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Additional resources for Algebra 3: algorithms in algebra [Lecture notes]
Bm To reduce the exponent occurring in the second term, we apply integration by parts, noting that D(b−(m−1) ) = −(m − 1) D(b) . bm 42 Symbolic integration This yields v · D( −b−(m−1) −v b−(m−1) )= + m−1 m−1 D(v) · b−(m−1) . m−1 The integral of a/bm then reduces to a −v = + m b (m − 1)bm−1 u + D(v)/(m − 1) . bm−1 So the exponent has been reduced by 1 at the cost of introducing an extra rational function. By repeating this process we can get down to exponent 1 in the denominator. , a suitable partial fraction expansion.
So v0 (θ) = a(θ), a polynomial expression in θ. From the remaining equality v(θ) p(θ) =c q(θ) v(θ) we deduce q(θ) = v(θ). But then p(θ) − cq(θ) = 0, so v(θ) = gcd(p(θ) − cq(θ) , q(θ)) showing that c is a zero of the resultant R(z) = resθ (p(θ) − zq(θ) , q(θ)). If R(d) = 0, then it follows that deg gcd(p(θ) − dq(θ) , q(θ)) > 0. Let g be an irreducible factor. Then g|q = v and g|p − dq = cv − dv = (c − d)v . Since gcd(v, v ) = 1, we conclude that c = d. 6 Using our previous results we can refine the theorem in the case the integral is elementary.
Single exponential extension) Let K be a differential field with field of constants C and let K(θ) be a transcendental exponential extension of K with the same field of constants. Suppose p(θ)/q(θ) satisfies a) gcd(p(θ), q(θ)) = 1 and θ does not divide q(θ); b) deg p(θ) < deg q(θ); c) q(θ) is monic and squarefree. Then p(θ) q(θ) is elementary if and only if the zeros of R(z) = Resθ (p(θ) − zq(θ) , q(θ)) are constants. 12 Again, there is an explicit form if the integral is elementary. Let θ /θ = u , then the explicit form is − ci deg(vi (θ))u + i ci log(vi ), i where the ci are the distinct roots of the resultant R(z), and vi = gcd(p(θ) − ci q(θ) , q(θ)).
Algebra 3: algorithms in algebra [Lecture notes] by Hans Sterk