Notes on Whittaker & Watson, Chapter XIV, part 3

Whittaker & Watson, A Course of Modern Analysis. Chapter XIV, The Hypergeometric Function.

14.6 Solution of Riemann’s equation by a contour integral.

With the substitution of the dependent variable $u$ by $$ u=(z-a)^\alpha(z-b)^\beta(z-c)^\gamma I, $$ Riemann’s equation becomes $$ Q(z)\frac{d^2I}{dz^2} + \Big[(\lambda-2)Q'(z) + R(z)\Big]\frac{dI}{dz} +\left(\frac12\left(\lambda-1\right)(\lambda-2)Q''(z)+(\lambda-1)R'(z)\right)=0, $$ where $$ \begin{cases}\lambda=1-\alpha-\beta-\gamma=\alpha'+\beta'+\gamma',\\ Q(z)=(z-a)(z-b)(z-c),\\ R(z)=\sum_{\text{cycl}}(\alpha'+\beta+\gamma)(z-b)(z-c). \end{cases} $$ Assuming the solution in the form $$ I=\int_C(t-a)^{\alpha'+\beta+\gamma-1}(t-b)^{\alpha+\beta'+\gamma-1}(t-c)^{\alpha+\beta+\gamma'-1}(z-t)^{-\alpha-\beta-\gamma}dt $$ the equation becomes $\int_C\frac{\partial V}{\partial t}dt=0$ with $V=(t-a)^{\alpha'+\beta+\gamma}(t-b)^{\alpha+\beta'+\gamma}(t-c)^{\alpha+\beta+\gamma'}(t-z)^{-(1+\alpha+\beta+\gamma)}$. Note that the integrand of $I$ and $V$ differ by a factor of $(t-a)(t-b)(t-c)(z-t)^{-1}$ which is single valued. Thus integrals in the form $$ (z-a)^\alpha(z-b)^\beta(z-c)^\gamma\int_C(t-a)^{\alpha'+\beta+\gamma-1}(t-b)^{\alpha+\beta'+\gamma-1}(t-c)^{\alpha+\beta+\gamma'-1}(z-t)^{-\alpha-\beta-\gamma}dt $$ are solutions to the equation if the contour $C$ is a closed loop in the Riemann surface of the multi-valued function $V$ or the integrand of $I$.

14.61 Determination of an integral which represents $P^{(\alpha)}$.

For $|z-a|<\min(|a-b|, |a-c|)$, consider the solution above integrated on the Pochhammer contour $(b+,c+,b-,c-)$ around $b,c$ such that $|z-a|<|t-a|$ on the contour. With appropriate choice of $\arg(z-a)<\pi$, we can expand $(z-b)^\beta$, $(z-c)^\gamma$, and $(t-z)^{-\alpha-\beta-\gamma}$ in power series of $(z-a)$ $$ (z-b)^\beta=(a-b)^\beta\left(1+\beta\frac{z-a}{a-b}+\cdots\right),\\ (z-c)^\gamma=(a-c)^\gamma\left(1+\gamma\frac{z-a}{a-c}+\cdots\right),\\ (t-z)^{-\alpha-\beta-\gamma}=(t-a)^{-\alpha-\beta-\gamma}\left(1+(\alpha+\beta+\gamma)\frac{a-z}{t-a}+\cdots\right). $$ Thus we have the solution $$ P^{(\alpha)}\cdot(a-b)^\beta(a-c)^\gamma\int_N^{(b+,c+,b-,c-)}(t-a)^{\alpha'+\beta+\gamma-1}(t-b)^{\alpha+\beta'+\gamma-1}(t-c)^{\alpha+\beta+\gamma'-1}dt, $$ and $P^{(\alpha')}$, $P^{(\beta)}$, $P^{(\beta')}$, $P^{(\gamma)}$, $P^{(\gamma')}$ can be found similarily.

14.7 Relations between contiguous hypergeometric functions.

Let $P(z)$ be a scalar multiple of one of the functions $P^{(\alpha)}$, $P^{(\alpha')}$, $P^{(\beta)}$, $P^{(\beta')}$, $P^{(\gamma)}$, $P^{(\gamma')}$ with exponents $l,m\in{\alpha,\alpha',\beta,\beta',\gamma,\gamma'}$. The function $P_{l+1,m-1}(z)$ with modified exponents $l+1,m-1$ is said to be contiguous to $P(z)$.

1º Given the contour integral form of $P(z)$ as in §14.6 above, consider the differentiation under the integral sign $$ \int_C\frac{\partial}{\partial t}\left[(t-a)^{\alpha'+\beta+\gamma}(t-b)^{\alpha+\beta'+\gamma-1}(t-c)^{\alpha+\beta+\gamma'-1}(t-z)^{-\alpha-\beta-\gamma}\right] dt=0 $$ which results in the first group of contiguous relations in the form $$ (\alpha'+\beta+\gamma)P(z)+(\alpha+\beta'+\gamma-1)P_{\alpha'+1,\beta'-1}(z)+ (\alpha+\beta+\gamma'-1)P_{\alpha'+1,\gamma'-1}(z)\\ = \frac{\alpha+\beta+\gamma}{z-b}P_{\beta+1,\gamma'-1}(z) = \frac{\alpha+\beta+\gamma}{z-c}P_{\gamma+1,\beta'-1}(z) $$ along with the analogous relations obtained by permutating $a,b,c$.

2º With $t-a=(t-b)+(b-a)$, the contour integral $P(z)$ can be split $$ P(z)=P_{\alpha'-1,\beta'+1}(z)+(b-a)P_{\alpha'-1}(z) = P_{\alpha'-1,\gamma'+1}(z)+(c-a)P_{\alpha'-1}(z) $$ where $P_{\alpha'-1}$ is the contour integral with only the exponent $\alpha'$ lowered without raising other exponents, thus not a Riemann’s $P$-function as its exponents do not sum to unity. Eliminating the term $P_{\alpha'-1}$, we have $$ (c-b)P(z) + (a-c)P_{\alpha'-1,\beta'+1}(z)+(b-a)P_{\alpha'-1,\gamma'+1}(z)=0. $$ 3º With $t-z=(t-a)-(z-a)$, we have $$ \begin{align*} & (z-a)^{-1}\left[P(z)-(z-b)^{-1}P_{\beta+1,\gamma'-1}(z)\right]\\ = {} & (z-b)^{-1}\left[P(z)-(z-c)^{-1}P_{\gamma+1,\alpha'-1}(z)\right]\\ = {} & (z-c)^{-1}\left[P(z)-(z-a)^{-1}P_{\alpha+1,\beta'-1}(z)\right]. \end{align*} $$