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186 CHAPTER 7. CALCULUS
concept.
Definition 86 A generalised (or Laurent) polynomial in θ over K is a sum
∑ n
i=−m a iθ i with a i ∈ K.
Lemma 11 (Decomposition Lemma (exponential)) f ∈ K(θ) can be written
uniquely as p+q/r, where p is a Laurent polynomial, q and r are polynomials
of K[θ] such that θ does not divide r, q and r are relatively prime, and the degree
of q is less than that of r. If f has an elementary integral over K, then each of
the terms of p, and also q/r, have an elementary integral over K.
Proof.
form
By Liouville’s Principle (Theorem 32), if f is integrable, it is of the
f = v ′ 0 +
n∑
i=1
c i v ′ i
v i
, (7.30 ter)
where v 0 ∈ K(θ), c i ∈ C, and v i ∈ K(c 1 , . . . , c n )[θ]. Write v 0 = p 0 + q0
r 0
, where
p 0 ∈ K[θ, 1/θ] is a Laurent polynomial, q 0 , r 0 ∈ K[θ] with deg(q 0 ) < deg(r 0 ) and
θ̸ |q 0 , and re-arrange the v i such that v 1 , . . . , v k ∈ K(c 1 , . . . , c n ), but v k+1 , . . . , v n
genuinely involve θ, and are monic. Furthermore, we can suppose that θ does
not divide any of these v i , since log θ = η (up to constants). Unlike Lemma
10, though, it is no longer the case that (log v i ) ′ (i > k) is a proper rational
expression. Let n i be the degree (in θ) of v i , then, recalling that we have
supposed that the v i are monic, v ′ i = n iη ′ θ ni + lower terms, and (v ′ i − n iη ′ v i )/v i
is a proper rational expression
Then we can re-arrange (7.30 ter) as
p + q k∑
r = c i v ′ n∑
p′ i
0 + + c i n i η ′ +
v
i=1 i
i=k+1
} {{ }
in K(c 1 , . . . , c n )[θ, 1/θ]
(
q0
n∑
c i (v ′ i − n iη ′ v i )
v i
) ′
+
r 0
i=k+1
} {{ }
proper rational expression
θ not dividing the denominator
, (7.36)
and the decomposition of the right-hand side proves the result.
This means that it is sufficient to integrate the polynomial part (which we
will do in Algorithm 35) and the rational part (which we will do in Algorithm
36) separately, and that failure in either part indicates that the whole expression
does not have an elementary integral. In other words, there is no crosscancellation
between these parts.
7.5.1 The Polynomial Part
In fact, this is simpler than the logarithmic case, since Lemma 11 says that each
summand a i θ i has to be integrable separately.The case i = 0 is just integration
in K, and all the others are cases of the Risch differential equation problem
(Definition 85).
This translates straightforwardly into Algorithm 35.