IV. Rational functions

1. It is proved in treatises on algebra[8] that any polynomial can be expressed in the form where are positive integers whose sum is , and , are constants; and that any rational function , whose denominator is , may be expressed in the form where are also constants. It follows that From this we conclude that the integral of any rational function is an elementary function which is rational save for the possible presence of logarithms of rational functions. In particular the integral will be rational if each of the numbers is zero: this condition is evidently necessary and sufficient. A necessary but not sufficient condition is that should contain no simple factors.

The integral of the general rational function may be expressed in a very simple and elegant form by means of symbols of differentiation. We may suppose for simplicity that the degree of is less than that of ; this can of course always be ensured by subtracting a polynomial from . Then where Now where is a polynomial; and so where But is also a polynomial, and the integral contains no polynomial term, since the degree of is less than that of . Thus must vanish identically, so that

For example

That is annihilated by the partial differentiations performed on it may be verified directly as follows. We obtain by picking out from the expansion the terms which involve positive powers of . Any such term is of the form where being the degree of . It follows that so that at least one of must be less than the corresponding one of .

It has been assumed above that if then The first equation means that and the second that . As it follows from the first that , what has really been assumed is that It is known that this equation is always true for if a circle can be drawn in the plane of whose centre is and within which the differential coefficients are continuous.

2. It appears from §1 that the integral of a rational function is in general composed of two parts, one of which is a rational function and the other a function of the form We may call these two functions the rational part and the transcendental part of the integral. It is evidently of great importance to show that the 'transcendental part' of the integral is really transcendental and cannot be expressed, wholly or in part, as a rational or algebraical function.

We are not yet in a position to prove this completely[9]; but we can take the first step in this direction by showing that no sum of the form (1) can be rational, unless every is zero.

Suppose, if possible, that where and are polynomials without common factor. Then

Suppose now that is a factor of . Then is divisible by and by no higher power of . Thus the right-hand side of (3), when expressed in its lowest terms, has a factor in its denominator. On the other hand the left-hand side, when expressed as a rational fraction in its lowest terms, has no repeated factor in its denominator. Hence , and so is a constant. We may therefore replace (2) by and (3) by Multiplying by , and making tend to , we see that .

3. The method of §1 gives a complete solution of the problem if the roots of can be determined; and in practice this is usually the case. But this case, though it is the one which occurs most frequently in practice, is from a theoretical point of view an exceedingly special case. The roots of are not in general explicit algebraical functions of the coefficients, and cannot as a rule be determined in any explicit form. The method of partial fractions is therefore subject to serious limitations. For example, we cannot determine, by the method of decomposition into partial fractions, such an integral as or even determine whether the integral is rational or not, although it is in reality a very simple function. A high degree of importance therefore attaches to the further problem of determining the integral of a given rational function so far as possible in an absolutely explicit form and by means of operations which are always practicable.

It is easy to see that a complete solution of this problem cannot be looked for.

Suppose for example that reduces to unity, and that is an equation of the fifth degree, whose roots are all distinct and not capable of explicit algebraical expression.

Then and it is only if at least two of the numbers are commensurable that any two or more of the factors can be associated so as to give a single term of the type , where is rational. In general this will not be the case, and so it will not be possible to express the integral in any finite form which does not explicitly involve the roots. A more precise result in this connection will be proved later (§6).

4. The first and most important part of the problem has been solved by Hermite, who has shown that the rational part of the integral can always be determined without a knowledge of the roots of , and indeed without the performance of any operations other than those of elementary algebra[10].

Hermite's method depends upon a fundamental theorem in elementary algebra[11] which is also of great importance in the ordinary theory of partial fractions, viz.:

'If and are two polynomials in which have no common factor, and any third polynomial, then we can determine two polynomials , , such that

Suppose that denoting polynomials which have only simple roots and of which no two have any common factor. We can always determine by elementary methods, as is shown in the elements of the theory of equations[12].

We can determine and so that and therefore so that By a repetition of this process we can express in the form and the problem of the integration of is reduced to that of the integration of a function where is a polynomial whose roots are all distinct. Since this is so, and its derived function have no common factor: we can therefore determine and so that Hence where Proceeding in this way, and reducing by unity at each step the power of which figures under the sign of integration, we ultimately arrive at an equation where is a rational function and a polynomial.

The integral on the right-hand side has no rational part, since all the roots of are simple (§2). Thus the rational part of is and it has been determined without the need of any calculations other than those involved in the addition, multiplication and division of polynomials[13].

5. (i) Let us consider, for example, the integral mentioned above (§3). We require polynomials such that where

In general, if the degrees of and are and , and that of does not exceed , we can suppose that the degrees of and do not exceed and respectively. For we know that polynomials and exist such that If is of degree not exceeding , we take , and if it is of higher degree we write where is of degree not exceeding . Similarly we write We have then In this identity or or both may vanish identically, and in any case we see, by equating to zero the coefficients of the powers of higher than the th, that vanishes identically. Thus is expressed in the form required.

The actual determination of the coefficients in and is most easily performed by equating coefficients. We have then linear equations in the same number of unknowns. These equations must be consistent, since we know that a solution exists[14].

If is of degree higher than , we must divide it by and express the remainder in the form required.

In this case we may suppose of degree 5 and of degree 6, and we find that Thus the rational part of the integral is and, since , there is no transcendental part.

(ii) The following problem is instructive: to find the conditions that may be rational, and to determine the integral when it is rational.

We shall suppose that is not a perfect square, as if it were the integral would certainly be rational. We can determine and so that and the integral becomes The condition that the integral should be rational is therefore .

Equating coefficients we find Hence we deduce and . The condition required is therefore that the two quadratics and should be harmonically related, and in this case

(iii) Another method of solution of this problem is as follows. If we write and use the bilinear substitution then the integral is reduced to one of the form and is rational if and only if . But this is the condition that the quadratic , corresponding to , should be harmonically related to the degenerate quadratic , corresponding to . The result now follows from the fact that harmonic relations are not changed by bilinear transformation.

It is not difficult to show, by an adaptation of this method, that is rational if all the quadratics are harmonically related to any one of those in the numerator. This condition is sufficient but not necessary.

(iv) As a further example of the use of the method (ii) the reader may show that the necessary and sufficient condition that where and are polynomials with no common factor, and has no repeated factor, should be rational, is that should be divisible by .

6. It appears from the preceding paragraphs that we can always find the rational part of the integral, and can find the complete integral if we can find the roots of . The question is naturally suggested as to the maximum of information which can be obtained about the logarithmic part of the integral in the general case in which the factors of the denominator cannot be determined explicitly. For there are polynomials which, although they cannot be completely resolved into such factors, can nevertheless be partially resolved. For example The factors of the first polynomial have rational coefficients: in the language of the theory of equations, the polynomial is reducible in the rational domain. The second polynomial is reducible in the domain formed by the adjunction of the single irrational to the rational domain[15].

We may suppose that every possible decomposition of of this nature has been made, so that Then we can resolve into a sum of partial fractions of the type and so we need only consider integrals of the type where no further resolution of is possible or, in technical language, is irreducible by the adjunction of any algebraical irrationality.

Suppose that this integral can be evaluated in a form involving only constants which can be expressed explicitly in terms of the constants which occur in . It must be of the form where the 's are constants and the 's polynomials. We can suppose that no has any repeated factor , where is a polynomial. For such a factor could be determined rationally in terms of the coefficients of , and the expression (1) could then be modified by taking out the factor from and inserting a new term . And for similar reasons we can suppose that no two 's have any factor in common.

Now or All the terms under the sign of summation are divisible by save the first, which is prime to . Hence must be divisible by : and similarly, of course, by . But, since is prime to , is divisible by . Thus must be a constant multiple of . But is ex hypothesi not resoluble into factors which contain only explicit algebraical irrationalities. Hence all the 's save one must reduce to constants, and so must be a constant multiple of , and where is a constant. Unless this is the case the integral cannot be expressed in a form involving only constants expressed explicitly in terms of the constants which occur in and .

Thus, for instance, the integral cannot, except in special cases[16], be expressed in a form involving only constants expressed explicitly in terms of and ; and the integral can in general be so expressed if and only if . We thus confirm an inference made before (§3) in a less accurate way.

Before quitting this part of our subject we may consider one further problem: under what circumstances is where is a constant and rational? Since the integral has no rational part, it is clear that must have only simple factors, and that the degree of must be less than that of . We may therefore use the formula The necessary and sufficient condition is that all the numbers should be commensurable. If e.g. then and must be commensurable, i.e. must be a rational number. If the denominator is given we can find all the values of which are admissible: for , where and are integers.

7. Our discussion of the integration of rational functions is now complete. It has been throughout of a theoretical character. We have not attempted to consider what are the simplest and quickest methods for the actual calculation of the types of integral which occur most commonly in practice. This problem lies outside our present range: the reader may consult

O. Stolz, Grundzüge der Differential- und Integralrechnung, vol. 1, ch. 7:

J. Tannery, Leçons d'algèbre et d'analyse, vol. 2, ch. 18:

Ch.-J. de la Vallée-Poussin, Cours d'analyse, ed. 3, vol. 1, ch. 5:

T. J. l'A. Bromwich, Elementary integrals (Bowes and Bowes, 1911):

G. H. Hardy, A course of pure mathematics, ed. 2, ch. 6.

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