APPLICATION OF CONTINUED FRACTIONS TO THE EVALUATION OF INTEGRALS OF BINOMIAL DIFFERENTIALS

V. E. KORNILOV

(Presented by the scientific seminar of the Departments of Higher Mathematics and Engineering-Computational Mathematics of TPI)

In the article the parameters \(\gamma>1,\ \gamma>\alpha>0,\ r>0,\ 0<\Theta<1,\ 0<\zeta<1;\ z\) is a complex variable. We introduce the notation

\[ (a)_k=a(a+1)\ldots(a+k-1)=\frac{\Gamma(a+k)}{\Gamma(a)},\qquad (a)_0=1. \]

1. Academician A. A. Markov ([1], p. 320) obtained for the hypergeometric series \(F(\alpha,1;\gamma;-z)\) the following expression in the form of a continued fraction:

\[ F(\alpha,1;\gamma;-z)=\frac{P_{2k}(z)}{Q_{2k}(z)}+R_{2k}(z), \]

where

\[ \frac{P_n(z)}{Q_n(z)}=\frac{1}{a_1+\ldots+\dfrac{1}{a_n}}, \]

\[ R_{2k}(z)=\sum_{n=k}^{\infty}\frac{a_{2n+2}}{Q_{2n}(z)\,Q_{2n+2}(z)}\quad ([2],\ \text{p. }34), \]

\[ a_1=1,\qquad a_{2k}=\frac{(\gamma-\alpha)_{k-1}(k-1)!(\gamma+2k-2)}{(\gamma)_{k-1}(\alpha)_k}\cdot\frac{1}{z}, \]

\[ a_{2k+1}=\frac{(\gamma)_{k-1}(\alpha)_k(\gamma+2k-1)}{(\gamma-\alpha)_k\cdot k!},\qquad k=1,2,\ldots \tag{1} \]

For the denominators of the convergents with even indices, the following recurrence relation is known ([2], p. 14):

\[ Q_{2k+2}(z)=\left(a_{2k+2}a_{2k+1}+\frac{a_{2k+2}}{a_{2k}}+1\right)Q_{2k}(z) -\frac{a_{2k+2}}{a_{2k}}Q_{2k-2}(z). \tag{2} \]

We further assume that

\[ Q_{2k}(z)=\sum_{n=0}^{k}C_k^n\frac{(\gamma+k-1)_n}{(\alpha)_n}\cdot\frac{1}{z^n},\qquad k=1,2,\ldots \tag{3} \]

After directly computing the continued fraction (1), we are convinced that equality (3) is valid for \(k=1,2\). On the basis of the method of mathematical induction it is not difficult to establish that, if the polynomials

will satisfy relation (2), then they will also be valid for \(k=3,4,\ldots\).

On the basis of (1)—(3) we obtain

\[ \begin{aligned} Q_{2k+2}(z)={}& \left[ \frac{(\gamma+2k-1)(\gamma+2k)}{(\gamma+k-1)(\alpha+k)}\cdot\frac{1}{z} + \frac{k(\gamma-\alpha+k-1)(\gamma+2k)} {(\gamma+k-1)(\alpha+k)(\gamma+2k-2)} +1 \right] \sum_{n=0}^{k} C_k^n\frac{(\gamma+k-1)_n}{(\alpha)_n}\cdot\frac{1}{z^n} \\ &- \frac{k(\gamma-\alpha+k-1)(\gamma+2k)} {(\gamma+k-1)(\alpha+k)(\gamma+2k-2)} \sum_{n=0}^{k-1} C_{k-1}^n\frac{(\gamma+k-2)_n}{(\alpha)_n}\cdot\frac{1}{z^n} \\ ={}& 1+\sum_{n=0}^{k-2} C_k^n\frac{(\gamma+k-1)_n}{(\alpha)_n}\cdot\frac{1}{z^{n+1}} \left[ \frac{(\gamma+2k-1)(\gamma+2k)}{(\gamma+k-1)(\alpha+k)} \right. \\ &\left. + \frac{(k-n)(\gamma-\alpha+k-1)(\gamma+2k)} {(\alpha+n)(\gamma+k-1)(\alpha+k)} + \frac{(k-n)(\gamma+k+n-1)} {(n+1)(\alpha+n)} \right] \\ &+ \left[ \frac{(\gamma+2k-1)(\gamma+2k)k(\gamma+k)_{k-2}} {(\alpha+k)(\alpha)_{k-1}} + \frac{(\gamma+k-1)_k}{(\alpha)_k} \right. \\ &\left. + \frac{k(\gamma-\alpha+k-1)(\gamma+2k)(\gamma+k)_{k-2}} {(\alpha+k)(\alpha)_k} \right]\frac{1}{z^k} + \frac{(\gamma+k)_{k+1}}{(\alpha)_{k+1}}\cdot\frac{1}{z^{k+1}} \\ ={}& 1+\sum_{n=0}^{k-2} C_k^n\frac{(\gamma+k-1)_n}{(\alpha)_n}\cdot\frac{1}{z^{n+1}}\cdot \frac{(\gamma+k+n-1)(k+1)(\gamma+n+k)} {(n+1)(\gamma+k-1)(\alpha+n)} \\ &+ (k+1)\frac{(\gamma+k)_k}{(\alpha)_k}\cdot\frac{1}{z^k} + \frac{(\gamma+k)_{k+1}}{(\alpha)_{k+1}}\cdot\frac{1}{z^{k+1}} \\ ={}& \sum_{n=0}^{k+1} C_{k+1}^n\frac{(\gamma+k)_n}{(\alpha)_n}\cdot\frac{1}{z^n}, \end{aligned} \]

thereby equality (3) is proved.

Next, according to (2), we obtain the numerator \(P_{2k}(z)\), which is transformed as follows:

\[ \begin{aligned} P_{2k}(z) &= \sum_{n=1}^{k} C_k^n\frac{(\gamma+k-1)_n}{(\alpha)_n}\cdot\frac{1}{z^n} \sum_{m=0}^{n-1}(-1)^m\frac{(\alpha)_m}{(\gamma)_m}z^m \\ &= C_k^1\frac{\gamma+k-1}{\alpha}\cdot\frac{1}{z} + C_k^2\frac{(\gamma+k-1)_2}{(\alpha)_2} \left[ \frac{1}{z^2}-\frac{\alpha}{\gamma}\cdot\frac{1}{z} \right] \\ &\quad+ C_k^3\frac{(\gamma+k-1)_3}{(\alpha)_3} \left[ \frac{1}{z^3}-\frac{\alpha}{\gamma}\cdot\frac{1}{z^2} +\frac{(\alpha)_2}{(\gamma)_2}\cdot\frac{1}{z} \right] +\cdots \\ &\quad+\cdots+ C_k^k\frac{(\gamma+k-1)_k}{(\alpha)_k} \left[ \frac{1}{z^k}-\frac{\alpha}{\gamma}\cdot\frac{1}{z^{k-1}} +\cdots+(-1)^{k-1}\frac{(\alpha)_{k-1}}{(\gamma)_{k-1}}\cdot\frac{1}{z} \right], \end{aligned} \]

\[ P_{2k}(z)= \sum_{n=1}^{k}\frac{1}{z^n} \sum_{m=0}^{k-n} C_k^{\,n+m}(-1)^m \frac{(\gamma+k-1)_{n+m}(\alpha)_m} {(\alpha)_{n+m}(\gamma)_m}. \tag{4} \]

Taking into account equalities (1), (3), and (4), we finally obtain

\[ F(\alpha,1;\gamma;-z)=\frac{P_{2k}(z)}{Q_{2k}(z)}+R_{2k}(z)= \]

\[ =\frac{\displaystyle \sum_{n=1}^{k}\frac{1}{z^{n}}\sum_{m=0}^{k-n} C_{k}^{\,n+m}(-1)^{m} \frac{(\gamma+k-1)_{n+m}}{(\alpha+m)_{n}(\gamma)_{m}}} {\displaystyle \sum_{n=0}^{k} C_{k}^{\,n}\frac{(\gamma+k-1)_{n}}{(\alpha)_{n}}\cdot\frac{1}{z^{n}}} +R_{2k}(z), \]

\[ |\arg(1+z)|<\pi,\qquad k=1,2,\ldots . \tag{5} \]

2. Since the limit of the ratio of the subsequent coefficient of the series \(F(\alpha,1;\gamma;-z)\) to the preceding one, for \(\gamma>\alpha>0\), tends to unity from the left, it follows, in view of ([2], pp. 5, 24), that all roots of the polynomial \(Q_{2k}(z)\) are located in the interval \((-\infty,-1)\), that is,

\[ Q_{2k}(z)=\prod_{n=1}^{k}\left(1+\frac{a_n}{z}\right), \]

\[ a_0=\infty>a_1>\ldots>a_k>1=a_{k+1}, \tag{6} \]

moreover

\[ Q_{2k+2}(z)=\prod_{n=1}^{k+1}\left(1+\frac{b_n}{z}\right), \]

\[ \infty>b_1>b_2>\ldots>b_{k+1}>1. \tag{7} \]

In view of the fact that the roots of the polynomial \(Q_{2k}(z)\) are located between the roots of the polynomial \(Q_{2k+2}(z)\) ([2], p. 14), we have

\[ a_0=\infty>b_1>a_1>\ldots>a_k>b_{k+1}>1=a_{k+1}. \tag{8} \]

Next, the polynomial (3) is transformed as

\[ Q_{2k}(z)=\frac{(\gamma+k-1)_k}{(\alpha)_k}\cdot \frac{1}{z^k}\sum_{n=0}^{k} C_k^n \frac{(\alpha+k-n)_n}{(\gamma+2k-1-n)_n}\cdot z^n = \]

\[ =\frac{(\gamma+k-1)_k}{(\alpha)_k}\cdot\frac{1}{z^k}\,q_{2k}(z), \tag{9} \]

and, along with formulas (6), (7), and (9),

\[ q_{2k}(z)=\prod_{n=1}^{k}\left(1+\frac{z}{a_n}\right), \tag{10} \]

\[ q_{2k+2}(z)=\prod_{n=1}^{k+1}\left(1+\frac{z}{b_n}\right). \tag{11} \]

Starting from formulas (8)—(11), it is not difficult to draw the following conclusions: for \(\operatorname{Re}(z)\geq 0\),

\[ \left|1+\frac{z}{a_{n-1}}\right|< \left|1+\frac{z}{b_n}\right|, \qquad n=1,2,\ldots,k+1, \]

\[ |q_{2k}(z)|<|q_{2k+2}(z)|<\ldots; \tag{12} \]

for \(-1<\operatorname{Re}(z)<0,\ \operatorname{Re}\left(\dfrac{1}{z}\right)\leqslant -1\),

\[ \left|1+\frac{z}{a_n}\right|<\left|1+\frac{z}{b_n}\right|,\quad n=1,2,\ldots,k+1, \]

\[ \left|q_{2k}(z)(1+z)\right|<\left|q_{2k+2}(z)\right|,\ldots; \tag{13} \]

for \(-N<\operatorname{Re}(z)\leqslant -1\)

\[ \left|q_{2k}(z)\cdot \operatorname{Im}(z)\right|<\left|q_{2k+2}(z)\right|,\ldots . \tag{14} \]

If we introduce the function

\[ \sigma(z)= \begin{cases} 1, & \text{for } 0\leqslant \operatorname{Re}(z)<4,\\[4pt] 1+z, & \text{for } -1<\operatorname{Re}(z)<0,\ \operatorname{Re}\left(\dfrac{1}{z}\right)\leqslant -1,\\[6pt] \operatorname{Im}(z), & \text{for } -N<\operatorname{Re}(z)\leqslant -1, \end{cases} \tag{15} \]

then inequalities (12)—(14) reduce to the following:

\[ \left|q_{2k}(z)\cdot \sigma(z)\right|<\left|q_{2k+2}(z)\right|, \]

\[ \left|q_{2k+2}(z)\cdot \sigma(z)\right|<\left|q_{2k+4}(z)\right|,\ldots;\quad -N<\operatorname{Re}(z)<4. \tag{16} \]

Next, according to (1), (9), and (16), we obtain

\[ |R_{2k}(z)|< \sum_{n=k}^{\infty} \frac{(\gamma-\alpha)_n\, n!\,(\alpha)_n\cdot |z|^{2n}} {(\gamma)_{2n}(\gamma+n-1)_n\, |q_{2n}(z)|\,|q_{2n+2}(z)|} \cdot \frac{\gamma+2n-1}{2n} < \]

\[ < \frac{|z|^{2k}}{2|\sigma(z)|\,|q_{2k}(z)|^2} \sum_{n=k}^{\infty} \frac{(\gamma-\alpha)_n\,(n-1)!\,(\alpha)_n} {(\gamma)_{2n-1}(\gamma+n-1)_n} \left|\frac{z}{\sigma(z)}\right|^{2n-2k}. \]

It is not difficult to verify that, for the infinite sum obtained, the ratio of a subsequent term to the preceding one is less than

\[ \left|\frac{z}{4\sigma(z)}\right|^2, \]

therefore

\[ |R_{2k}(z)|< \frac{8|\sigma(z)|}{|4\sigma(z)|^2-|z|^2} \cdot \frac{1}{|Q_{2k}(z)|^2} \cdot \frac{(\gamma-\alpha)_k\,(k-1)!}{(\alpha)_k(\gamma)_{k-1}}, \]

\[ -N<\operatorname{Re}(z)<4,\quad \left|\frac{z}{4\sigma(z)}\right|^2<1. \tag{17} \]

1. The integrals \(I\) are expressed by means of the hypergeometric function ([3], p. 308)

\[ I=\int_0^z \frac{t^{r\Theta-1}\,dt}{(a+bt^r)^{1-\zeta}} = \frac{z^{r\Theta}(a+bz^r)^\zeta}{ar\Theta} F\left(\Theta+\zeta;\,1;\,1+\Theta;\,-\frac{bz^r}{a}\right), \]

\[ r>0,\quad 0<\Theta<1,\quad 0<\zeta<1. \tag{18} \]

Taking (5) and (18) into account, we obtain:

\[ I=\int_0^z t^{r\Theta-1}(a+bt^r)^{\zeta-1}\,dt = \frac{z^{r\Theta}(a+bz^r)^\zeta}{ar\Theta}\times \]

\[ \times \frac{ \displaystyle \sum_{n=1}^{\kappa}\left(\frac{a}{bz^{r}}\right)^{n} \sum_{m=0}^{\kappa-n} C_{\kappa}^{\,n+m} \frac{(-1)^m(\Theta+k)_{n+m}}{(\Theta+\zeta+m)_n(1+\Theta)_m} }{ \displaystyle \sum_{n=0}^{\kappa} C_{\kappa}^{\,n} \frac{(\Theta+k)_n}{(\Theta+\zeta)_n} \left(\frac{a}{bz^r}\right)^n } + \frac{z^{r\Theta}(a+bz^r)^{\zeta}}{ar\Theta} R_{2k}\left(\frac{bz^r}{a}\right). \tag{19} \]

If one takes into account the analytic continuation of the hypergeometric function in the neighborhood of the points \(\left(-\frac{a}{b}\right)^{1/r}\) and \(\infty\) ([3], p. 308), we also obtain

\[ I= \frac{a^{\Theta+\zeta-1}}{(-b)^{\Theta}r}\cdot B(\Theta,\zeta) + \int_{\left(-\frac{a}{b}\right)^{1/r}}^{z} t^{r\Theta-1}(a+bt^r)^{\zeta-1}\,dt = \]

\[ = \frac{a^{\Theta+\zeta-1}}{(-b)^{\Theta}r}\cdot B(\Theta,\zeta) - \frac{z^{r\Theta}(a+bz^r)^{\zeta}}{ar\zeta} \left[ \frac{ P_{2k}\left(-\frac{bz^r}{a}-1\right) }{ Q_{2k}\left(-\frac{bz^r}{a}-1\right) } + R_{2k}\left(-\frac{bz^r}{a}-1\right) \right], \quad \alpha=\Theta+\zeta,\ \gamma=1+\zeta, \tag{20} \]

\[ I= \frac{a^{\Theta+\zeta-1}}{b^{\Theta}r}\cdot B(\Theta,1-\Theta-\zeta) - \int_z^{\infty} t^{r\Theta-1}(a+bt^r)^{\zeta-1}\,dt = \]

\[ = \frac{a^{\Theta+\zeta-1}}{b^{\Theta}r}\cdot B(\Theta,1-\Theta-\zeta) + \frac{z^{r(\Theta-1)}(a+bz^r)}{br(1-\Theta-\zeta)} \left[ \frac{ P_{2k}\left(\frac{a}{bz^r}\right) }{ Q_{2k}\left(\frac{a}{bz^r}\right) } + R_{2k}\left(\frac{a}{bz^r}\right) \right], \quad \alpha=1-\Theta,\ \gamma=2-\Theta-\zeta,\ \Theta+\zeta<1. \tag{21} \]

In equalities (20) and (21) the appropriate fractions are computed according to formula (5), and the estimate of the remainder terms in absolute value is carried out according to formula (17).

REFERENCES

1. A. A. Markov. Selected Works on the Theory of Continued Fractions and the Theory of Functions Least Deviating from Zero. Moscow–Leningrad, Gostekhizdat, 1948.

2. T. I. Stieltjes. Studies on Continued Fractions. Moscow, ONTI, 1936.

3. I. M. Ryzhik. Tables of Integrals, Sums, Series, and Products. Moscow–Leningrad, Gostekhizdat, 1948.