Gegenbauer多項式のLaplace型積分表示

今回は次を示す.

以下の等式が成り立つ.
$\begin{aligned} C_{n}^{(a)} (\cos θ) & = \frac{(2 a)_{n}}{n!} \frac{Γ (a + \frac{1}{2})}{\sqrt{π} Γ (a)} \int_{0}^{π} (\cos θ + i \sin θ \cos ϕ)^{n} \sin^{2 a - 1} ϕ d ϕ \end{aligned}$

まず, ベータ積分を計算することによって,
$\begin{aligned} \int_{0}^{π} (\cos θ + i \sin θ \cos ϕ)^{n} \sin^{2 a - 1} ϕ d ϕ & = \int_{- π / 2}^{π / 2} (\cos θ - i \sin θ \sin ϕ)^{n} \cos^{2 a - 1} ϕ d ϕ \\ = \int_{0}^{π / 2} ((\cos θ - i \sin θ \sin ϕ)^{n} + (\cos θ - i \sin θ \sin ϕ)^{n}) \cos^{2 a - 1} ϕ d ϕ \\ = 2 \sum_{0 \leq k} (- 1)^{k} (\binom{n}{2 k}) \cos^{n - 2 k} θ \sin^{2 k} θ \int_{0}^{π / 2} \sin^{2 k} ϕ \cos^{2 a - 1} ϕ d ϕ \\ = \sum_{0 \leq k} (- 1)^{k} (\binom{n}{2 k}) \cos^{n - 2 k} θ \sin^{2 k} θ \frac{Γ (k + \frac{1}{2}) Γ (a)}{Γ (k + a + \frac{1}{2})} \\ = \frac{\sqrt{π} Γ (a)}{Γ (a + \frac{1}{2})} \sum_{0 \leq k} (- 1)^{k} (\binom{n}{2 k}) \cos^{n - 2 k} θ \sin^{2 k} θ \frac{{(\frac{1}{2})}_{k}}{{(a + \frac{1}{2})}_{k}} \\ = \frac{\sqrt{π} Γ (a)}{Γ (a + \frac{1}{2})} \sum_{0 \leq k} (- 1)^{k} \frac{n!}{4^{k} k! {(a + \frac{1}{2})}_{k} (n - 2 k)!} \cos^{n - 2 k} θ \sin^{2 k} θ \end{aligned}$
を得る. $n$ が偶数のとき, $n = 2 m$ として,
$\begin{aligned} \sum_{0 \leq k} (- 1)^{k} \frac{n!}{4^{k} k! {(a + \frac{1}{2})}_{k} (n - 2 k)!} \cos^{n - 2 k} θ \sin^{2 k} θ & = \sum_{0 \leq k} (- 1)^{m - k} \frac{(2 m)!}{4^{m - k} (m - k)! {(a + \frac{1}{2})}_{m - k} (2 k)!} \cos^{2 k} θ \sin^{2 m - 2 k} θ \\ = \frac{(- 1)^{m} (2 m)!}{4^{m} m! {(a + \frac{1}{2})}_{m}} \sum_{0 \leq k} (- 1)^{k} \frac{{(- m, \frac{1}{2} - m - a)}_{k}}{k! {(\frac{1}{2})}_{k}} \cos^{2 k} θ \sin^{2 m - 2 k} θ \end{aligned}$
超幾何関数のPfaff変換によって,
$\begin{aligned} \sum_{0 \leq k} (- 1)^{k} \frac{{(- m, \frac{1}{2} - m - a)}_{k}}{k! {(\frac{1}{2})}_{k}} \cos^{2 k} θ \sin^{2 m - 2 k} θ & = \sum_{0 \leq k} \frac{(- m, a + m)_{k}}{k! {(\frac{1}{2})}_{k}} \cos^{2 k} θ \\ = \frac{(- 1)^{m} m!}{(a)_{m}} C_{2 m}^{(a)} (\cos θ) \end{aligned}$
よってこのとき,
$\begin{aligned} \int_{0}^{π} (\cos θ + i \sin θ \cos ϕ)^{n} \sin^{2 a - 1} ϕ d ϕ & = \frac{(2 m)!}{(2 a)_{2 m}} \frac{\sqrt{π} Γ (a)}{Γ (a + \frac{1}{2})} C_{2 m}^{(a)} (\cos θ) \end{aligned}$
$n$ が奇数のとき, $n = 2 m + 1$ として,
$\begin{aligned} \sum_{0 \leq k} (- 1)^{k} \frac{n!}{4^{k} k! {(a + \frac{1}{2})}_{k} (n - 2 k)!} \cos^{n - 2 k} θ \sin^{2 k} θ & = \sum_{0 \leq k} (- 1)^{m - k} \frac{(2 m + 1)!}{4^{m - k} (m - k)! {(a + \frac{1}{2})}_{m - k} (2 k + 1)!} \cos^{2 k + 1} θ \sin^{2 m - 2 k} θ \\ = \frac{(- 1)^{m} (2 m + 1)!}{4^{m} m! {(a + \frac{1}{2})}_{m}} \cos θ \sum_{0 \leq k} (- 1)^{k} \frac{{(- m, \frac{1}{2} - m - a)}_{k}}{k! {(\frac{3}{2})}_{k}} \cos^{2 k} θ \sin^{2 m - 2 k} θ \end{aligned}$
超幾何関数のPfaff変換によって,
$\begin{aligned} \cos θ \sum_{0 \leq k} (- 1)^{k} \frac{{(- m, \frac{1}{2} - m - a)}_{k}}{k! {(\frac{3}{2})}_{k}} \cos^{2 k} θ \sin^{2 m - 2 k} θ & = \cos θ \sum_{0 \leq k} \frac{(- m, a + m + 1)_{k}}{k! {(\frac{3}{2})}_{k}} \cos^{2 k} θ \\ = \frac{(- 1)^{m} m!}{2 (a)_{m + 1}} C_{2 m + 1}^{(a)} (\cos θ) \end{aligned}$
よってこのとき,
$\begin{aligned} \int_{0}^{π} (\cos θ + i \sin θ \cos ϕ)^{n} \sin^{2 a - 1} ϕ d ϕ & = \frac{(2 m + 1)!}{(2 a)_{2 m + 1}} \frac{\sqrt{π} Γ (a)}{Γ (a + \frac{1}{2})} C_{2 m + 1}^{(a)} (\cos θ) \end{aligned}$
となり, いずれの場合も定理が示された.

Legendre多項式

$a = \frac{1}{2}$ とすることによって以下のLegendre多項式のLaplace型積分表示を得る.

以下の等式が成り立つ.
$\begin{aligned} P_{n} (\cos θ) & = \frac{1}{π} \int_{0}^{π} (\cos θ + i \sin θ \sin ϕ)^{n} d ϕ \end{aligned}$

Hermite多項式

Hermite多項式にもLaplace型積分表示があるので示しておく.

$\begin{aligned} H_{n} (x) & = \frac{2^{n}}{\sqrt{π}} \int_{- \infty}^{\infty} (x + i t)^{n} e^{- t^{2}} d t \end{aligned}$

$\begin{aligned} \int_{- \infty}^{\infty} (x + i t)^{n} e^{- t^{2}} d t & = \int_{0}^{\infty} ((x + i t)^{n} + (x - i t)^{n}) e^{- t^{2}} d t \\ = 2 \sum_{0 \leq k} (- 1)^{k} (\binom{n}{2 k}) x^{n - 2 k} \int_{0}^{\infty} t^{2 k} e^{- t^{2}} d t \\ = \sqrt{π} \sum_{0 \leq k} (- 1)^{k} (\binom{n}{2 k}) x^{n - 2 k} {(\frac{1}{2})}_{k} \\ = \sqrt{π} \sum_{0 \leq k} (- 1)^{k} \frac{n!}{4^{k} k! (n - 2 k)!} x^{n - 2 k} \\ = \frac{\sqrt{π}}{2^{n}} H_{n} (x) \end{aligned}$