Gegenbauer多項式とBessel関数

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Gegenbauer多項式に重み関数を掛けてFourier変換するとBessel関数になることを示す.

$\begin{aligned} \int_{- 1}^{1} (1 - x^{2})^{a - \frac{1}{2}} C_{n}^{(a)} (x) e^{i x y} d x & = \frac{2 π i^{n} Γ (2 a + n)}{n! Γ (a) (2 y)^{a}} J_{n + a} (y) \end{aligned}$

$n = 2 k$ のとき, $y^{2 m}$ の係数を比較した等式
$\begin{aligned} \int_{- 1}^{1} x^{2 m} (1 - x^{2})^{a - \frac{1}{2}} C_{2 k}^{(a)} (x) d x & = \frac{2 π (2 m)! Γ (2 a + 2 k)}{2^{2 m + 2 a} (m - k)! (2 k)! Γ (a) Γ (a + m + k + 1)} \end{aligned}$
を示せばよい. 項別積分によって,
$\begin{aligned} \int_{- 1}^{1} x^{2 m} (1 - x^{2})^{a - \frac{1}{2}} C_{2 k}^{(a)} (x) d x & = \frac{(- 1)^{k} (a)_{k}}{k!} \int_{- 1}^{1} x^{2 m} (1 - x^{2})^{a - \frac{1}{2}} \sum_{0 \leq j} \frac{(- k, a + k)_{j}}{j! {(\frac{1}{2})}_{j}} x^{2 j} d x \\ = \frac{(- 1)^{k} (a)_{k}}{k!} \sum_{0 \leq j} \frac{(- k, a + k)_{j}}{j! {(\frac{1}{2})}_{j}} \frac{Γ (m + j + \frac{1}{2}) Γ (a + \frac{1}{2})}{Γ (a + m + j + 1)} \\ = \frac{(- 1)^{k} (a)_{k}}{k!} \frac{Γ (m + \frac{1}{2}) Γ (a + \frac{1}{2})}{Γ (a + m + 1)}_{3} F_{2} [\begin{array}{c} - k, a + k, m + \frac{1}{2} \\ \frac{1}{2}, a + m + 1 \end{array}; 1] \end{aligned}$
ここで, Saalschützの和公式より,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} - k, a + k, m + \frac{1}{2} \\ \frac{1}{2}, a + m + 1 \end{array}; 1] & = \frac{{(\frac{1}{2} + a, - m)}_{k}}{{(\frac{1}{2}, a + m + 1)}_{k}} \end{aligned}$
であるから, Legendreの倍角公式を用いて整理すると,
$\begin{aligned} \frac{(- 1)^{k} (a)_{k}}{k!} \frac{Γ (m + \frac{1}{2}) Γ (a + \frac{1}{2})}{Γ (a + m + 1)}_{3} F_{2} [\begin{array}{c} - k, a + k, m + \frac{1}{2} \\ \frac{1}{2}, a + m + 1 \end{array}; 1] \\ = \frac{2^{2 k} (a)_{k}}{(2 k)!} \frac{Γ (m + \frac{1}{2}) Γ (a + \frac{1}{2} + k)}{Γ (a + m + k + 1)} \frac{m!}{(m - k)!} \\ = \frac{2 π (2 m)! Γ (2 a + 2 k)}{2^{2 a + 2 m} (m - k)! (2 k)! Γ (a) Γ (a + m + k + 1)} \end{aligned}$
を得る. $n = 2 k + 1$ のとき, $y^{2 m + 1}$ の係数を比較した等式
$\begin{aligned} \int_{- 1}^{1} x^{2 m + 1} (1 - x^{2})^{a - \frac{1}{2}} C_{2 k + 1}^{(a)} (x) d x & = \frac{2 π (2 m + 1)! Γ (2 a + 2 k + 1)}{2^{2 m + 2 a + 1} (m - k)! (2 k + 1)! Γ (a) Γ (a + m + k + 2)} \end{aligned}$
を示せば良い. 項別積分によって,
$\begin{aligned} \int_{- 1}^{1} x^{2 m + 1} (1 - x^{2})^{a - \frac{1}{2}} C_{2 k + 1}^{(a)} (x) d x & = \frac{2 (- 1)^{k} (a)_{k + 1}}{k!} \int_{- 1}^{1} x^{2 m + 1} (1 - x^{2})^{a - \frac{1}{2}} \sum_{0 \leq j} \frac{(- k, a + k + 1)_{j}}{j! {(\frac{3}{2})}_{j}} x^{2 j + 1} d x \\ = \frac{2 (- 1)^{k} (a)_{k + 1}}{k!} \sum_{0 \leq j} \frac{(- k, a + k + 1)_{j}}{j! {(\frac{3}{2})}_{j}} \frac{Γ (m + j + \frac{3}{2}) Γ (a + \frac{1}{2})}{Γ (a + m + j + 2)} \\ = \frac{2 (- 1)^{k} (a)_{k + 1}}{k!} \frac{Γ (m + \frac{3}{2}) Γ (a + \frac{1}{2})}{Γ (a + m + 2)}_{3} F_{2} [\begin{array}{c} - k, a + k + 1, m + \frac{3}{2} \\ \frac{3}{2}, a + m + 2 \end{array}; 1] \end{aligned}$
ここで, Saalschützの和公式より,
$\begin{aligned} _{3} F_{2} [\begin{array}{c} - k, a + k + 1, m + \frac{3}{2} \\ \frac{3}{2}, a + m + 2 \end{array}; 1] & = \frac{{(a + \frac{1}{2}, - m)}_{k}}{{(\frac{3}{2}, a + m + 2)}_{k}} \end{aligned}$
であるから, Legendreの倍角公式を用いて整理すると
$\begin{aligned} \frac{2 (- 1)^{k} (a)_{k + 1}}{k!} \frac{Γ (m + \frac{3}{2}) Γ (a + \frac{1}{2})}{Γ (a + m + 2)}_{3} F_{2} [\begin{array}{c} - k, a + k + 1, m + \frac{3}{2} \\ \frac{3}{2}, a + m + 2 \end{array}; 1] \\ = \frac{2^{2 k + 1} (a)_{k + 1}}{(2 k + 1)!} \frac{Γ (m + \frac{3}{2}) Γ (a + \frac{1}{2} + k)}{Γ (a + m + k + 2)} \frac{m!}{(m - k)!} \\ = \frac{2 π (2 m + 1)! Γ (2 a + 2 k + 1)}{2^{2 a + 2 m + 1} (2 k + 1)! (m - k)! Γ (a) Γ (a + m + k + 2)} \end{aligned}$
となって示される.

実部と虚部に分けると以下を得る.

$\begin{aligned} \int_{0}^{1} (1 - x^{2})^{a - \frac{1}{2}} C_{2 n}^{(a)} (x) \cos x y d x & = \frac{(- 1)^{n} π Γ (2 a + 2 n) J_{2 n + a} (y)}{(2 n)! Γ (a) (2 y)^{a}} \\ \int_{0}^{1} (1 - x^{2})^{a - \frac{1}{2}} C_{2 n + 1}^{(a)} (x) \sin x y d x & = \frac{(- 1)^{n} π Γ (2 a + 2 n + 1) J_{2 n + a + 1} (y)}{(2 n + 1)! Γ (a) (2 y)^{a}} \end{aligned}$

これによって, Gegenbauer多項式で展開された関数を変換してBessel関数で展開された関数に移すことができる.

Legendre多項式

$a = \frac{1}{2}$ とすることによって, Legendre多項式の場合を得る.

$\begin{aligned} \int_{- 1}^{1} P_{n} (x) e^{i x y} d x & = i^{n} \sqrt{\frac{2 π}{y}} J_{n + \frac{1}{2}} (y) \end{aligned}$
実部と虚部を分けると,
$\begin{aligned} \int_{0}^{1} P_{2 n} (x) \cos x y d x & = (- 1)^{n} \sqrt{\frac{π}{2 y}} J_{2 n + \frac{1}{2}} (y) \\ \int_{0}^{1} P_{2 n + 1} (x) \sin x y d x & = (- 1)^{n} \sqrt{\frac{π}{2 y}} J_{2 n + \frac{3}{2}} (y) \end{aligned}$