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Contents - Student subdomain for University of Bath

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7.2. INTEGRATION OF RATIONAL EXPRESSIONS 173<br />

where the integral on the right-hand resolves itself as purely a sum <strong>of</strong> logarithms,<br />

i.e. is the ∑ n ∑ ni<br />

i=1 j=i β i,j,1log(x − α i,j ) term. Then a standard argument from<br />

Galois theory shows that s 1 , t 1 , s 2 and t 2 do not involve any <strong>of</strong> the α i , i.e.<br />

that the decomposition (7.13) can be written without introducing any algebraic<br />

numbers. If we could actually obtain this decomposition without introducing<br />

these algebraic numbers, we would have gone a long way to solving objection 2<br />

above.<br />

We can per<strong>for</strong>m a square-free decomposition (Definition 32) <strong>of</strong> r as ∏ ri i,<br />

and then a partial fraction decomposition to write<br />

q<br />

r = ∑ q i<br />

r i i<br />

(7.14)<br />

and, since each term is a rational expression and there<strong>for</strong>e integrable, it suffices<br />

to integrate (7.14) term-by-term.<br />

Now r i and r i ′ are relatively prime, so, by Bezout’s Identity (2.10), there are<br />

polynomials a and b satisfying ar i + br i ′ = 1. There<strong>for</strong>e<br />

∫ ∫<br />

qi qi (ar i + br i ′<br />

ri<br />

i =<br />

)<br />

ri<br />

i (7.15)<br />

∫ ∫<br />

qi a qi br i<br />

′ =<br />

r i−1 +<br />

r i (7.16)<br />

i<br />

i<br />

∫ ∫<br />

qi a (qi b/(i − 1)) ′ ( ) ′ qi b/(i − 1)<br />

= +<br />

−<br />

(7.17)<br />

r i−1<br />

i<br />

(<br />

qi b/(i − 1)<br />

= −<br />

r i−1<br />

i<br />

r i−1<br />

i<br />

)<br />

+<br />

r i−1<br />

i<br />

∫<br />

qi a + (q i b/(i − 1)) ′<br />

, (7.18)<br />

r i−1<br />

i<br />

and we have reduced the exponent <strong>of</strong> r i by one. When programming this method<br />

one may need to take care <strong>of</strong> the fact that, while qi<br />

is a proper rational expression,<br />

q ib<br />

r i−1<br />

i<br />

may not be, but the excess is precisely compensated <strong>for</strong> by the other<br />

term in (7.18).<br />

Hence, at the cost <strong>of</strong> a square-free decomposition and a partial fraction decomposition,<br />

but not a factorization, we have found the rational part <strong>of</strong> the<br />

integral, i.e. per<strong>for</strong>med the decomposition <strong>of</strong> (7.14). In fact, we have done<br />

term will have been split into summands cor-<br />

somewhat better, since the ∫ s 2<br />

t2<br />

responding to the different r i .<br />

r i i<br />

7.2.3 The Ostrogradski–Horowitz Algorithm<br />

Although quite simple, Hermite’s method still needs square-free decomposition<br />

and partial fractions. Horowitz [Hor69, Hor71] there<strong>for</strong>e proposed to computer<br />

algebraists the following method, which was in fact already known [Ost45], but<br />

largely <strong>for</strong>gotten 2 in the west. It follows from (7.18) that, in the notation <strong>of</strong><br />

name.<br />

2 The author had found no references to it, but apparently it had been taught under that

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