2 - a[3]sin(phi)^2)/4] Of course there might be a clever substitution or something. Here's some Mathematica code you can play around with BEGIN-MATHEMATICA-CODE{{{ Angles[n_] := Map[Subscript[\[Phi], #] &, Range[1, n]] Hyperspherical[1, nmax_] := Cos[\[Theta]] Hyperspherical[2, 2] := Sin[\[Theta]] Hyperspherical[n_, nmax_] := Sin[\[Theta]]*Cos[Subscript[\[Phi], n - 1]]*\!\( \*UnderoverscriptBox[\(\[Product]\), \(k = 1\), \(n - 2\)]\(Sin[ \*SubscriptBox[\(\[Phi]\), \(k\)]]\)\) /; n < nmax Hyperspherical[n_, nmax_] := Sin[\[Theta]]*\!\( \*UnderoverscriptBox[\(\[Product]\), \(k = 1\), \(n - 2\)]\(Sin[ \*SubscriptBox[\(\[Phi]\), \(k\)]]\)\) /; n == nmax HypersphericalRules[nmax_] := Table[Subscript[x, k] -> Hyperspherical[k, nmax], {k, 1, nmax}] OneTest[n_] := (\!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)] \*SubsuperscriptBox[\(x\), \(k\), \(2\)]\)) /. HypersphericalRules[n] // FullSimplify Gaussian[n_] := Exp[-(1/2) \!\( \*UnderoverscriptBox[\(\[Sum]\), \(k = 1\), \(n\)]\( \*SubscriptBox[\(a\), \(k\)] \*SubsuperscriptBox[\(x\), \(k\), \(2\)]\)\)] SphericalGaussian[n_] := Gaussian[n] /. HypersphericalRules[n] SalaminIntegral[f_, n_] := Fold[Integrate[#1, {#2, 0, \[Pi]}] &, Integrate[f, {\[Theta], 0, 2 \[Pi]}], Angles[n - 1]] Salamin[n_] := SalaminIntegral[SphericalGaussian[n], n - 1] }}} END-MATHEMATICA-CODE Maybe you can come up with something with this. :) -Robert On Wed, Mar 23, 2011 at 10:58 PM, Henry Baker <hbaker1@pipeline.com> wrote:
Is this one of those situations where the even & odd dimensional cases have a different form?
At 06:03 PM 3/23/2011, Eugene Salamin wrote:
Can anyone find a closed form expression for the integral of
exp[(-1/2) (a[1] x[1]^2 + a[2] x[2]^2 + ... + a[n] x[n]^2)]
over the unit (n-1)-sphere? In the n=2 case, we have
a x^2 + b y^2 = a cos(t)^2 + b sin(t)^2 = (1/2) ((a+b) + (a-b) cos(2t)), and integrating over t in [0, 2 pi] gives
2 pi exp((-1/4) (a+b)) BesselI(0, (1/4) (a-b)).
-- Gene
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun