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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 8,600 of 10,432    |
|    clicliclic@freenet.de to paulandrewbird@gmail.com    |
|    Re: Can someone solve this improper inte    |
|    01 Jul 14 17:12:38    |
      paulandrewbird@gmail.com schrieb:       >       > Hi, I'm trying to solve the integral:       >       > integral( exp( a*x^4+4*b*x^3*y+6*c*x^2*y^2+4*d*x*y^3+e*y^4)       > ,x=-infty..infty, y=-infty..finty)       >       > I tried on Wolfram alpha but it was too complicated for it.       >       > I'm guessing it's going to be in the form 1/P(a,b,c,d,e)^(1/8) where P       > is a 4th degree polynomial in the coefficients of the quartic.       >       > My guess is that P is the discriminant of the polynomial but that is       > just a guess.       >       > Is there any good free software that can solve this? Can Mathematica       > or Maple solve this? Can you?       >              I cannot solve your integral, but find your conjecture must be wrong.       The integral remains finite when the quartic discriminant vanishes;       taking a=-1, b=0, c=-2, d=0, e=-1 for instance, it numerically evaluates       to 1.98670, whereas the quartic discriminant              1/27*(4*(12*a*e - 3*b*d + c^2)^3        - (72*a*c*e - 27*a*d^2 - 27*b^2*e + 9*b*c*d - 2*c^3)^2)              vanishes for these coefficients. More generally, since the integrand is       stricly positive everywhere, so must be the integral, and its symbolic       evaluation must be positive and finite over the entire (a,b,c,d,e)       domain where it converges.              Martin.              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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