"sk maidul haque" <skmaidulhaque@gmail.com> wrote in message <ij0f6b$jgl$1@fred.mathworks.com>...
> I do not know whether this is the right place to pose this problem.
> I have come across one definite integral which I am not able to resolve. Kindly let me know if you have any idea to solve it or suggest me any blog where I can request for suggestion.
> Integration has to be carried out of the function (a0*x^n+a1*x^(n1)+ a2*x^(n2)+ a3*x^(n3)+ ...+1)^(1). It is not possible to factorize the polynomial since a0, a1,a2...are all unknown. I want the final answer in terms of a0,a1,a2,a3.
> Looking forward for your suggestion.
You get a mess even just for going to 2:
MAPLE> `assuming`([int(1/(sum(a[n]*x^n, n = 1 .. 2)+1), x = c .. d, AllSolutions)], [c <= d, real]);
piecewise(c < d, piecewise(And(c < (1/2)*(a[1]+Re((a[1]^24*a[2])^(1/2)))/a[2], c < (1/2)*(a[1]+Re((a[1]^24*a[2])^(1/2)))/a[2], (1/2)*(a[1]+Re((a[1]^24*a[2])^(1/2)))/a[2] < d, (1/2)*(a[1]+Re((a[1]^24*a[2])^(1/2)))/a[2] < d), undefined, And(c < (1/2)*(a[1]+Re((a[1]^24*a[2])^(1/2)))/a[2], (1/2)*(a[1]+Re((a[1]^24*a[2])^(1/2)))/a[2] < d), undefined, And(c < (1/2)*(a[1]+Re((a[1]^24*a[2])^(1/2)))/a[2], (1/2)*(a[1]+Re((a[1]^24*a[2])^(1/2)))/a[2] < d), undefined, piecewise(c = (1/2)*(a[1]+(a[1]^24*a[2])^(1/2))/a[2], infinity/signum(a[1]^24*a[2])^(1/2), c = (1/2)*(a[1]+(a[1]^24*a[2])^(1/2))/a[2], infinity/signum(a[1]^24*a[2])^(1/2), 2*arctanh((2*c*a[2]+a[1])/(a[1]^24*a[2])^(1/2))/(a[1]^24*a[2])^(1/2))+piecewise(d = (1/2)*(a[1]+(a[1]^24*a[2])^(1/2))/a[2], infinity/signum(a[1]^24*a[2])^(1/2), d = (1/2)*(a[1]+(a[1]^24*a[2])^(1/2))/a[2],
infinity/signum(a[1]^24*a[2])^(1/2), 2*arctanh((2*d*a[2]+a[1])/(a[1]^24*a[2])^(1/2))/(a[1]^24*a[2])^(1/2))), d = c, 0)
The cubic is much more tractable:
sum(ln(d_R)/(a[1]+2*a[2]*_R+3*a[3]*_R^2), _R = RootOf(a[1]*_Z+a[2]*_Z^2+a[3]*_Z^3+1))(sum(ln(c_R)/(a[1]+2*a[2]*_R+3*a[3]*_R^2), _R = RootOf(a[1]*_Z+a[2]*_Z^2+a[3]*_Z^3+1)))
The quartic... my system was still choking on that one after 20 or so hours of computation.
I would say that these two answers together are enough to establish that although *potentially* there is a closed form solution for each of the maximum n, that there is no general extensible solution, at least not without some deep calculus theorem or other.
If the pattern continues like it does for the cubic, there would be RootOf() polynomials of degree 5 or more, and there are no general closed form solutions for the roots of such polynomials.
