Extracting the Algebraic Part of Integrals of Binomial Differentials
Abstract Generated abstract
This paper studies the extraction of algebraic parts from integrals of binomial differentials by expressing them through the hypergeometric function F(alpha, 1; gamma; z). It derives and proves several recurrence and expansion relations for this function, then applies them to obtain reduction formulas for six classes of binomial differential integrals, separating finite algebraic sums from residual integrals. The paper identifies parameter cases in which these integrals become algebraic, and further gives conditions under which the remaining integrals can be expressed in elementary functions using Chebyshev substitutions and standard integral forms.
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EXTRACTION OF THE ALGEBRAIC PART OF INTEGRALS OF BINOMIAL DIFFERENTIALS
V. G. KORNILOV
(Presented by the scientific seminar of the departments of higher mathematics and engineering-computational mathematics of TPI)
The article examines the question of extracting the algebraic part in integrals of binomial differentials. In doing so, certain relations for the hypergeometric function \(F(\alpha, 1; \gamma; z)\) are used.
It is known that the function
\[ F(\alpha, 1; \gamma; z)=\sum_{k=0}^{\infty}\frac{(\alpha)_k}{(\gamma)_k}z^k,\quad \gamma\ne 0,-1,\ldots;\ |z|<1, \tag{1} \]
where
\[ (\alpha)_k=\alpha(\alpha+1)\ldots(\alpha+k-1)=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)},\quad (\alpha)_0=1, \tag{2} \]
is expressed by means of the integral ([1], p. 308)
\[ F(\alpha, 1; \gamma; z)= \frac{(\gamma-1)z^{1-\gamma}}{(1-z)^{\alpha-\gamma+1}} \int_0^z z^{\gamma-2}(1-z)^{\alpha-\gamma}\,dz,\quad \gamma>1 \tag{3} \]
and satisfies the following relations:
\[ F(\alpha, 1; \gamma; z)= \sum_{k=0}^{n-1}\frac{(\alpha)_k}{(\gamma)_k}z^k+ \frac{(\alpha)_n}{(\gamma)_n}z^nF(\alpha+n,1;\gamma+n;z), \tag{4} \]
\[ F(\alpha, 1; \gamma; z)= -(\gamma-1)\sum_{k=0}^{n-1} \frac{(\alpha)_k}{(\alpha-\gamma+1)_{k+1}}(1-z)^k+ \frac{(\alpha)_n}{(\alpha-\gamma+1)_n}(1-z)^nF(\alpha+n,1;\gamma;z), \tag{5} \]
\[ F(\alpha, 1; \gamma; z)= \frac{1}{1-z}\sum_{k=0}^{n-1}(-1)^k \frac{(\gamma-\alpha)_k}{(\gamma)_k} \left(\frac{z}{1-z}\right)^k+ \]
\[ +(-1)^n\frac{(\gamma-\alpha)_n}{(\gamma)_n} \left(\frac{z}{1-z}\right)^n F(\alpha,1;\gamma+n;z). \tag{6} \]
To prove equalities (5), (6), we consider the recurrence relations
\[ F(\alpha,1;\gamma;z)=-\frac{\gamma-1}{\alpha-\gamma+1} +\frac{\alpha}{\alpha-\gamma+1}(1-z)F(\alpha+1,1;\gamma;z), \tag{7} \]
\[ (1-z)F(\alpha,1;\gamma;z)=1-\frac{\gamma-\alpha}{\gamma}\,zF(\alpha,1;\gamma+1;z), \tag{8} \]
which are easily verified by substituting the series (1) and comparing the coefficients of equal powers of \(z\) on the left- and right-hand sides of the equalities under consideration. For example, in equality (7) the coefficient of \(z^n\) is as follows:
\[ \frac{(\alpha)_n}{(\gamma)_n} = -\frac{\alpha}{\alpha-\gamma+1} \left[ \frac{(\alpha+1)_n}{(\gamma)_n} - \frac{(\alpha+1)_{n-1}}{(\gamma)_{n-1}} \right] = \]
\[ = \frac{\alpha(\alpha+1)_{n-1}[(\alpha+n)-(\gamma+n-1)]} {(\alpha-\gamma+1)(\gamma)_n} = \frac{(\alpha)_n}{(\gamma)_n}. \]
The equalities (4) and (8) are proved in an analogous way.
Equality (5) for \(n=1\) coincides with relation (7). Suppose that equality (5) holds; then, replacing \(F(\alpha+n,1;\gamma;z)\) by formula (7), we obtain:
\[ F(\alpha,1;\gamma;z) = -(\gamma-1)\sum_{k=0}^{n-1} \frac{(\alpha)_k}{(\alpha-\gamma+1)_{k-1}}(1-z)^k + \]
\[ + \frac{(\alpha)_n}{(\alpha-\gamma+1)_n}(1-z)^n \left[ -\frac{\gamma-1}{\alpha-\gamma+n+1} + \right. \]
\[ \left. +\frac{\alpha+n}{\alpha-\gamma+n+1}(1-z)F(\alpha+n+1,1;\gamma;z) \right] = \]
\[ = -(\gamma-1)\sum_{k=0}^{n} \frac{(\alpha)_k}{(\alpha-\gamma+1)_{k+1}}(1-z)^k + \]
\[ + \frac{(\alpha)_{n+1}}{(\alpha-\gamma+1)_{n+1}} (1-z)^{n+1}F(\alpha+n+1,1;\gamma;z), \]
which proves equality (5). Equality (6) is proved analogously with the aid of relation (8).
If we take into account the circumstance that the integral of the binomial differential (3) is expressed by means of the function (1), which in addition satisfies relations (4)—(6), then the latter are used to extract the algebraic part of the following integrals. For formulas (9)—(14), \(n=0,1,\ldots;\ m=0,1,\ldots;\ r>0,\ 0\leq\theta<1,\ 0\leq\zeta<1\).
\[ \int x^{r(1-\theta-\zeta+n)-1}(a+bx^r)^{\zeta-1+m}\,dx = \]
\[ = \frac{1}{br} \sum_{k=0}^{n-1} \frac{(-1)^k(1-\theta-\zeta+n-k)_k} {(-\theta+n+m-k)_{k+1}} \cdot \frac{a^k}{b^k} \frac{x^{r(-\theta-\zeta+n-k)}}{(a+bx^r)^{\zeta-m}} + \]
\[ + \frac{(-1)^n(1-\theta-\zeta)_n}{ra^{-n}b^n} \sum_{k=0}^{m-1} \frac{(\zeta+m-k)_k} {(-\theta+m-k)_{n+k+1}} \cdot a^k \frac{x^{r(1-\theta-\zeta)}}{(a+bx^r)^{\zeta-m+k+1}} + \]
\[ + (-1)^n\frac{a^{n+m}}{b^n} \cdot \frac{(1-\theta-\zeta)_n(\zeta)_m}{(1-\theta)_{n+m}} \int x^{r(1-\theta-\zeta)-1}(a+bx^r)^{\zeta-1}\,dx. \tag{9} \]
\[ \begin{aligned} &\int x^{-r(-\theta+n+m)-1}(a+bx^r)^{\zeta+m-1}\,dx=\\ &= -\frac{1}{ar}\sum_{k=0}^{n-1} \frac{(-1)^k(1-\theta-\zeta+n-k)_k}{(-\theta+n+m-k)_{k+1}}\cdot \frac{b^k}{a^k}\cdot \frac{(a+bx^r)^{\zeta+m}}{x^{r(-\theta+m+n-k)}}\\ &\quad -\frac{(-1)^n(1-\theta-\zeta)_n}{a^n b^{-n}r} \sum_{k=0}^{m-1} \frac{(\zeta+m-k)_k}{(-\theta+m-k)_{n+k+1}}\, b^k\frac{(a+bx^r)^{\zeta+m-k-1}}{x^{r(-\theta+m-k)}}\\ &\quad +(-1)^n\frac{b^{n+m}}{a^n}\, \frac{(1-\theta-\zeta)_n(\zeta)_m}{(1-\theta)_{n+m}} \int x^{r\theta-1}(a+bx^r)^{\zeta-1}\,dx . \tag{10} \end{aligned} \]
\[ \begin{aligned} &\int x^{r(\zeta+m)-1}(a+bx^r)^{\theta-n-m-1}\,dx=\\ &= -\frac{1}{ar}\sum_{k=0}^{n-1} \frac{(1-\theta-\zeta+n-k)_k}{(-\theta+n+m-k)_{k+1}}\cdot \frac{1}{a^k}\, \frac{x^{r(\zeta+m)}}{(a+bx^r)^{-\theta+n+m-k}}\\ &\quad -\frac{(1-\theta-\zeta)_n}{a^n br} \sum_{k=0}^{m-1} \frac{(\zeta+m-k)_k}{(-\theta+m-k)_{n+k+1}}\cdot \frac{1}{b^k}\cdot \frac{x^{r(\zeta+m-k-1)}}{(a+bx^r)^{-\theta+m-k}}\\ &\quad +\frac{(1-\theta-\zeta)_n(\zeta)_m}{a^n b^m(1-\theta)_{n+m}} \int x^{r\zeta-1}(a+bx^r)^{\theta-1}\,dx . \tag{11} \end{aligned} \]
\[ \begin{aligned} &\int x^{r(\theta-n)-1}(a+bx^r)^{\zeta-m-1}\,dx=\\ &= -\frac{1}{ar}\sum_{k=0}^{n-1} (-1)^k \frac{(1-\theta-\zeta+n+m-k)_k}{(-\theta+n-k)_{k+1}}\cdot \frac{b^k}{a^k}\cdot \frac{x^{r(\theta-n+k)}}{(a+bx^r)^{-\zeta+m}}\\ &\quad +\frac{(-1)^n a^{-n}b^n}{ar(1-\theta)_n} \sum_{k=0}^{m-1} \frac{(1-\theta-\zeta+m-k)_{k+n}}{(-\zeta+m-k)_{k+1}}\cdot \frac{1}{a^k}\cdot \frac{x^{r\theta}}{(a+bx^r)^{-\zeta+m-k}}\\ &\quad +\frac{(-1)^n b^n}{a^{n+m}}\cdot \frac{(1-\theta-\zeta)_{n+m}}{(1-\theta)_n(1-\zeta)_m} \int x^{r\theta-1}(a+bx^r)^{\zeta-1}\,dx . \tag{12} \end{aligned} \]
\[ \begin{aligned} &\int x^{r(1-\theta-\zeta+n+m)-1}(a+bx^r)^{\zeta-n-1}\,dx=\\ &= -\frac{1}{br}\sum_{k=0}^{n-1} \frac{(1-\theta-\zeta+n+m-k)_k}{(-\zeta+n-k)_{k+1}}\cdot \frac{1}{b^k}\cdot \frac{x^{r(-\theta-\zeta+n+m-k)}}{(a+bx^r)^{-\zeta+n-k}}\\ &\quad +\frac{b^{-n-1}}{r(1-\zeta)_n} \sum_{k=0}^{m-1} \frac{(-1)^k(1-\theta-\zeta+m-k)_{n+k}}{(-\theta+m-k)_{k+1}}\cdot \frac{a^k}{b^k}\cdot \frac{x^{r(-\theta-\zeta+m-k)}}{(a+bx^r)^{-\zeta}}\\ &\quad +\frac{(-1)^m a^m}{b^{n+m}}\cdot \frac{(1-\theta-\zeta)_{n+m}}{(1-\zeta)_n(1-\theta)_m} \int x^{r(1-\theta-\zeta)-1}(a+bx^r)^{\zeta-1}\,dx . \tag{13} \end{aligned} \]
\[ \begin{aligned} &\int x^{r(\theta-n)-1}(a+bx^r)^{-\theta-\zeta+n+m}\,dx=\\ &= -\frac{1}{r}\sum_{k=0}^{n-1} \frac{(1-\theta-\zeta+n+m-k)_k}{(-\theta+n-k)_{k+1}}\cdot b^k\, \frac{(a+bx^r)^{-\theta-\zeta+n+m-k}}{x^{r(-\theta+n-k)}}\\ &\quad +\frac{b^n}{r(1-\theta)_n} \sum_{k=0}^{m-1} \frac{(1-\theta-\zeta+m-k)_{n+k}}{(-\zeta+m-k)_{k+1}}\cdot a^k\cdot \frac{x^{r\theta}}{(a+bx^r)^{\theta+\zeta-m+k}}\\ &\quad +a^m b^n \frac{(1-\theta-\zeta)_{n+m}}{(1-\theta)_n(1-\zeta)_m} \int x^{r\theta-1}(a+bx^r)^{-\theta-\zeta}\,dx . \tag{14} \end{aligned} \]
Theorem 1. The expression of integrals of binomial differentials in algebraic functions is possible for the integrals (9)—(11) in two cases: a) \(\zeta = 0\), b) \(\theta + \zeta = 1\); for the integrals (12)—(14) only in the case \(\theta + \zeta = 1\).
Proof. The integrals appearing on the right-hand sides of equalities (9)—(14) contain a factor equal to zero precisely in the cases listed, and on the right-hand sides of these equalities there remain only algebraic functions.
Theorem 2. The expression of the integrals appearing on the right-hand sides of equalities (9)—(14) in elementary functions is possible in the following cases:
1) for the integrals (9)—(14), when \(\theta = 0\) and for rational values of \(\zeta\), and for the integral on the right-hand side of equality (11), moreover, for rational values of \(r\);
2) for the integrals (12)—(14), when \(\zeta = 0\) and for rational values of \(\theta\) and \(r\).
Proof. After applying to each of the listed integrals appearing on the right-hand sides of equalities (9)—(14) the corresponding Chebyshev substitution ([2], p. 52), they are transformed into integrals of the form
\[ \int t^{m-1}(a_1 + b_1 t^n)^{-1}\,dt, \]
where \(m = 1,2,\ldots;\ n = 1,2,\ldots;\ m \leq n\), whose representations in elementary functions are given in tables of integrals ([3], pp. 78—79).
REFERENCES
-
I. M. Ryzhik. Tables of integrals, sums, series, and products. Moscow—Leningrad, Gostekhizdat, 1948.
-
G. M. Fikhtengol'ts. Course of differential and integral calculus. Vol. II, Moscow, Fizmatgiz, 1962.
-
I. S. Gradshteyn, I. M. Ryzhik. Tables of integrals, sums, series, and products. Moscow, Fizmatgiz, 1962.