Contents - Student subdomain for University of Bath
Contents - Student subdomain for University of Bath
Contents - Student subdomain for University of Bath
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184 CHAPTER 7. CALCULUS<br />
Figure 7.1: Algorithm 33: IntLog–Polynomial<br />
Algorithm 33 (IntLog–Polynomial)<br />
Input: p = ∑ n<br />
i=0 a iθ i ∈ K[θ].<br />
Output: An expression <strong>for</strong> ∫ pdx, or failed if not elementary<br />
Ans:=0<br />
<strong>for</strong> i := n, n − 1, . . . , 1 do<br />
c i := ∫ a idx # integration in K: (7.34)<br />
if c i =failed or c i /∈ K[θ]<br />
return failed<br />
Write c i = b i + (i + 1)b i+1 : b i ∈ K and b i+1 constant<br />
Ans:=Ans+b i θ i + b i+1 θ i+1<br />
a i−1 := a i−1 − iθ ′ b i # correction from (7.35)<br />
c 0 := ∫ a 0dx<br />
# integration in K<br />
if c 0 =failed<br />
return failed<br />
Ans:=Ans+c 0<br />
7.4.2 The Rational Expression Part<br />
Now <strong>for</strong> the rational expression part, where we have to integrate a proper rational<br />
expression, and the integral will be a proper rational expression plus a sum<br />
<strong>of</strong> logarithms with constant coefficients — see (*) in Algorithm 34. The proper<br />
rational expression part is determined by an analogue <strong>of</strong> Hermite’s algorithm,<br />
and (7.18) is still valid.<br />
The Trager–Rothstein algorithm is still valid, with that additional clause<br />
that the roots <strong>of</strong> P (λ), which don’t depend on θ since P (λ) is a resultant,<br />
actually be constants. To see why this is necessary, consider ∫ 1<br />
log xdx. Here<br />
q 1 = 1, r 1 = θ, and P (λ) = Res(1−λ/x, θ, θ) = 1−λ/x (made monic, i.e. λ−x).<br />
This would suggest a contribution to the integral <strong>of</strong> x log log x, which indeed<br />
gives log x as one term on differentiation, but also a term <strong>of</strong> log log x, which is<br />
not allowed, since it is not present in the integrand. Note that ∫ 1<br />
x log xdx gives<br />
P = λ − 1 and an integral <strong>of</strong> log log x, which is correct.<br />
7.5 Integration <strong>of</strong> Exponential Expressions<br />
Throughout this section, we let θ = θ n be a (transcendental) exponential over<br />
K = C(x, θ 1 , . . . , θ n−1 ), so that θ ′ = η ′ θ. We should note that this choice<br />
is somewhat arbitrary, since θ −1 = θ satisfies θ ′ = −η ′ θ and K(θ) ≡ K(θ).<br />
Hence negative powers <strong>of</strong> θ are just as legitimate as positive powers, and this<br />
translates into a difference in the next result: rather than writing expressions as<br />
“polynomial + proper rational expressions”, we will make use <strong>of</strong> the following