Bessel関数に関するGegenbauerの加法定理

Gegenbauer多項式は
$$\begin{array}{rl}{C}_n^{(a)}(x)& =\frac{(2a{)}_n}{n!}{}_2{F}_1[\begin{array}{c}-n,2a+n\\ a+\frac{1}{2}\end{array};\frac{1-x}{2}]\end{array}$$
によって与えられる. 前の記事 でGegenbauerの加法定理
$$\begin{array}{rl}& C_n^{(a)}(st+x\sqrt{1-s^2}\sqrt{1-t^2})\\ & =\sum _{k=0}^n\frac{2a+2k-1}{2a-1}\frac{(n-k)!(a{)}_k^2}{(2a{)}_{n+k}}(4\sqrt{1-s^2}\sqrt{1-t^2}{)}^k{C}_{n-k}^{(a+k)}(s)C_{n-k}^{(a+k)}(t)C_k^{(a-\frac{1}{2})}(x)\end{array}$$
を示した. 定義から,
$$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\frac{n!}{(2a{)}_n}{C}_n(1-\frac{x^2}{2n^2})& =\sum _{0\le k}\frac{(-1{)}^k}{k!{(a+\frac{1}{2})}_k}{\left(\frac{x}{2}\right)}^{2k}\\ & =\mathrm{\Gamma }(a+\frac{1}{2}){\left(\frac{x}{2}\right)}^{\frac{1}{2}-a}{J}_{a-\frac{1}{2}}(x)\end{array}$$
である. ここで, $J_a(x)$はBessel関数であり,
$$\begin{array}{rl}{J}_a(x)& =\sum _{0\le n}\frac{(-1{)}^n}{n!\mathrm{\Gamma }(a+n+1)}{\left(\frac{x}{2}\right)}^{2n+a}\end{array}$$
によって定義される. Gegenbauerの加法定理において, $s=1-\frac{u^2}{2n^2},t=1-\frac{v^2}{2n^2}$としてから, $\frac{n!}{(2a{)}_n}$を掛けて$n\to \mathrm{\infty }$とすると, 左辺は
$$\begin{array}{rl}& \underset{n\to \mathrm{\infty }}{lim}\frac{n!}{(2a{)}_n}{C}_n^{(a)}(1-\frac{u^2}{2n^2}-\frac{v^2}{2n^2}+\frac{uvx}{n^2}\sqrt{1-\frac{u^2}{4n^2}}\sqrt{1-\frac{v^2}{4n^2}})\\ & =\mathrm{\Gamma }(a+\frac{1}{2}){\left(\frac{\sqrt{u^2+v^2-2uvx}}{2}\right)}^{\frac{1}{2}-a}{J}_{a-\frac{1}{2}}(\sqrt{u^2+v^2-2uvx})\end{array}$$
となり, 右辺は
$$\begin{array}{rl}& \underset{n\to \mathrm{\infty }}{lim}\frac{n!}{(2a{)}_n}\sum _{k=0}^n\frac{2a+2k-1}{2a-1}\frac{(n-k)!(a{)}_k^2}{(2a{)}_{n+k}}{\left(\frac{4uv}{n^2}\sqrt{1-\frac{u^2}{4n^2}}\sqrt{1-\frac{v^2}{4n^2}}\right)}^k\\ & \cdot C_{n-k}^{(a+k)}(1-\frac{u^2}{2n^2})C_{n-k}^{(a+k)}(1-\frac{v^2}{2n^2})C_k^{(a-\frac{1}{2})}(x)\\ & =\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^n\frac{2a+2k-1}{2a-1}\frac{n!(2a{)}_{n+k}(a{)}_k^2}{n^{2k}(2a{)}_n(n-k)!(2a{)}_{2k}^2}{\left(4uv\sqrt{1-\frac{u^2}{4n^2}}\sqrt{1-\frac{v^2}{4n^2}}\right)}^k\\ & \cdot \frac{(n-k)!}{(2a+2k{)}_{n-k}}{C}_{n-k}^{(a+k)}(1-\frac{u^2}{2n^2})\frac{(n-k)!}{(2a+2k{)}_{n-k}}{C}_{n-k}^{(a+k)}(1-\frac{v^2}{2n^2})C_k^{(a-\frac{1}{2})}(x)\\ & =\sum _{k=0}^{\mathrm{\infty }}\frac{2a+2k-1}{2a-1}\frac{(a{)}_k^2}{(2a{)}_{2k}^2}{\left(4uv\right)}^k\\ & \cdot \mathrm{\Gamma }{(a+k+\frac{1}{2})}^2{\left(\frac{uv}{4}\right)}^{\frac{1}{2}-a-k}{J}_{a+k-\frac{1}{2}}(u)J_{a+k-\frac{1}{2}}(v)C_k^{(a-\frac{1}{2})}(x)\\ & =\mathrm{\Gamma }(a+\frac{1}{2})\mathrm{\Gamma }(a-\frac{1}{2})\sum _{k=0}^{\mathrm{\infty }}(a+k-\frac{1}{2}){\left(\frac{uv}{4}\right)}^{\frac{1}{2}-a}{J}_{a+k-\frac{1}{2}}(u)J_{a+k-\frac{1}{2}}(v)C_k^{(a-\frac{1}{2})}(x)\end{array}$$
よって$a\mapsto a+\frac{1}{2}$とすれば, 以下を得る.

Gegenbauer多項式の直交性
$$\begin{array}{rl}{\int }_{-1}^1{C}_n^{(a)}(x)C_m^{(a)}(x)(1-x^2{)}^{a-\frac{1}{2}}dx& =\frac{\sqrt{\pi }\mathrm{\Gamma }(a+\frac{1}{2})}{\mathrm{\Gamma }(a)}\frac{(2a{)}_n}{n!(n+a)}{\delta }_{n,m}\end{array}$$
を考えれば, これは
$$\begin{array}{rl}& {\int }_{-1}^1\frac{J_a(\sqrt{u^2+v^2-2uvx})}{(\sqrt{u^2+v^2-2uvx}{)}^a}{C}_n^{(a)}(x)(1-x^2{)}^{a-\frac{1}{2}}dx\\ & =\sqrt{\pi }\mathrm{\Gamma }(a+\frac{1}{2})\frac{(2a{)}_n}{n!}{\left(\frac{2}{uv}\right)}^a{J}_{a+n}(u)J_{a+n}(v)\end{array}$$
と同値である. 特に$n=0$とすると,
$$\begin{array}{r}{\int }_{-1}^1\frac{J_a(\sqrt{u^2+v^2-2uvx})}{(\sqrt{u^2+v^2-2uvx}{)}^a}(1-x^2{)}^{a-\frac{1}{2}}dx=\sqrt{\pi }\mathrm{\Gamma }(a+\frac{1}{2}){\left(\frac{2}{uv}\right)}^a{J}_a(u)J_a(v)\end{array}$$
を得る.